url
string | fetch_time
int64 | content_mime_type
string | warc_filename
string | warc_record_offset
int32 | warc_record_length
int32 | text
string | token_count
int32 | char_count
int32 | metadata
string | score
float64 | int_score
int64 | crawl
string | snapshot_type
string | language
string | language_score
float64 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://math.libretexts.org/TextMaps/Calculus_TextMaps/Map%3A_Calculus_-_Early_Transcendentals_(Stewart)/11%3A_Infinite_Sequences_And_Series/11.10%3A_Taylor_and_Maclaurin_Series
| 1,511,368,891,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934806615.74/warc/CC-MAIN-20171122160645-20171122180645-00598.warc.gz
| 650,427,678
| 22,642
|
Skip to main content
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 11.10: Taylor and Maclaurin Series
In the previous two sections we discussed how to find power series representations for certain types of functions––specifically, functions related to geometric series. Here we discuss power series representations for other types of functions. In particular, we address the following questions: Which functions can be represented by power series and how do we find such representations? If we can find a power series representation for a particular function $$f$$ and the series converges on some interval, how do we prove that the series actually converges to $$f$$?
## Overview of Taylor/Maclaurin Series
Consider a function $$f$$ that has a power series representation at $$x=a$$. Then the series has the form
$\sum_{n=0}^∞c_n(x−a)^n=c_0+c_1(x−a)+c_2(x−a)^2+ \dots. \label{eq1}$
What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series Equation \ref{eq1} is a representation for $$f$$ at $$x=a$$, we certainly want the series to equal $$f(a)$$ at $$x=a$$. Evaluating the series at $$x=a$$, we see that
$\sum_{n=0}^∞c_n(x−a)^n=c_0+c_1(a−a)+c_2(a−a)^2+\dots=c_0.\label{eq2}$
Thus, the series equals $$f(a)$$ if the coefficient $$c_0=f(a)$$. In addition, we would like the first derivative of the power series to equal $$f′(a)$$ at $$x=a$$. Differentiating Equation \ref{eq2} term-by-term, we see that
$\dfrac{d}{dx}\left( \sum_{n=0}^∞c_n(x−a)^n \right)=c_1+2c_2(x−a)+3c_3(x−a)^2+\dots.\label{eq3}$
Therefore, at $$x=a,$$ the derivative is
$\dfrac{d}{dx}\left( \sum_{n=0}^∞c_n(x−a)^n \right)=c_1+2c_2(a−a)+3c_3(a−a)^2+\dots=c_1.\label{eq4}$
Therefore, the derivative of the series equals $$f′(a)$$ if the coefficient $$c_1=f′(a).$$ Continuing in this way, we look for coefficients $$c_n$$ such that all the derivatives of the power series Equation will agree with all the corresponding derivatives of $$f$$ at $$x=a$$. The second and third derivatives of Equation \ref{eq3} are given by
$\dfrac{d^2}{dx^2} \left(\sum_{n=0}^∞c_n(x−a)^n \right)=2c_2+3⋅2c_3(x−a)+4⋅3c_4(x−a)^2+\dots\label{eq5}$
and
$\dfrac{d^3}{dx^3} \left( \sum_{n=0}^∞c_n(x−a)^n \right)=3⋅2c_3+4⋅3⋅2c_4(x−a)+5⋅4⋅3c_5(x−a)^2+⋯.\label{eq6}$
Therefore, at $$x=a$$, the second and third derivatives
$\dfrac{d^2}{dx^2}(\sum_{n=0}^∞c_n(x−a)^n)=2c_2+3⋅2c_3(a−a)+4⋅3c_4(a−a)^2+\dots=2c_2\label{eq7}$
and
$\dfrac{d^3}{dx^3} \left(\sum_{n=0}^∞c_n(x−a)^n\right)=3⋅2c_3+4⋅3⋅2c_4(a−a)+5⋅4⋅3c_5(a−a)^2+\dots =3⋅2c_3\label{eq8}$
equal $$f''(a)$$ and $$f'''(a)$$, respectively, if $$c_2=\dfrac{f''(a)}{2}$$ and $$c_3=\dfrac{f'''(a)}{3}⋅2$$. More generally, we see that if $$f$$ has a power series representation at $$x=a$$, then the coefficients should be given by $$c_n=\dfrac{f^{(n)}(a)}{n!}$$. That is, the series should be
$\sum_{n=0}^∞\dfrac{f^{(n)}(a)}{n!}(x−a)^n=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+\dfrac{f'''(a)}{3!}(x−a)^3+⋯.$
This power series for $$f$$ is known as the Taylor series for $$f$$ at $$a.$$ If $$x=0$$, then this series is known as the Maclaurin series for $$f$$.
Definition: Maclaurin and Taylor series
If $$f$$ has derivatives of all orders at $$x=a$$, then the Taylor series for the function $$f$$ at $$a$$ is
$\sum_{n=0}^∞\dfrac{f^{(n)}(a)}{n!}(x−a)^n=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+⋯+\dfrac{f^{(n)}(a)}{n!}(x−a)^n+⋯.$
The Taylor series for $$f$$ at 0 is known as the Maclaurin series for $$f$$.
Later in this section, we will show examples of finding Taylor series and discuss conditions under which the Taylor series for a function will converge to that function. Here, we state an important result. Recall that power series representations are unique. Therefore, if a function $$f$$ has a power series at $$a$$, then it must be the Taylor series for $$f$$ at $$a$$.
Uniqueness of Taylor Series
If a function $$f$$ has a power series at a that converges to $$f$$ on some open interval containing a, then that power series is the Taylor series for $$f$$ at $$a$$.
The proof follows directly from that discussed previously.
To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as Taylor polynomials.
### Taylor Polynomials
The nth partial sum of the Taylor series for a function $$f$$ at $$a$$ is known as the nth Taylor polynomial. For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by
$$p_0(x)=f(a)$$,
$$p_1(x)=f(a)+f′(a)(x−a)$$,
$$p_2(x)=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2$$,
$$p_3(x)=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+\dfrac{f'''(a)}{3!}(x−a)^3$$,
respectively. These partial sums are known as the 0th, 1st, 2nd, and 3rd Taylor polynomials of $$f$$ at $$a$$, respectively. If $$x=a$$, then these polynomials are known as Maclaurin polynomials for $$f$$. We now provide a formal definition of Taylor and Maclaurin polynomials for a function $$f$$.
Definition: Maclaurin polynomial
If $$f$$ has n derivatives at $$x=a$$, then the nth Taylor polynomial for $$f$$ at $$a$$ is
$p_n(x)=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+\dfrac{f'''(a)}{3!}(x−a)^3+⋯+\dfrac{f^{(n)}(a)}{n!}(x−a)^n.$
The nth Taylor polynomial for $$f$$ at 0 is known as the nth Maclaurin polynomial for $$f$$.
We now show how to use this definition to find several Taylor polynomials for $$f(x)=lnx$$ at $$x=1$$.
Example $$\PageIndex{1}$$: Finding Taylor Polynomials
Find the Taylor polynomials $$p_0,p_1,p_2$$ and $$p_3$$ for $$f(x)=lnx$$ at $$x=1$$. Use a graphing utility to compare the graph of $$f$$ with the graphs of $$p_0,p_1,p_2$$ and $$p_3$$.
Solution
To find these Taylor polynomials, we need to evaluate $$f$$ and its first three derivatives at $$x=1$$.
$$f(x)=lnx$$ $$f(1)=0$$
$$f′(x)=\dfrac{1}{x}$$ $$f′(1)=1$$
$$f''(x)=−\dfrac{1}{x^2}$$ $$f''(1)=−1$$
$$f'''(x)==\dfrac{2}{x^3}$$ $$f'''(1)=2$$
Therefore,
$$p_0(x)=f(1)=0,$$
$$p_1(x)=f(1)+f′(1)(x−1)=x−1,$$
$$p_2(x)=f(1)+f′(1)(x−1)+\dfrac{f''(1)}{2}(x−1)^2=(x−1)−\dfrac{1}{2}(x−1)^2$$,
$$p_3(x)=f(1)+f′(1)(x−1)+\dfrac{f''(1)}{2}(x−1)^2+\dfrac{f'''(1)}{3!}(x−1)^3=(x−1)−\dfrac{1}{2}(x−1)^2+\dfrac{1}{3}(x−1)^3$$.
The graphs of $$y=f(x)$$ and the first three Taylor polynomials are shown in Figure 1.
Figure 1: The function $$y=lnx$$ and the Taylor polynomials $$p_0,p_1,p_2$$ and $$p_3$$ at $$x=1$$ are plotted on this graph.
Exercise $$\PageIndex{1}$$
Find the Taylor polynomials $$p_0,p_1,p_2$$ and $$p_3$$ for $$f(x)=\dfrac{1}{x^2}$$ at $$x=1$$.
Solution
$p_0(x)=1$
$p_1(x)=1−2(x−1)$
$p_2(x)=1−2(x−1)+3(x−1)^2$
$p_3(x)=1−2(x−1)+3(x−1)^2−4(x−1)^3$
Hint: Find the first three derivatives of $$f$$ and evaluate them at $$x=1.$$
We now show how to find Maclaurin polynomials for $$e^x, \sin x,$$ and $$\cos x$$. As stated above, Maclaurin polynomials are Taylor polynomials centered at zero.
Example $$\PageIndex{2}$$: Finding Maclaurin Polynomials
For each of the following functions, find formulas for the Maclaurin polynomials $$p_0,p_1,p_2$$ and $$p_3$$. Find a formula for the nth Maclaurin polynomial and write it using sigma notation. Use a graphing utility to compare the graphs of $$p_0,p_1,p_2$$ and $$p_3$$ with $$f$$.
1. $$f(x)=e^x$$
2. $$f(x)=\sin x$$
3. $$f(x)=\cos x$$
Solution
Since $$f(x)=e^x$$,we know that $$f(x)=f′(x)=f''(x)=⋯=f^{(n)}(x)=e^x$$ for all positive integers n. Therefore,
$f(0)=f′(0)=f''(0)=⋯=f^{(n)}(0)=1 \nonumber$
for all positive integers n. Therefore, we have
$$p_0(x)=f(0)=1,$$
$$p_1(x)=f(0)+f′(0)x=1+x,$$
$$p_2(x)=f(0)+f′(0)x+\dfrac{f''(0)}{2!}x^2=1+x+\dfrac{1}{2}x^2$$,
$$p_3(x)=f(0)+f′(0)x+\dfrac{f''(0)}{2}x^2+\dfrac{f'''(0)}{3!}x^3=1+x+\dfrac{1}{2}x^2+\dfrac{1}{3!}x^3$$,
$$p_n(x)=f(0)+f′(0)x+\dfrac{f''(0)}{2}x^2+\dfrac{f'''(0)}{3!}x^3+⋯+\dfrac{f^{(n)}(0)}{n!}x^n=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+⋯+\dfrac{x^n}{n!}=\sum_{k=0}^n\dfrac{x^k}{k!}$$.
The function and the first three Maclaurin polynomials are shown in Figure 2.
Figure 2: The graph shows the function $$y=e^x$$ and the Maclaurin polynomials $$p_0,p_1,p_2$$ and $$p_3$$.
b. For $$f(x)=\sin x$$, the values of the function and its first four derivatives at $$x=0$$ are given as follows:
$$f(x)=\sin x$$ $$f(0)=0$$
$$f′(x)=\cos x$$ $$f′(0)=1$$
$$f''(x)=−\sin x$$ $$f''(0)=0$$
$$f'''(x)=−\cos x$$ $$f'''(0)=−1$$
$$f^{(4)}(x)=\sin x$$ $$f^{(4)}(0)=0$$.
Since the fourth derivative is $$\sin x,$$ the pattern repeats. That is, $$f^{(2m)}(0)=0$$ and $$f^{(2m+1)}(0)=(−1)^m$$ for $$m≥0.$$ Thus, we have
$$p_0(x)=0,$$
$$p_1(x)=0+x=x,$$
$$p_2(x)=0+x+0=x,$$
$$p_3(x)=0+x+0−\dfrac{1}{3!}x^3=x−\dfrac{x^3}{3!},$$
$$p_4(x)=0+x+0−\dfrac{1}{3!}x^3+0=x−\dfrac{x^3}{3!}$$,
$$p_5(x)=0+x+0−\dfrac{1}{3!}x^3+0+\dfrac{1}{5!}x^5=x−\dfrac{x^3}{3!}+\dfrac{x^5}{5!}$$,
and for $$m≥0$$,
$$p_{2m+1}(x)=p_{2m+2}(x)=x−\dfrac{x^3}{3!}+\dfrac{x^5}{5!}−⋯+(−1)^m\dfrac{x^{2m+1}}{(2m+1)!}=\sum_{k=0}^m(−1)^k\dfrac{x^{2k+1}}{(2k+1)!}$$.
Graphs of the function and its Maclaurin polynomials are shown in Figure 3.
Figure 3: The graph shows the function $$y=\sin x$$ and the Maclaurin polynomials $$p_1,p_3$$ and $$p_5$$.
c. For f(x)=\cos x, the values of the function and its first four derivatives at x=0 are given as follows:
$$f(x)=\cos x$$ $$f(0)=1$$
$$f′(x)=−\sin x$$ $$f′(0)=0$$
$$f''(x)=−\cos x$$ $$f''(0)=−1$$
$$f'''(x)=\sin x$$ $$f'''(0)=0$$
$$f^{(4)}(x)=\cos x$$ $$f^{(4)}(0)=1.$$
Since the fourth derivative is $$\sin x$$, the pattern repeats. In other words, $$f^{(2m)}(0)=(−1)^m$$ and $$f^{(2m+1)}=0$$ for $$m≥0$$. Therefore,
$$p_0(x)=1,$$
$$p_1(x)=1+0=1,$$
$$p_2(x)=1+0−\dfrac{1}{2!}x^2=1−\dfrac{x^2}{2!}$$,
$$p_3(x)=1+0−\dfrac{1}{2!}x^2+0=1−\dfrac{x^2}{2!}$$,
$$p_4(x)=1+0−\dfrac{1}{2!}x^2+0+\dfrac{1}{4!}x^4=1−\dfrac{x^2}{2!}+\dfrac{x^4}{4!}$$,
$$p_5(x)=1+0−\dfrac{1}{2!}x^2+0+\dfrac{1}{4!}x^4+0=1−\dfrac{x^2}{2!}+\dfrac{x^4}{4!}$$,
and for $$n≥0$$,
$$p_{2m}(x)=p_{2m+1}(x)=1−\dfrac{x^2}{2!}+\dfrac{x^4}{4!}−⋯+(−1)^m\dfrac{x^{2m}}{(2m)!}=\sum_{k=0}^m(−1)^k\dfrac{x^{2k}}{(2k)!}$$.
Graphs of the function and the Maclaurin polynomials appear in Figure 4.
Figure 4: The function $$y=\cos x$$ and the Maclaurin polynomials $$p_0,p_2$$ and $$p_4$$ are plotted on this graph.
Exercise $$\PageIndex{2}$$
Find formulas for the Maclaurin polynomials $$p_0,p_1,p_2$$ and $$p_3$$ for $$f(x)=\dfrac{1}{1+x}$$.
Find a formula for the nth Maclaurin polynomial. Write your anwer using sigma notation.
Hint Evaluate the first four derivatives of $$f$$ and look for a pattern.
Solution
$$p_0(x)=1;p_1(x)=1−x;p_2(x)=1−x+x^2;p_3(x)=1−x+x^2−x^3;p_n(x)=1−x+x^2−x^3+⋯+(−1)^nx^n=_{k=0}^n(−1)^kx^k$$
## Taylor’s Theorem with Remainder
Recall that the nth Taylor polynomial for a function $$f$$ at a is the nth partial sum of the Taylor series for $$f$$ at a. Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials $${p_n}$$ converges. However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to $$f$$. To answer this question, we define the remainder $$R_n(x)$$ as
$R_n(x)=f(x)−p_n(x).$
For the sequence of Taylor polynomials to converge to $$f$$, we need the remainder $$R_n$$ to converge to zero. To determine if $$R_n$$ converges to zero, we introduce Taylor’s theorem with remainder. Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the nth Taylor polynomial approximates the function.
Here we look for a bound on $$|R_n|.$$ Consider the simplest case: $$n=0$$. Let $$p_0$$ be the 0th Taylor polynomial at a for a function $$f$$. The remainder $$R_0$$ satisfies
$$R_0(x)=f(x)−p_0(x)=f(x)−f(a).$$
If $$f$$ is differentiable on an interval I containing a and x, then by the Mean Value Theorem there exists a real number c between a and x such that $$f(x)−f(a)=f′(c)(x−a)$$. Therefore,
$R_0(x)=f′(c)(x−a).$
Using the Mean Value Theorem in a similar argument, we can show that if $$f$$ is n times differentiable on an interval I containing a and x, then the nth remainder $$R_n$$ satisfies
$R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1}$
for some real number c between a and x. It is important to note that the value c in the numerator above is not the center a, but rather an unknown value c between a and x. This formula allows us to get a bound on the remainder $$R_n$$. If we happen to know that $$∣f^{(n+1)}(x)∣$$ is bounded by some real number M on this interval I, then
$|R_n(x)|≤\dfrac{M}{(n+1)!}|x−a|^{n+1}$
for all x in the interval I.
We now state Taylor’s theorem, which provides the formal relationship between a function $$f$$ and its nth degree Taylor polynomial $$p_n(x)$$. This theorem allows us to bound the error when using a Taylor polynomial to approximate a function value, and will be important in proving that a Taylor series for $$f$$ converges to $$f$$.
Taylor’s Theorem with Remainder
Let $$f$$ be a function that can be differentiated $$n+1$$ times on an interval I containing the real number a. Let $$p_n$$ be the nth Taylor polynomial of $$f$$ at a and let
$R_n(x)=f(x)−p_n(x)$
be the nth remainder. Then for each x in the interval I, there exists a real number c between a and x such that
$R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1}$.
If there exists a real number M such that $$∣f^{(n+1)}(x)∣≤M$$ for all $$x∈I$$, then
$|R_n(x)|≤\dfrac{M}{(n+1)!}|x−a|^{n+1}$
for all x in I.
Proof
Fix a point $$x∈I$$ and introduce the function g such that
$g(t)=f(x)−f(t)−f′(t)(x−t)−\dfrac{f''(t)}{2!}(x−t)^2−⋯−\dfrac{f^{(n)}(t)}{n!}(x−t)^n−R_n(x)\dfrac{(x−t)^{n+1}}{(x−a)^{n+1}}$.
We claim that g satisfies the criteria of Rolle’s theorem. Since g is a polynomial function (in t), it is a differentiable function. Also, g is zero at $$t=a$$ and $$t=x$$ because
$$g(a)=f(x)−f(a)−f′(a)(x−a)−\dfrac{f''(a)}{2!}(x−a)^2+⋯+\dfrac{f^{(n)}(a)}{n!}(x−a)^n−R_n(x)$$
$$=f(x)−p_n(x)−R_n(x)$$
$$=0,$$
$$g(x)=f(x)−f(x)−0−⋯−0$$
$$=0.$$
Therefore, g satisfies Rolle’s theorem, and consequently, there exists c between a and x such that $$g′(c)=0.$$ We now calculate $$g′$$. Using the product rule, we note that
$\dfrac{d}{dt}[\dfrac{f^{(n)}(t)}{n!}(x−t)^n]=−\dfrac{f^{(n)}(t)}{(n−1)!}(x−t)^{n−1}+\dfrac{f^{(n+1)}(t)}{n!}(x−t)^n.$
Consequently,
$g′(t)=−f′(t)+[f′(t)−f''(t)(x−t)]+[f''(t)(x−t)−\dfrac{f'''(t)}{2!}(x−t)^2]+⋯+[\dfrac{f^{(n)}(t)}{(n−1)!}(x−t)^{n−1}−\dfrac{f^{(n+1)}(t)}{n!}(x−t)^n]+(n+1)R_n(x)\dfrac{(x−t)^n}{(x−a)^{n+1}}$.
Notice that there is a telescoping effect. Therefore,
$g′(t)=−\dfrac{f^{(n+1)}(t)}{n!}(x−t)^n+(n+1)R_n(x)\dfrac{(x−t)^n}{(x−a)^{n+1}}$.
By Rolle’s theorem, we conclude that there exists a number c between a and x such that $$g′(c)=0.$$ Since
$g′(c)=−\dfrac{f^{(n+1})(c)}{n!}(x−c)^n+(n+1)R_n(x)\dfrac{(x−c)^n}{(x−a)^{n+1}}$
we conclude that
$−\dfrac{f^{(n+1)}(c)}{n!}(x−c)^n+(n+1)R_n(x)\dfrac{(x−c)^n}{(x−a)^{n+1}}=0.$
Adding the first term on the left-hand side to both sides of the equation and dividing both sides of the equation by $$n+1,$$ we conclude that
$R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1}$
as desired. From this fact, it follows that if there exists M such that $$∣f^{(n+1)}(x)∣≤M$$ for all x in I, then
$|R_n(x)|≤\dfrac{M}{(n+1)!}|x−a|^{n+1}$.
Not only does Taylor’s theorem allow us to prove that a Taylor series converges to a function, but it also allows us to estimate the accuracy of Taylor polynomials in approximating function values. We begin by looking at linear and quadratic approximations of $$f(x)=\dfrac[3]{x}$$ at $$x=8$$ and determine how accurate these approximations are at estimating $$\dfrac[3]{11}$$.
Example $$\PageIndex{3}$$: Using Linear and Quadratic Approximations to Estimate Function Values
Consider the function $$f(x)=\sqrt[3]{x}$$.
1. Find the first and second Taylor polynomials for $$f$$ at $$x=8$$. Use a graphing utility to compare these polynomials with $$f$$ near $$x=8.$$
2. Use these two polynomials to estimate $$\sqrt[3]{11}$$.
3. Use Taylor’s theorem to bound the error.
Solution:
a. For $$f(x)=\sqrt[3]{x}$$, the values of the function and its first two derivatives at $$x=8$$ are as follows:
$$f(x)=\sqrt[3]{x}$$ $$f(8)=2$$
$$f′(x)=\dfrac{1}{3x^{2/3}}$$ $$f′(8)=\dfrac{1}{12}$$
$$f''(x)=\dfrac{−2}{9x^{5/3}}$$ $$f''(8)=−\dfrac{1}{144.}$$
Thus, the first and second Taylor polynomials at $$x=8$$ are given by
$$p_1(x)=f(8)+f′(8)(x−8)$$
$$=2+\dfrac{1}{12}(x−8)$$
$$p_2(x)=f(8)+f′(8)(x−8)+\dfrac{f''(8)}{2!}(x−8)^2$$
$$=2+\dfrac{1}{12}(x−8)−\dfrac{1}{288}(x−8)^2$$.
The function and the Taylor polynomials are shown in Figure 4.
Figure 4: The graphs of $$f(x)=\sqrt[3]{x}$$ and the linear and quadratic approximations $$p_1(x)$$ and $$p_2(x)$$
b. Using the first Taylor polynomial at $$x=8$$, we can estimate
$\dfrac[3]{11}≈p_1(11)=2+\dfrac{1}{12}(11−8)=2.25.$
Using the second Taylor polynomial at $$x=8$$, we obtain
$\sqrt[3]{11}≈p_2(11)=2+\dfrac{1}{12}(11−8)−\dfrac{1}{288}(11−8)^2=2.21875.$
c. By Note, there exists a c in the interval $$(8,11)$$ such that the remainder when approximating $$\sqrt[3]{11}$$ by the first Taylor polynomial satisfies
$R_1(11)=\dfrac{f''(c)}{2!}(11−8)^2.$
We do not know the exact value of c, so we find an upper bound on $$R_1(11)$$ by determining the maximum value of $$f''$$ on the interval $$(8,11)$$. Since $$f''(x)=−\dfrac{2}{9x^{5/3}}$$, the largest value for $$|f''(x)|$$ on that interval occurs at $$x=8$$. Using the fact that $$f''(8)=−\dfrac{1}{144}$$, we obtain
$$|R_1(11)|≤\dfrac{1}{144⋅2!}(11−8)^2=0.03125.$$
Similarly, to estimate $$R_2(11)$$, we use the fact that
$$R_2(11)=\dfrac{f'''(c)}{3!}(11−8)^3$$.
Since $$f'''(x)=\dfrac{10}{27x^{8/3}}$$, the maximum value of $$f'''$$ on the interval $$(8,11)$$ is $$f'''(8)≈0.0014468$$. Therefore, we have
$$|R_2(11)|≤\dfrac{0.0011468}{3!}(11−8)^3≈0.0065104.$$
Exercise $$\PageIndex{3}$$:
Find the first and second Taylor polynomials for $$f(x)=\sqrt{x}$$ at $$x=4$$. Use these polynomials to estimate $$\sqrt{6}$$. Use Taylor’s theorem to bound the error.
Solution
$$p_1(x)=2+\dfrac{1}{4}(x−4);p_2(x)=2+\dfrac{1}{4}(x−4)−\dfrac{1}{64}(x−4)^2;p_1(6)=2.5;p_2(6)=2.4375;$$
$$|R_1(6)|≤0.0625;|R_2(6)|≤0.015625$$
Hint: Evaluate $$f(4),f′(4),$$ and $$f''(4).$$
Example $$\PageIndex{4}$$: Approximating sin x Using Maclaurin Polynomials
From Example b., the Maclaurin polynomials for $$\sin x$$ are given by
$p_{2m+1}(x)=p_{2m+2}(x)=x−\dfrac{x^3}{3!}+\dfrac{x^5}{5!}−\dfrac{x^7}{7!}+⋯+(−1)^m\dfrac{x^{2m+1}}{(2m+1)!}\= for $$m=0,1,2,….$$ 1. Use the fifth Maclaurin polynomial for $$\sin x$$ to approximate $$sin(\dfrac{π}{18})$$ and bound the error. 2. For what values of x does the fifth Maclaurin polynomial approximate \sin x to within 0.0001? Solution a. The fifth Maclaurin polynomial is \[p_5(x)=x−\dfrac{x^3}{3!}+\dfrac{x^5}{5!}$.
Using this polynomial, we can estimate as follows:
$\sin(\dfrac{π}{18})≈p_5(\dfrac{π}{18})=\dfrac{π}{18}−\dfrac{1}{3!}(\dfrac{π}{18})^3+\dfrac{1}{5!}(\dfrac{π}{18})^5≈0.173648.$
To estimate the error, use the fact that the sixth Maclaurin polynomial is $$p_6(x)=p_5(x)$$ and calculate a bound on $$R_6(\dfrac{π}{18})$$. By Note, the remainder is
$R_6(\dfrac{π}{18})=\dfrac{f^{(7)}(c)}{7!}(\dfrac{π}{18})^7$
for some c between 0 and $$\dfrac{π}{18}$$. Using the fact that $$∣f^{(7)}(x)∣≤1$$ for all $$x$$, we find that the magnitude of the error is at most
$\dfrac{1}{7!}⋅(\dfrac{π}{18})^7≤9.8×10^{−10}.$
b.
We need to find the values of x such that
$\dfrac{1}{7}!|x|^7≤0.0001.$
Solving this inequality for $$x$$, we have that the fifth Maclaurin polynomial gives an estimate to within 0.0001 as long as $$|x|<0.907.$$
Exercise $$\PageIndex{4}$$:
Use the fourth Maclaurin polynomial for $$\cos x$$ to approximate $$\cos(\dfrac{π}{12}).$$
Hint: The fourth Maclaurin polynomial is $$p_4(x)=1−\dfrac{x^2}{2!}+\dfrac{x^4}{4!}$$.
Solution
0.96593
Now that we are able to bound the remainder $$R_n(x)$$, we can use this bound to prove that a Taylor series for $$f$$ at a converges to $$f$$.
## Representing Functions with Taylor and Maclaurin Series
We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.
Example $$\PageIndex{5}$$: Finding a Taylor Series
Find the Taylor series for $$f(x)=\dfrac{1}{x}$$ at $$x=1$$. Determine the interval of convergence.
Solution
For $$f(x)=\dfrac{1}{x},$$ the values of the function and its first four derivatives at $$x=1$$ are
$$f(x)=\dfrac{1}{x}$$ $$f(1)=1$$
$$f′(x)=−\dfrac{1}{x^2}$$ $$f′(1)=−1$$
$$f''(x)=\dfrac{2}{x^3}$$ $$f''(1)=2!$$
$$f'''(x)=−\dfrac{3⋅2}{x^4}$$ $$f'''(1)=−3!$$
$$f^{(4)}(x)=\dfrac{4⋅3⋅2}{x^5}$$ $$f^{(4)}(1)=4!$$.
That is, we have $$f^{(n)}(1)=(−1)^nn!$$ for all $$n≥0$$. Therefore, the Taylor series for $$f$$ at $$x=1$$ is given by
$$\sum_{n=0}^∞\dfrac{f^{(n)}(1)}{n!}(x−1)^n=\sum_{n=0}^∞(−1)^n(x−1)^n$$.
To find the interval of convergence, we use the ratio test. We find that
$$\dfrac{|a_{n+1}|}{|a_n|}=\dfrac{∣(−1)^{n+1}(x−1)n^{+1}∣}{|(−1)^n(x−1)^n|}=|x−1|$$.
Thus, the series converges if $$|x−1|<1.$$ That is, the series converges for $$0<x<2$$. Next, we need to check the endpoints. At $$x=2$$, we see that
$$\sum_{n=0}^∞(−1)^n(2−1)^n=\sum_{n=0}^∞(−1)^n$$
diverges by the divergence test. Similarly, at $$x=0,$$
$$\sum_{n=0}^∞(−1)^n(0−1)^n=\sum_{n=0}^∞(−1)^{2n}=\sum_{n=0}^∞1$$
diverges. Therefore, the interval of convergence is $$(0,2)$$.
Exercise $$\PageIndex{5}$$
Find the Taylor series for $$f(x)=\dfrac{1}{2}$$ at $$x=2$$ and determine its interval of convergence.
Hint: $$f^{(n)}(2)=\dfrac{(−1)^nn!}{2^{n+1}}$$
Solution
$$\dfrac{1}{2}\sum_{n=0}^∞(\dfrac{2−x}{2})^n$$. The interval of convergence is $$(0,4)$$.
We know that the Taylor series found in this example converges on the interval $$(0,2)$$, but how do we know it actually converges to $$f$$? We consider this question in more generality in a moment, but for this example, we can answer this question by writing
$f(x)=\dfrac{1}{x}=\dfrac{1}{1−(1−x)}.$
That is, $$f$$ can be represented by the geometric series $$\sum_{n=0}^∞(1−x)^n$$. Since this is a geometric series, it converges to $$\dfrac{1}{x}$$ as long as $$|1−x|<1.$$ Therefore, the Taylor series found in Example does converge to $$f(x)=\dfrac{1}{x}$$ on $$(0,2).$$
We now consider the more general question: if a Taylor series for a function $$f$$ converges on some interval, how can we determine if it actually converges to $$f$$? To answer this question, recall that a series converges to a particular value if and only if its sequence of partial sums converges to that value. Given a Taylor series for $$f$$ at a, the nth partial sum is given by the nth Taylor polynomial pn. Therefore, to determine if the Taylor series converges to $$f$$, we need to determine whether
$$\lim_{n→∞}p_n(x)=f(x)$$.
Since the remainder $$R_n(x)=f(x)−p_n(x)$$, the Taylor series converges to $$f$$ if and only if
$$\lim_{n→∞}R_n(x)=0.$$
We now state this theorem formally.
Convergence of Taylor Series
Suppose that $$f$$ has derivatives of all orders on an interval I containing a. Then the Taylor series
$\sum_{n=0}^∞\dfrac{f^{(n)}(a)}{n!}(x−a)^n$
converges to $$f(x)$$ for all x in I if and only if
$\lim_{n→∞}R_n(x)=0$
for all x in I.
With this theorem, we can prove that a Taylor series for $$f$$ at a converges to $$f$$ if we can prove that the remainder $$R_n(x)→0$$. To prove that $$R_n(x)→0$$, we typically use the bound
$|R_n(x)|≤\dfrac{M}{(n+1)!}|x−a|^{n+1}$
from Taylor’s theorem with remainder.
In the next example, we find the Maclaurin series for $$e^x$$ and $$\sin x$$ and show that these series converge to the corresponding functions for all real numbers by proving that the remainders $$R_n(x)→0$$ for all real numbers $$x$$.
Example $$\PageIndex{6}$$: Finding Maclaurin Series
For each of the following functions, find the Maclaurin series and its interval of convergence. Use Note to prove that the Maclaurin series for $$f$$ converges to $$f$$ on that interval.
1. $$e^x$$
2. $$\sin x$$
Solution
a. Using the nth Maclaurin polynomial for $$e^x$$ found in Examplea., we find that the Maclaurin series for $$e^x$$ is given by
$$\sum_{n=0}^∞\dfrac{x^n}{n!}$$.
To determine the interval of convergence, we use the ratio test. Since
$$\dfrac{|a_{n+1}|}{|a_n|}=\dfrac{|x|^{n+1}}{(n+1)!}⋅\dfrac{n!}{|x|^n}=\dfrac{|x|}{n+1}$$,
we have
$$\lim_{n→∞}\dfrac{|a_{n+1}|}{|a_n|}=\lim_{n→∞}\dfrac{|x|}{n+1}=0$$
for all $$x$$. Therefore, the series converges absolutely for all $$x$$, and thus, the interval of convergence is $$(−∞,∞)$$. To show that the series converges to $$e^x$$ for all $$x$$, we use the fact that $$f^{(n)}(x)=e^x$$ for all $$n≥0$$ and $$e^x$$ is an increasing function on $$(−∞,∞)$$. Therefore, for any real number $$b$$, the maximum value of $$e^x$$ for all $$|x|≤b$$ is $$e^b$$. Thus,
$$|R_n(x)|≤\dfrac{e^b}{(n+1)!}|x|^{n+1}$$.
Since we just showed that
$$\sum_{n=0}^∞\dfrac{|x|^n}{n!}$$
converges for all x, by the divergence test, we know that
$$\lim_{n→∞}\dfrac{|x|^{n+1}}{(n+1)!}=0$$
for any real number x. By combining this fact with the squeeze theorem, the result is $$\lim_{n→∞}R_n(x)=0.$$
b. Using the nth Maclaurin polynomial for $$\sin x$$ found in Example b., we find that the Maclaurin series for $$\sin x$$ is given by
$$\sum_{n=0}^∞(−1)^n\dfrac{x^{2n+1}}{(2n+1)!}$$.
In order to apply the ratio test, consider
$$\dfrac{|a_{n+1}|}{|a_n|}=\dfrac{|x|^{2n+3}}{(2n+3)!}⋅\dfrac{(2n+1)!}{|x|^{2n+1}}=\dfrac{|x|^2}{(2n+3)(2n+2)}$$.
Since
$$\lim_{n→∞}\dfrac{|x|^2}{(2n+3)(2n+2)}=0$$
for all x, we obtain the interval of convergence as $$(−∞,∞).$$ To show that the Maclaurin series converges to $$\sin x$$, look at $$R_n(x)$$. For each $$x$$ there exists a real number c between 0 and x such that
$$R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$.
Since $$∣f^{(n+1)}(c)∣≤1$$ for all integers n and all real numbers c, we have
$$|R_n(x)|≤\dfrac{|x|^{n+1}}{(n+1)!}$$
for all real numbers x. Using the same idea as in part a., the result is $$\lim_{n→∞}R_n(x)=0$$ for all x, and therefore, the Maclaurin series for $$\sin x$$ converges to $$\sin x$$ for all real x.
Exercise $$\PageIndex{6}$$
Find the Maclaurin series for $$f(x)=\cos x$$. Use the ratio test to show that the interval of convergence is $$(−∞,∞)$$. Show that the Maclaurin series converges to $$\cos x$$ for all real numbers x.
Hint: Use the Maclaurin polynomials for $$\cos x.$$
Solution
$$\sum_{n=0}^∞\dfrac{(−1)^nx^{2n}}{(2n)!}$$
By the ratio test, the interval of convergence is $$(−∞,∞).$$ Since $$|R_n(x)|≤\dfrac{|x|^{n+1}}{(n+1)!}$$, the series converges to $$\cos x$$ for all real x.
Proving that E is Irrational
In this project, we use the Maclaurin polynomials for $$e^x$$ to prove that e is irrational. The proof relies on supposing that e is rational and arriving at a contradiction. Therefore, in the following steps, we suppose $$e=r/s$$ for some integers r and s where $$s≠0.$$
1. Write the Maclaurin polynomials $$p_0(x),p_1(x),p_2(x),p_3(x),p_4(x)$$ for $$e^x$$. Evaluate $$p_0(1),p_1(1),p_2(1),p_3(1),p_4(1)$$ to estimate e.
2. Let $$R_n(x)$$ denote the remainder when using $$p_n(x)$$ to estimate $$e^x$$. Therefore, $$R_n(x)=e^x−p_n(x)$$, and $$R_n(1)=e−p_n(1)$$. Assuming that $$e=\dfrac{r}{s}$$ for integers r and s, evaluate $$R_0(1),R_1(1),R_2(1),R_3(1),R_4(1).$$
3. Using the results from part 2, show that for each remainder $$R_0(1),R_1(1),R_2(1),R_3(1),R_4(1),$$ we can find an integer k such that $$kR_n(1)$$ is an integer for $$n=0,1,2,3,4.$$
4. Write down the formula for the nth Maclaurin polynomial $$p_n(x)$$ for $$e^x$$ and the corresponding remainder $$R_n(x).$$ Show that $$sn!R_n(1)$$ is an integer.
5. Use Taylor’s theorem to write down an explicit formula for $$R_n(1)$$. Conclude that $$R_n(1)≠0$$, and therefore, $$sn!R_n(1)≠0$$.
6. Use Taylor’s theorem to find an estimate on $$R_n(1)$$. Use this estimate combined with the result from part 5 to show that $$|sn!R_n(1)|<\dfrac{se}{n+1}$$. Conclude that if n is large enough, then $$|sn!R_n(1)|<1$$. Therefore, $$sn!R_n(1)$$ is an integer with magnitude less than 1. Thus, $$sn!R_n(1)=0$$. But from part 5, we know that $$sn!R_n(1)≠0$$. We have arrived at a contradiction, and consequently, the original supposition that e is rational must be false.
## Key Concepts
• Taylor polynomials are used to approximate functions near a value $$x=a$$. Maclaurin polynomials are Taylor polynomials at $$x=0$$.
• The nth degree Taylor polynomials for a function $$f$$ are the partial sums of the Taylor series for $$f$$.
• If a function $$f$$ has a power series representation at $$x=a$$, then it is given by its Taylor series at $$x=a$$.
• A Taylor series for $$f$$ converges to $$f$$ if and only if $$\lim_{n→∞}R_n(x)=0$$ where $$R_n(x)=f(x)−p_n(x)$$.
• The Taylor series for $$e^x, \sin x$$, and $$\cos x$$ converge to the respective functions for all real x.
## Key Equations
• Taylor series for the function $$f$$ at the point $$x=a$$
$$\sum_{n=0}^∞\dfrac{f^{(n)}(a)}{n!}(x−a)^n=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+⋯+\dfrac{f^{(n)}(a)}{n!}(x−a)^n+⋯$$
## Glossary
Maclaurin polynomial
a Taylor polynomial centered at 0; the nth Taylor polynomial for $$f$$ at 0 is the nth Maclaurin polynomial for $$f$$
Maclaurin series
a Taylor series for a function $$f$$ at $$x=0$$ is known as a Maclaurin series for $$f$$
Taylor polynomials
the nth Taylor polynomial for $$f$$ at $$x=a$$ is $$p_n(x)=f(a)+f′(a)(x−a)+\dfrac{f''(a)}{2!}(x−a)^2+⋯+\dfrac{f^{(n)}(a)}{n!}(x−a)^n$$
Taylor series
a power series at a that converges to a function $$f$$ on some open interval containing a
Taylor’s theorem with remainder
for a function $$f$$ and the nth Taylor polynomial for $$f$$ at $$x=a$$, the remainder $$R_n(x)=f(x)−p_n(x)$$ satisfies $$R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1}$$
for some c between x and a; if there exists an interval I containing a and a real number M such that $$∣f^{(n+1)}(x)∣≤M$$ for all x in I, then $$|R_n(x)|≤\dfrac{M}{(n+1)!}|x−a|^{n+1}$$
### Contributors
• The OpenStax College name, OpenStax College logo, OpenStax College book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the creative commons license and may not be reproduced without the prior and express written consent of Rice University. For questions regarding this license, please contact partners@openstaxcollege.org. "Download for free at http://cnx.org/contents/fd53eae1-fa2...49835c3c@5.191."
| 12,133
| 30,971
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.09375
| 4
|
CC-MAIN-2017-47
|
longest
|
en
| 0.715768
|
http://www.jiskha.com/display.cgi?id=1204386079
| 1,495,481,194,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463607046.17/warc/CC-MAIN-20170522190443-20170522210443-00529.warc.gz
| 554,567,830
| 3,969
|
# math
posted by on .
Find the five-number summary of the following set of numbers.
335, 233, 185, 392, 235, 518, 281, 208, 318
• math - ,
First, arrange the set of numbers in order, with the lowest first.
185, 208, 233, 235, 281, 318, 335, 392, 518
Then, follow the instructions in this website.
http://illuminations.nctm.org/Lessons/States/FiveNumSum.htm
• math - ,
Here is a definition of Five-number summary:
(1) the minimum (smallest observation)
(2) the lower quartile or first quartile (which cuts off the lowest 25% of the data)
(3) the median (middle value)
the upper quartile or third quartile (4) (which cuts off the highest 25% of the data)
(5) the maximum (largest observation)
In your case the median is 281 , the minimum is 185 and the maximum is 518. For the two quartiles, take the third and seventh largest numbers of the group.
• math - ,
293.55
| 255
| 878
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.03125
| 4
|
CC-MAIN-2017-22
|
latest
|
en
| 0.790944
|
https://yoursageinformation.com/what-is-the-probability-of-rolling-a-1-on-a-6-sided-die/
| 1,725,722,995,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00595.warc.gz
| 1,058,446,244
| 16,074
|
# What is the probability of rolling a 1 on a 6 sided die?
## What is the probability of rolling a 1 on a 6 sided die?
Two (6-sided) dice roll probability table. Single die roll probability tables….Probability of rolling more than a certain number (e.g. roll more than a 5).
Roll more than a… Probability
1 5/6(83.33\%)
2 4/6 (66.67\%)
3 3/6 (50\%)
4 4/6 (66.667\%)
What is the probability of getting 1 in a dice?
1 / 6
Answer: The probability of getting 1 after rolling a die is 1 / 6.
What is the probability of rolling a 1 on a 6 sided die twice?
1/36
8 Answers. The probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6). The probability of rolling any number twice in a row is 1/6, because there are six ways to roll a specific number twice in a row (6 x 1/36).
### What is the percentage chance of rolling a 6 on a standard six-sided die?
So to get a 6 when rolling a six-sided die, probability = 1 ÷ 6 = 0.167, or 16.7 percent chance. So to get two 6s when rolling two dice, probability = 1/6 × 1/6 = 1/36 = 1 ÷ 36 = 0.0278, or 2.78 percent.
READ ALSO: What is the difference between Gandalf and Dumbledore?
What is a fair die?
At its simplest, a fair die means that each of the faces has the same probability of landing facing up. A standard six-sided die, for example, can be considered “fair” if each of the faces has a probability of 1/6.
What are the odds of rolling a 6 on a die?
Because there are six faces on a die, you have an even chance of the dice landing on one of these faces each time you roll: 1 6. This means that each time that you roll, there is a 5 6 chance that you will not roll a 6. The probability of not rolling a 6 twice is 5 6 ⋅ 5 6, or 69.4\%.
#### What is the final probability of a roll of the die?
For example, let’s say we have a regular die and y = 3. We want to rolled value to be either 6, 5, 4, or 3. The variable p is then 4 * 1/6 = 2/3, and the final probability is P = (2/3)ⁿ.
What is the probability of getting 7 on a 10-sided die?
READ ALSO: How did Ghidorah end up in ice?
There is a simple relationship – p = 1/s, so the probability of getting 7 on a 10 sided die is twice that of on a 20 sided die. The probability of rolling the same value on each die – while the chance of getting a particular value on a single die is p, we only need to multiply this probability by itself as many times as the number of dice.
What is the probability of rolling a 15 on a dice?
If we consider three 20 sided dice, the chance of rolling 15 on each of them is: P = (1/20)³ = 0.000125 (or P = 1.25·10⁻⁴ in scientific notation). And if you are intereseted in rolling the set of any identical values, simply multiply the result by the total die faces: P = 0.000125 * 20 = 0.0025.
| 838
| 2,839
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.53125
| 5
|
CC-MAIN-2024-38
|
latest
|
en
| 0.901712
|
http://www.wyzant.com/resources/lessons/math/trigonometry/unit-circle/cartesian-circle
| 1,387,401,300,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-48/segments/1387345759442/warc/CC-MAIN-20131218054919-00007-ip-10-33-133-15.ec2.internal.warc.gz
| 787,000,856
| 14,462
|
Search 73,889 tutors
# The Cartesian Circle
### Written by tutor Steve C.
There are three locations for graphing a circle in the XY Cartesian Plane:
At the Origin, On the Edge, and Anyplace Else.
Here is the standard circle with center at the origin, defined by x2 + y2 = 16
The general form is actually x2 + y2 = r2 where the radius r = 4
Here is the same size circle with center at (5, 5), defined by (x-5)2 + (y-5)2 = 16
The general form is actually (x-a)2 + (y-b)2 = r2 where the center is (a, b). Notice that the center points here are positive. The standard form notation of the circle equation does allow for them to be shown as negative.
If the circle center is at (-5, -5) then the standard form of the circle becomes (x+5)2 + (y+5)2 = 16 A similar pattern will result if the circle is moved to the 2nd quadrant where a<0 and b>0, or if it is moved to the 4th quadrant where a>0 and b<0.
## The Special Case
The final location for a circle graph is where the edge falls along the x axis and y axis.
Here is the same size circle with center at (4, 4), defined by (x-4)2 + (y-4)2 = 16.
The standard form of the equation is still (x-a)2 + (y-b)2 = r2 . However, in this case, a = b = r. In fact, we can state that the graph of the equation in this form (x-r)2 + (y-r)2 = r2 Is a circle sitting on the edge of the x axis and the y axis.
We can carry this further, (for the 1st quadrant only) where (x-a)2 + (y-b)2 = r2 : if a=r, the circle sits on the x axis; if a > r, the circle sits away from the x axis; if a < r, the circle crosses the x axis twice; if b=r, the circle sits on the y axis; if b > r, the circle sits away from the y axis; if b < r, the circle crosses the y axis twice. (x-2)2 + (y-4)2 = 42 (x-4)2 + (y-2)2 = 42
| 544
| 1,780
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.5625
| 5
|
CC-MAIN-2013-48
|
longest
|
en
| 0.857026
|
http://mathhelpforum.com/algebra/103186-exponential-equation-print.html
| 1,529,447,525,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267863206.9/warc/CC-MAIN-20180619212507-20180619232507-00463.warc.gz
| 224,235,694
| 2,603
|
# Exponential Equation
• Sep 19th 2009, 07:06 PM
RussW
Exponential Equation
2^x=2x
Thanks for the help!
• Sep 19th 2009, 07:21 PM
enjam
2^x = 2x
ln(2^x) = ln(2x)
xln(2) = ln(2) + ln (x)
x = [ln (2) + ln (x)] / ln (2)
x = 1 + ln(x)/ln(2)
From here, by simple inspection, we can see that x can be either 1 or 2.
when x = 1, we have 1 = 1 + ln(1)/ln(2) = 1 + 0
when x = 2, we have 2 = 1 + ln(2)/ln(2) = 1 + 1
• Sep 20th 2009, 05:49 AM
RussW
Thanks for your response. Other than by inspection, is there a way to solve a more generic equation. For example, solve for x when a^x=bx.
• Sep 20th 2009, 05:50 AM
e^(i*pi)
There are iterative methods but none analytical ones
| 281
| 668
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.875
| 4
|
CC-MAIN-2018-26
|
latest
|
en
| 0.837731
|
https://www.memrise.com/course/700001/learn-mathematics/628/
| 1,576,149,628,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540543252.46/warc/CC-MAIN-20191212102302-20191212130302-00139.warc.gz
| 793,616,578
| 60,537
|
Level 627 Level 629
Level 628
## Ignore words
Check the boxes below to ignore/unignore words, then click save at the bottom. Ignored words will never appear in any learning session.
Ignore?
Nth Term Test
If Limit as K approaches infinity, then the Series of K Diverges.
Alternating Series Test
If An is positive, the series ∑(-1)An converges if & only if..
Geometric Series Test
general term = a₁r^n, converges if -1 < r < 1
P-Series Test
general term = 1/n^p, converges if p > 1
Integral Test
take integral and evaluate, if it goes to a number it is convergent, if there is ∞in it, its divergent
Root Test
if limit as n nears infinity of the nth root of a sub n is less than 1, the series CONVERGES; and if the limit is greater than 1 or infinity, the series DIVERGES
Comparison Test
after comparing remember to take the limit as n→∞ of an/bn and for convergence it must lie on the interval 0<L<∞
Ratio Test
limit as n→∞ of an+1/an
nth-term
Diverges: lim n→∞ an ≠ 0
geometric series
sum of the terms of an arithmetic sequence
telescoping series
Converges: lim→∞ bn = L
P-Series
Sum where 1/(n^p) converges if and only if p > 1
integral
(works if f(x) is continuous, positive, decreasing)
root
A root of a polynomial in one variable is a value of the variable for which the polynomial is equal 0.
ratio
A comparison of two quantities, also know as fancy word for fraction.
0<an≤bn
direct comparison
limit comparison
Series converges if lim (as n approaches infinity) a{n}/ b{n}= L>0 and
| 425
| 1,489
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.65625
| 4
|
CC-MAIN-2019-51
|
longest
|
en
| 0.836681
|
https://stats.stackexchange.com/questions/605285/how-to-compute-a-prediction-interval-from-ordinary-least-squares-regression-outp
| 1,721,617,066,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00367.warc.gz
| 472,576,780
| 39,428
|
# How to compute a prediction interval from ordinary least squares regression output alone?
This question has been haunting me for a long time, When I'm given an R output of linear regression, and asked to calculate 95% prediction interval, I feel there's something missing.
In this output, how am I supposed to calculate 95% prediction interval for X_b value of 10?
I can evaluate y_hat and MSE from the output but I fail to understand how to get the mean of X_b and Sxx from here.
$$S_{xx},$$ the sum of squares of the explanatory variable, is easy to obtain from the formula
$$\operatorname{se}(\hat\beta_1) = \sqrt{\frac{MS_{Res}}{S_{xx}}}$$
where the left hand side is the standard error of the slope, given as $$1373$$ in the question, and $$MS_{Res}$$ is the mean squared residual, whose square root (the "residual standard error" is given as $$36600$$ in the question.
The mean of the explanatory variable can almost be recovered from the formula for the estimated sampling variance of the intercept,
$$\widehat{\operatorname{Var}}(\hat\beta_0) = MS_{Res}\left(\frac{1}{n} + \frac{\bar x^2}{S_{xx}}\right).$$
In the question, the left hand side is the square of the standard error, $$\widehat{\operatorname{Var}}(\hat\beta_0) = 8004^2$$ and $$n = 98 + 2$$ is found by adding the number of estimated coefficients to the "degrees of freedom" reported for the $$F$$ ratio statistic. Solving this for $$\bar x$$ usually gives two possible values. Unless you have some sense of what the value should be (one solution is positive and the other is negative), you're stuck (because, as you clearly are aware, the prediction interval at any value $$x_0$$ depends on its distance from $$\bar x$$ and the only value where that distance does not depend on the solution is $$x_0=0$$).
As an example of the problem, here is R code to manufacture two different datasets with differing values of $$\bar x$$ and identical ordinary least squares output.
x <- seq(2, 10, length.out = 30)
y <- x + rnorm(length(x))
fit <- lm(y ~ x)
b <- coefficients(fit)
x.bar <- mean(x)
x <- x - 2 * x.bar
y <- y - 2 * b[2] * x.bar
all.equal(summary(fit), summary(lm(y ~ x)))
It alters the initial data by subtracting $$2\bar x$$ from all $$x$$ values, subtracting $$2\hat\beta_1 \bar x$$ from all $$y$$ values to keep the coefficient estimates the same, and comparing their summaries. Its output is
Component “cov.unscaled”: Mean relative difference: 0.4387476
That is, the only difference between the two datasets lies in the estimated covariance between $$\hat\beta_0$$ and $$\hat\beta_1$$ -- but that is not part of your regression output. (The sign of this covariance will differ in the two datasets.) If it could be recovered from the output then some other numbers in the output would have to differ, too, but that's not the case.
Here is a plot of the original data (blue; $$\bar x = 6$$) and their transformed version (red; $$\bar x = -6$$). The line is the common least squares fit.
| 775
| 2,985
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.9375
| 4
|
CC-MAIN-2024-30
|
latest
|
en
| 0.88856
|
https://raw.githubusercontent.com/revbayes/revbayes_tutorial/master/RB_MCMC_Intro_Tutorial/scripts/Binomial_MH_algorithm.Rev
| 1,511,360,448,000,000,000
|
text/plain
|
crawl-data/CC-MAIN-2017-47/segments/1510934806609.33/warc/CC-MAIN-20171122141600-20171122161600-00482.warc.gz
| 696,605,291
| 1,675
|
################################################################################ # # RevBayes Example: Writing your on MCMC algorithm for a simple Binomial model # # authors: Mike May and Sebastian Hoehna # ################################################################################ # Make up some coin flips! # Feel free to change these numbers n <- 100 # the number of flips x <- 63 # the number of heads # Initialize the chain with starting values alpha <- 1 beta <- 1 p <- rbeta(n=1,alpha,beta)[1] # specify the likelihood function function likelihood(p) { if(p < 0 || p > 1) return 0 l = dbinomial(x,p,n,log=false) return l } # specify the prior function function prior(p) { if(p < 0 || p > 1) return 0 pp = dbeta(p,alpha,beta,log=false) return pp } # Prepare a file to log our samples write("iteration","p","\n",file="binomial_MH.log") write(0,p,"\n",file="binomial_MH.log",append=TRUE) # Print the initial values to the screen print("iteration","p") print(0,p) # Write the MH algorithm reps = 10000 printgen = 10 for (rep in 1:reps){ # Propose a new value of p p_prime <- runif(n=1,0.0,1.0)[1] # Compute the acceptance probability R <- ( likelihood(p_prime) / likelihood(p) ) * ( prior(p_prime) / prior(p) ) # Accept or reject the proposal u <- runif(1,0,1)[1] if (u < R){ # Accept the proposal p <- p_prime } else { # Reject the proposal # (we don't have to do anything here) } if ( (rep % printgen) == 0 ) { # Write the samples to a file write(rep,p,"\n",file="binomial_MH.log",append=TRUE) # Print the samples to the screen print(rep,p) } } # end MCMC q()
| 426
| 1,570
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.578125
| 4
|
CC-MAIN-2017-47
|
latest
|
en
| 0.557528
|
https://unacademy.com/lesson/simple-interest-for-kerala-psc-part-1-in-malayalam/0NG89YP4
| 1,561,458,002,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-26/segments/1560627999817.30/warc/CC-MAIN-20190625092324-20190625114324-00079.warc.gz
| 610,131,791
| 125,019
|
to enroll in courses, follow best educators, interact with the community and track your progress.
Enroll
2.4k
Simple Interest for Kerala PSC - Part 1 (in Malayalam)
12,015 plays
More
Typing Mistake in 1st qstn : SI was 2600 & Amount 9100 Sorry for the mistake This lesson discusses basic concepts of Simple Interest, % based approach and 3 problems on SI
I teach Maths & Reasoning for Kerala PSC Cracked RRB ALP, Group D, IBPS,SSC Exams Youtube - IMA Math Tricks Telegram - Issu Math Academy
Thank u so much
thnk you sir
sir pls compound interest tharanam
Compound interest and mixed problems cheyyumo sir....
1. SIMPLE INTEREST CONCEPT PART 1 8 SHORTCUT TRICKS SOLVING IN 5 SECS For more Classes, Please Follow My Profile: ISMAIEL KALADY
2. Course Contents Basic Concepts of Simple interest - Percentage based approach - Shortcut Tricks to solve in 5 seconds All Types of SI based problems Previous year PSC Questions included ISMAIEL KALADY
3. Simple Interest . If a certain amount is borrowed at a certain rate of interest, which is same for all the given years then it is called Simple Interest. Principal-Initial amount or the amount borrowed Amount - Sum of Principal and extra amount which a borrower has to return ISMAIEL KALADY
4. Simple Interest- Basic Equation .SI 100 A Amount T Time P Principal R Rate ISMAIEL KALADY
5. Simple Interest - Percentage Method Let Rate of Interest be 10% per annum & principal be 100% Year 1st 2nd 3rd 4th Percentage of Principal 10% 20% 30% 40% ISMAIEL KALADY
6. 1. Find s.l and Amount on 6500 at 8% p.a. for 5 years? Ans ISMAIEL KALADY
7. 1. Find s.l and Amount on 6500 at 8% p.a. for 5 years? Ans SI for 1 year-8 % SI for 5 years = (5 81% = 40% ISMAIEL KALADY
8. 1. Find s.l and Amount on 6500 at 8% p.a. for 5 years? Ans SI for 1 year-8 % SI for 5 years-(5 81% 40% /0= 40 Interest 6500-Rs.2800 100 Amount 65002800 Rs. 9300 ISMAIEL KALADY
9. 4000 will become 2. At what rate % of s.l A) 12% Ans 5600 in 5 years? B)896 C)4% D) 10% ISMAIEL KALADY
10. 2. At what rate % of S1, 4000 will become 5600 in 5 years? A)12% B)8% Ans C)4% D) 10% Interest = 5600-4000 = Rs. 1600 ISMAIEL KALADY
11. 2. At what rate % of S1, 4000 will become 5600 in 5 years? A)12% B)8% Ans C)4% D) 10% 0 Interest = 5600-4000 = Rs. 1600 1600 Total % for 5 years-4000 100 = 40% 40 _R0 Rate of Interest for 1 year = 5-8 % ISMAIEL KALADY
12. 6800 at the 3. In how much time as S.I, 5000 will become rate of 12%? A)1 years B)3 years C)2 years D)5 years Ans Interest = 6800-5000 = Rs. 1800 ISMAIEL KALADY
| 860
| 2,527
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.5
| 4
|
CC-MAIN-2019-26
|
latest
|
en
| 0.822334
|
https://metanumbers.com/35448
| 1,627,654,182,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00591.warc.gz
| 402,313,130
| 11,019
|
35448
35,448 (thirty-five thousand four hundred forty-eight) is an even five-digits composite number following 35447 and preceding 35449. In scientific notation, it is written as 3.5448 × 104. The sum of its digits is 24. It has a total of 6 prime factors and 32 positive divisors. There are 10,080 positive integers (up to 35448) that are relatively prime to 35448.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 24
• Digital Root 6
Name
Short name 35 thousand 448 thirty-five thousand four hundred forty-eight
Notation
Scientific notation 3.5448 × 104 35.448 × 103
Prime Factorization of 35448
Prime Factorization 23 × 3 × 7 × 211
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 8862 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 35,448 is 23 × 3 × 7 × 211. Since it has a total of 6 prime factors, 35,448 is a composite number.
Divisors of 35448
1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168, 211, 422, 633, 844, 1266, 1477, 1688, 2532, 2954, 4431, 5064, 5908, 8862, 11816, 17724, 35448
32 divisors
Even divisors 24 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 32 Total number of the positive divisors of n σ(n) 101760 Sum of all the positive divisors of n s(n) 66312 Sum of the proper positive divisors of n A(n) 3180 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 188.276 Returns the nth root of the product of n divisors H(n) 11.1472 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 35,448 can be divided by 32 positive divisors (out of which 24 are even, and 8 are odd). The sum of these divisors (counting 35,448) is 101,760, the average is 3,180.
Other Arithmetic Functions (n = 35448)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 10080 Total number of positive integers not greater than n that are coprime to n λ(n) 420 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3771 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 10,080 positive integers (less than 35,448) that are coprime with 35,448. And there are approximately 3,771 prime numbers less than or equal to 35,448.
Divisibility of 35448
m n mod m 2 3 4 5 6 7 8 9 0 0 0 3 0 0 0 6
The number 35,448 is divisible by 2, 3, 4, 6, 7 and 8.
Classification of 35448
• Arithmetic
• Abundant
Expressible via specific sums
• Polite
• Practical
• Non-hypotenuse
Base conversion (35448)
Base System Value
2 Binary 1000101001111000
3 Ternary 1210121220
4 Quaternary 20221320
5 Quinary 2113243
6 Senary 432040
8 Octal 105170
10 Decimal 35448
12 Duodecimal 18620
20 Vigesimal 48c8
36 Base36 rco
Basic calculations (n = 35448)
Multiplication
n×i
n×2 70896 106344 141792 177240
Division
ni
n⁄2 17724 11816 8862 7089.6
Exponentiation
ni
n2 1256560704 44542563835392 1578944802836975616 55970435370965111635968
Nth Root
i√n
2√n 188.276 32.8496 13.7214 8.12678
35448 as geometric shapes
Circle
Diameter 70896 222726 3.9476e+09
Sphere
Volume 1.86579e+14 1.57904e+10 222726
Square
Length = n
Perimeter 141792 1.25656e+09 50131
Cube
Length = n
Surface area 7.53936e+09 4.45426e+13 61397.7
Equilateral Triangle
Length = n
Perimeter 106344 5.44107e+08 30698.9
Triangular Pyramid
Length = n
Surface area 2.17643e+09 5.24939e+12 28943.2
Cryptographic Hash Functions
md5 bc2ed059241324f16bb08d020e634321 0abb6eb0a181687a3b4ce99e272ae9f439372097 eced292f8ad7cf17fe655139e8bd4ee6ca14c37e2f2f3258202f3ec139d112d5 0c41f4abf5291d8c66e5a7101b291d4e25a505a03d2065b112ddcf3ab6cb161fbf9a01d40ec0e72e89083182621db4c0ec1e2163b75a1b126739fc18ae26bd1b 40399cab1147a55f697249653662078a6912c8ee
| 1,533
| 4,216
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.734375
| 4
|
CC-MAIN-2021-31
|
latest
|
en
| 0.796814
|
https://docs.moodle.org/39/en/index.php?title=Calculated_multichoice_question_type&oldid=131909
| 1,597,369,532,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439739134.49/warc/CC-MAIN-20200814011517-20200814041517-00564.warc.gz
| 283,827,604
| 10,344
|
Calculated multichoice question type
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Calculated multichoice questions are like multichoice questions with the additional property that the elements to select can include formula results from numeric values that are selected randomly from a set when the quiz is taken. They use the same wildcards than Calculated questions and their wildcards can be shared with other Calculated multichoice or regular Calculated questions.
The main difference is that text and the formula can be included in the answer choice as {=...}.
In this example we want the student to see text in the answer along with the answer. Given a question written by the teacher as:
```Calculate the area of a rectangle where l = {A} cm and h = {B}cm.
```
The correct answer choice text written by the teacher would be:
```The rectangle's area is {={A}*{B}} cm2.
```
The correct answer's choice will display as:
```The rectangle area is 10.0 cm2
```
The variables picked by the dataset in the example were {A} = 4.0 {B} = 2.5 .
You will also need to provide distractors - additional incorrect options presented to the student to choose from. In this example with rectangle's area, example formulas for incorrect answers could be
```The rectangle's area is {={A}*{B}-{B}} cm2.
```
and
```The rectangle's area is {={A}*{B}+{A}} cm2.
```
Showing a formula as a choice
In this example, we want the student to demonstrate they know how to correctly factor out a binomial equation. We want every student to have a unique problem to solve.
For example the teacher enters the question as:
```Given the binomial equation 3x2+5xy+2y2, where x = {A} and y={B} how would you simplify it before solving it?
```
The correct choice would be written:
```This polynomial can be reduced to (3*{A}+2*{B})({A}+{B}).
```
This choice would display as:
```This polynomial can be reduced to (3+4)(1+2).
```
| 482
| 1,946
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.953125
| 4
|
CC-MAIN-2020-34
|
latest
|
en
| 0.937426
|
https://byjus.com/question-answer/what-are-the-condition-s-for-a-2-4-x-d-2-bx-c-0-5/
| 1,643,021,829,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320304528.78/warc/CC-MAIN-20220124094120-20220124124120-00006.warc.gz
| 216,136,688
| 24,595
|
Question
# What are the condition(s) for (a2−4)xd−2+bx+c=0 to be a quadratic equation in x?
A
a is ANY real number other than 2 and 2
B
a, b, c are ALL real numbers where a cannot be 4 and 4
C
d=4
D
a, b, c are real
Solution
## The correct options are A a is ANY real number other than 2 and −2 C d=4 D a, b, c are real The standard form of any quadratic equation is ax2+bx+c=0, provided a, b, c are real numbers and a≠0 [a is called the leading coefficient]. So, in the above question, let us look at the conditions one by one. 1) Coefficients should be real, which means, (a2−4), b and c should be real which follows that a, b and c are real. [If (a2−4) is real, this follows that a2 is real and hence, a is real] 2) Degree of the polynomial should be 2. In this case, the degree is given by d−2. So, d−2=4, which means d=4. 3) Leading coefficient is non-zero a real number. So, (a2−4) should not be zero. ⇒a2−4≠0 ⇒a2≠4 ⇒a≠2,−2 So, the correct options are A, C and D. Mathematics
Suggest Corrections
0
Similar questions
View More
People also searched for
View More
| 353
| 1,081
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.9375
| 4
|
CC-MAIN-2022-05
|
latest
|
en
| 0.864519
|
https://ibizasailing.net/let-s-zxci/69beda-weibull-distribution-in-excel
| 1,618,802,097,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038863420.65/warc/CC-MAIN-20210419015157-20210419045157-00600.warc.gz
| 421,774,900
| 10,116
|
How To Get My Cbs Report In Singapore, Battle Of May Island, School In Minecraft, Ca Covid Watchlist, St Augustine Weddings, Lake Forest High School Website, Maltese Falcon Monologue, Mitchell Mcclenaghan Current Teams, Explain Sentence And Its Types, " /> How To Get My Cbs Report In Singapore, Battle Of May Island, School In Minecraft, Ca Covid Watchlist, St Augustine Weddings, Lake Forest High School Website, Maltese Falcon Monologue, Mitchell Mcclenaghan Current Teams, Explain Sentence And Its Types, "/>
### Single Blog Title
This is a single blog caption
The Weibull distribution’s two parameters allow it to reflect all these possibilities. The Weibull distribution can approximate many other distributions: normal, exponential and so on. Let's understand these arguments with the function syntax as explained below. We provide you with A - Z of Excel Functions and Formulas, solved examples for Beginners, Intermediate, Advanced and up to Expert Level. Depending on the parameters' values, the Weibull distribution can approximate an exponential, a normal or a skewed distribution. The Weibull distribution's virtually limitless versatility is matched by Excel's countless capabilities. We can now use Excel’s Solver to find the values of α and β which maximize LL(α, β). The Weibull distribution is a special case of the generalized extreme value distribution.It was in this connection that the distribution was first identified by Maurice Fréchet in 1927. After copying the example to a blank worksheet, select the range A5:A104 starting with the formula cell. ; The shape parameter, k. is the Weibull shape factor.It specifies the shape of a Weibull distribution and takes on a value of between 1 and 3. It takes the value and the two parameters named alpha and beta along with the type of distribution required(cdf or pdf). Weibull Distribution. How to create an interactive graph in Excel in Minutes of the Weibull Distribution - both the PDF and CDF. The other, … repeat Example 1 of Method of Moments: Weibull Distribution using the MLE approach). The Weibull curve is called a "bathtub curve," because it descends in the beginning (infant mortality); flattens out in the middle and ascends toward the end of life. Use this distribution in reliability analysis, such as calculating a device's mean time to failure. The Weibull distribution's virtually limitless versatility is matched by Excel's countless capabilities. WEIBULL.DIST function: Description, Usage, Syntax, Examples and Explanation Excel How Tos, Shortcuts, Tutorial, Tips and Tricks on Excel Office. WEIBULL.DIST Function in Excel. It has become widely used, especially in the reliability field. 100 Weibull deviates based on Mersenne-Twister algorithm for which the parameters above Note The formula in the example must be entered as an array formula. WEIBULL.DIST is a statistical function which returns the weibull distribution at a particular value. An astute data analyst who understands the theory behind a given analysis can often get results from Excel that others might assume require specialized statistical software. WorksheetFunction.Weibull method (Excel) 05/25/2019; 2 minutes to read; o; k; O; S; J; In this article. The Weibull distribution's popularity resulted from its ability to be used with small sample sizes and its flexibility. Example 1: Find the parameters of the Weibull distribution which best fit the data in range A4:A15 of Figure 1 (i.e. The scale parameter, c, is the Weibull scale factor in m/s; a measure for the characteristic wind speed of the distribution. One parameter, Alpha, determines how wide or narrow the distribution is. A small value for k signifies very variable winds, while constant winds are characterised by a larger k. The Weibull distribution's strength is its versatility. Returns the Weibull distribution. The Weibull distribution is a continuous distribution that was publicized by Waloddi Weibull in 1951. Excel ’ s Solver to find the values of α and β which maximize LL α! Reflect all these possibilities k signifies very variable winds, while constant winds are by. By Waloddi Weibull in 1951 ability to be used with small sample sizes and its.! While constant winds are characterised by a larger k. Weibull distribution can approximate many distributions... A skewed distribution it to reflect all these possibilities the two parameters named alpha and beta with... The characteristic wind speed of the Weibull distribution can approximate weibull distribution in excel other distributions: normal, exponential and on! Speed of the Weibull distribution at a particular value A104 starting with the type of distribution required cdf! Variable winds, while constant winds are characterised by a larger k. Weibull distribution is a device 's mean to... K signifies very variable winds, while constant winds are characterised by a k.... Required ( cdf or pdf ) scale parameter, c, is the distribution. 'S popularity resulted from its ability to be used with small sample sizes and its flexibility: normal, and... Normal or a skewed weibull distribution in excel worksheet, select the range A5: A104 starting with type. Which returns the Weibull distribution m/s ; a measure for the characteristic wind speed of the Weibull can! A104 starting with the type of distribution required ( cdf or pdf ) Excel 's countless capabilities these... Weibull.Dist is a statistical function which returns the Weibull distribution at a particular value parameters ' values, Weibull... In Excel in Minutes of the Weibull distribution at a particular value the function syntax as explained below is continuous. Blank worksheet, select the range A5: A104 starting with the formula cell larger k. Weibull distribution at particular. Distribution is … how to create an interactive graph in Excel in Minutes of the Weibull factor... Approximate an exponential, a normal or a skewed distribution along with the type of distribution weibull distribution in excel ( cdf pdf... Distribution ’ s two parameters named alpha and beta along with the function syntax as below... Mle approach ) at a particular value become widely used, especially in the reliability field by... Winds, while constant winds are characterised by a larger k. Weibull distribution can approximate many distributions... Continuous distribution that was publicized by Waloddi Weibull in 1951 starting with the function syntax as below., β ) value for k signifies very variable winds, while constant winds are characterised by larger. Which returns the Weibull distribution 's virtually limitless versatility is matched by 's..., determines how wide or narrow the distribution is distribution at a particular value, β ) device... The MLE approach ) at a particular value: A104 starting with the formula.... The parameters ' values, the Weibull distribution ’ s Solver to the!, such as calculating a device 's mean time to failure depending on the parameters ',. Solver to find the values of α and β which maximize LL ( α, β ) distribution approximate. Moments: Weibull distribution can approximate an exponential, a normal or a skewed distribution - both the pdf cdf! Matched by Excel 's countless capabilities Solver to find the values of α and β which maximize (... Named alpha and beta along with the function syntax as explained below the... The characteristic wind speed of the distribution now use Excel ’ s Solver to find values. Factor in m/s ; a measure for the characteristic wind speed of the.. Narrow the distribution the MLE approach ) a measure for the characteristic wind of. Allow it to reflect all these possibilities, the Weibull distribution is distribution. Use Excel ’ s Solver to weibull distribution in excel the values of α and β which LL! The Weibull distribution at a particular value to find the values of α β. A measure for the characteristic wind speed of the distribution Weibull scale in... Use Excel ’ s Solver to find the values of α and β which LL... All these possibilities statistical function which returns the Weibull distribution 's popularity resulted from ability! These possibilities ability to be used with small sample sizes and its flexibility to the...
| 1,701
| 8,187
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.734375
| 4
|
CC-MAIN-2021-17
|
latest
|
en
| 0.847522
|
http://fsharpforfunandprofit.com/posts/tuples/
| 1,508,837,976,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187828356.82/warc/CC-MAIN-20171024090757-20171024110757-00704.warc.gz
| 140,484,763
| 9,937
|
##### Part of the "Understanding F# types" series (more)
We’re ready for our first extended type – the tuple.
Let’s start by stepping back again and looking at a type such as “int”. As we hinted at before, rather than thinking of “int” as a abstract thing, you can think of it as concrete collection of all its possible values, namely the set {…,-3, -2, -1, 0, 2, 3, …}.
So next, imagine two copies of this “int” collection. We can “multiply” them together by taking the Cartesian product of them; that is, making a new list of objects by picking every possible combination of the two “int” lists, as shown below:
As we have already seen, these pairs are called tuples in F#. And now you can see why they have the type signature that they do. In this example, the “int times int” type is called “`int * int`”, and the star symbol means “multiply” of course! The valid instances of this new type are all the pairs: (-2,2),(-1,0), (2,2) and so on.
Let’s see how they might be used in practice:
``````let t1 = (2,3)
let t2 = (-2,7)
``````
Now if you evaluate the code above you will see that the types of t1 and t2 are `int*int` as expected.
``````val t1 : int * int = (2, 3)
val t2 : int * int = (-2, 7)
``````
This “product” approach can be used to make tuples out of any mixture of types. Here is one for “int times bool”.
And here is the usage in F#. The tuple type above has the signature “`int*bool`”.
``````let t3 = (2,true)
let t4 = (7,false)
// the signatures are:
val t3 : int * bool = (2, true)
val t4 : int * bool = (7, false)
``````
Strings can be used as well, of course. The universe of all possible strings is very large, but conceptually it is the same thing. The tuple type below has the signature “`string*int`”.
Test the usage and signatures:
``````let t5 = ("hello",42)
let t6 = ("goodbye",99)
// the signatures are:
val t5 : string * int = ("hello", 42)
val t6 : string * int = ("goodbye", 99)
``````
And there is no reason to stop at multiplying just two types together. Why not three? Or four? For example, here is the type `int * bool * string`.
Test the usage and signatures:
``````let t7 = (42,true,"hello")
// the signature is:
val t7 : int * bool * string = (42, true, "hello")
``````
## Generic tuples
Generics can be used in tuples too.
The usage is normally associated with functions:
``````let genericTupleFn aTuple =
let (x,y) = aTuple
printfn "x is %A and y is %A" x y
``````
And the function signature is:
``````val genericTupleFn : 'a * 'b -> unit
``````
which means that “`genericTupleFn`” takes a generic tuple `('a * 'b)` and returns a `unit`
## Tuples of complex types
Any kind of type can be used in a tuple: other tuples, classes, function types, etc. Here are some examples:
``````// define some types
type Person = {First:string; Last:string}
type Complex = float * float
type ComplexComparisonFunction = Complex -> Complex -> int
// define some tuples using them
type PersonAndBirthday = Person * System.DateTime
type ComplexPair = Complex * Complex
type ComplexListAndSortFunction = Complex list * ComplexComparisonFunction
type PairOfIntFunctions = (int->int) * (int->int)
``````
Some key things to know about tuples are:
• A particular instance of a tuple type is a single object, similar to a two-element array in C#, say. When using them with functions they count as a single parameter.
• Tuple types cannot be given explicit names. The “name” of the tuple type is determined by the combination of types that are multiplied together.
• The order of the multiplication is important. So `int*string` is not the same tuple type as `string*int`.
• The comma is the critical symbol that defines tuples, not the parentheses. You can define tuples without the parentheses, although it can sometimes be confusing. In F#, if you see a comma, it is probably part of a tuple.
These points are very important – if you don’t understand them you will get confused quite quickly!
And it is worth re-iterating the point made in previous posts: don’t mistake tuples for multiple parameters in a function.
``````// a function that takes a single tuple parameter
// but looks like it takes two ints
let addConfusingTuple (x,y) = x + y
``````
## Making and matching tuples
The tuple types in F# are somewhat more primitive than the other extended types. As you have seen, you don’t need to explicitly define them, and they have no name.
It is easy to make a tuple – just use a comma!
``````let x = (1,2)
let y = 1,2 // it's the comma you need, not the parentheses!
let z = 1,true,"hello",3.14 // create arbitrary tuples as needed
``````
And as we have seen, to “deconstruct” a tuple, use the same syntax:
``````let z = 1,true,"hello",3.14 // "construct"
let z1,z2,z3,z4 = z // "deconstruct"
``````
When pattern matching like this, you must have the same number of elements, otherwise you will get an error:
``````let z1,z2 = z // error FS0001: Type mismatch.
// The tuples have differing lengths
``````
If you don’t need some of the values, you can use the “don’t care” symbol (the underscore) as a placeholder.
``````let _,z5,_,z6 = z // ignore 1st and 3rd elements
``````
As you might guess, a two element tuple is commonly called a “pair” and a three element tuple is called a “triple” and so on. In the special case of pairs, there are functions `fst` and `snd` which extract the first and second element.
``````let x = 1,2
fst x
snd x
``````
They only work on pairs. Trying to use `fst` on a triple will give an error.
``````let x = 1,2,3
fst x // error FS0001: Type mismatch.
// The tuples have differing lengths of 2 and 3
``````
## Using tuples in practice
Tuples have a number of advantages over other more complex types. They can be used on the fly because they are always available without being defined, and thus are perfect for small, temporary, lightweight structures.
### Using tuples for returning multiple values
It is a common scenario that you want to return two values from a function rather than just one. For example, in the `TryParse` style functions, you want to return (a) whether the value was parsed and (b) if parsed, what the parsed value was.
Here is an implementation of `TryParse` for integers (assuming it did not already exist, of course):
``````let tryParse intStr =
try
let i = System.Int32.Parse intStr
(true,i)
with _ -> (false,0) // any exception
//test it
tryParse "99"
tryParse "abc"
``````
Here’s another simple example that returns a pair of numbers:
``````// return word count and letter count in a tuple
let wordAndLetterCount (s:string) =
let words = s.Split [|' '|]
let letterCount = words |> Array.sumBy (fun word -> word.Length )
(words.Length, letterCount)
//test
wordAndLetterCount "to be or not to be"
``````
### Creating tuples from other tuples
As with most F# values, tuples are immutable and the elements within them cannot be assigned to. So how do you change a tuple? The short answer is that you can’t – you must always create a new one.
Say that you need to write a function that, given a tuple, adds one to each element. Here’s an obvious implementation:
``````let addOneToTuple aTuple =
let (x,y,z) = aTuple
(x+1,y+1,z+1) // create a new one
// try it
``````
This seems a bit long winded – is there a more compact way? Yes, because you can deconstruct a tuple directly in the parameters of a function, so that the function becomes a one liner:
``````let addOneToTuple (x,y,z) = (x+1,y+1,z+1)
// try it
``````
### Equality
Tuples have an automatically defined equality operation: two tuples are equal if they have the same length and the values in each slot are equal.
``````(1,2) = (1,2) // true
(1,2,3,"hello") = (1,2,3,"bye") // false
(1,(2,3),4) = (1,(2,3),4) // true
``````
Trying to compare tuples of different lengths is a type error:
``````(1,2) = (1,2,3) // error FS0001: Type mismatch
``````
And the types in each slot must be the same as well:
``````(1,2,3) = (1,2,"hello") // element 3 was expected to have type
// int but here has type string
(1,(2,3),4) = (1,2,(3,4)) // elements 2 & 3 have different types
``````
Tuples also have an automatically defined hash value based on the values in the tuple, so that tuples can be used as dictionary keys without problems.
``````(1,2,3).GetHashCode()
``````
### Tuple representation
And as noted in a previous post, tuples have a nice default string representation, and can be serialized easily.
``````(1,2,3).ToString()
``````
| 2,319
| 8,551
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.6875
| 4
|
CC-MAIN-2017-43
|
latest
|
en
| 0.908057
|
https://www.jiskha.com/questions/755793/need-to-make-a-venn-diagram-using-3-circles-a-survey-on-subject-being-taken-by-250
| 1,590,415,743,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00294.warc.gz
| 722,422,634
| 5,892
|
math
need to make a venn diagram using 3 circles. A survey on subject being taken by 250 students at a certain college revealed the following info:
1. 90 were taking math
2. 145 were taking history
3. 88 were taking english
4. 25 were taking math and history
5. 38 were taking history and english
6. 59 were taking math and english
7. 15 were taking all three subjects
1. 👍 0
2. 👎 0
3. 👁 581
1. Each of your overlapping circles should be labeled
Math
History
English
Put the numbers in each circle to represent those students taking just one subject.
You'll have 25 in the overlapping sections of math and English.
http://www.classroomjr.com/printable-blank-venn-diagrams/
1. 👍 0
2. 👎 0
2. Ms Sue when you said 25 in the overlapping section of math and english did mean math and history/
1. 👍 0
2. 👎 0
3. Yes, I'm sorry. 25 students are taking math and history.
Thanks for catching my mistake.
1. 👍 0
2. 👎 0
4. Is 100 student did a surveys 47 Math, 57 Chemistry, 69,English 29 in English an Math, 25 English and chemistry, 37 Math and chemistry, 22 in all
How student are you taking Math but not English an Chemistry
1. 👍 0
2. 👎 1
Similar Questions
1. Math
i don't get it, i have got homework to create a venn diagram of natural and whole numbers aren't they both the same thing? how do i make a venn diagram for that?? it also says that what numbers belong outside the circle? 2.
asked by Michelle White on September 12, 2012
2. Unions and Intersections of Sets (Algebra)
Part 1: Answer the following question: In a survey of students about favorite sports, the results include 22 who like tennis, 25 who like football, 9 who like tennis and football, 17 who like tennis and baseball, 20 who like
asked by Gabby on October 22, 2013
3. Math
Part 1: Answer the following question: In a survey of students about favorite sports, the results include 22 who like tennis, 25 who like football, 9 who like tennis and football, 17 who like tennis and baseball, 20 who like
asked by Anonymous on October 23, 2013
4. Algebra
Part 1: Answer the following question: In a survey of students about favorite sports, the results include 22 who like tennis, 25 who like football, 9 who like tennis and football, 17 who like tennis and baseball, 20 who like
asked by Anonymous on October 23, 2013
5. social studies
Can somebody help me on this? I don't know what to answer. Venn Diagram Write details that tell how the subjects are different in the outer circles. Write details that tell how the subjects are alike where the circles overlap.
asked by Anonymous on October 16, 2014
1. Math
Part 1: Answer the following question: In a survey of students about favorite sports, the results include 22 who like tennis, 25 who like football, 9 who like tennis and football, 17 who like tennis and baseball, 20 who like
asked by Anonymous on October 23, 2013
2. Math
i don't get it, i have got homework to create a venn diagram of natural and whole numbers aren't they both the same thing? how do i make a venn diagram for that??
asked by DemiDL on September 12, 2012
How do you use a venn diagram to find the GCF and LCM of two or more factors. This is what a venn diagram looks like: ()()
asked by Angelina on October 4, 2012
4. Math
The Venn diagram below shows the dinner orders for a local restaurant. This was submitted before, but no one answered. Thanks! How many people did NOT order steak for dinner? Steak, chicken & fish were order. In the Venn circles,
asked by Jane on January 7, 2012
5. math
The Venn diagram below shows the dinner orders for a local restaurant. How many people did NOT order steak for dinner? Steak, chicken & fish were order. In the Venn circles, the total that did not order steak was 56. There is 43
asked by Jane on January 6, 2012
More Similar Questions
| 1,022
| 3,802
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.578125
| 4
|
CC-MAIN-2020-24
|
longest
|
en
| 0.970273
|
https://www.smartkeeda.com/Quantitative_Aptitude/Arithmetic/Approximations_Quiz/newest/all/passage/Approximations_Quiz_18/
| 1,627,479,804,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046153729.44/warc/CC-MAIN-20210728123318-20210728153318-00622.warc.gz
| 1,063,218,537
| 20,156
|
# Approximation Questions and Answers for SBI Clerk Pre 2021, IBPS Clerk, CET, RBI Assistant, LIC at Smartkeeda
Directions: What value will come in place of question mark (?) in the following questions: (You are not expected to calculate the exact value)
Important for :
1
7(6 - ?) × 119 ÷ √1225 = √289 × 20.98 × 7.03 ÷ 3.02 ÷ 4.95
» Explain it
A
7(6 - ?) × 119 ÷ √1225 = √289 × 20.98 × 7.03 ÷ 3.02 ÷ 4.95
7(6 - ?) × 119 ÷ 35 = 17 × 21 × 7 ÷ 3 ÷ 5
7(6 - ?) = 17 × 21 × 7 ÷ 3 ÷ 5 ÷ 119 × 35
7(6 - ?) = 7 × 7
7(6 - ?) = 72
6 - ? = 2
? = 6 - 2 = 4
Hence, option A is correct.
2
(128.78 + 307.24 + 111.8) – (246.35 – 786.47) = ?2
» Explain it
A
(128.78 + 307.24 + 111.8) – (246.35 – 786.47) = ?2
(129 + 307 + 112) – (246 – 786) = ?2
548 – (– 540) = ?2
?2 = 548 + 540
?2 = 1088 ≈ 1089
? = 33
Hence, option A is correct.
3
54.7% of 1190 ÷ (√676 + ∛ 1728.11 – √361) = ?2 ÷ 57.02
» Explain it
C
54.7% of 1190 ÷ (√676 + ∛ 1728.11 – √361) = ?2 ÷ 57.02
1190 × 55/ 100 ÷ (26 + 12 – 19) = ?2 ÷ 57
654.5 ÷ 19 = ?2 ÷ 57
655 ÷ 19 × 57 = ?2
?2 = 655 × 3 = 1965
? = 44.32 ≈ 44
Hence, option C is correct.
4
» Explain it
B
(100 / 1300) × 208 + {(17 × 304) / 323) = ?1/2 +5
16 + 16 – 5 = ?1/2
?1/2 = 27
? = 729 ≈ 730
Hence, option B is correct.
5
9.09% of 11.11% of 2430 = ?3 ÷ √289 + √( ∛4096 )
» Explain it
D
9.09% of 11.11% of 2430 = ?3 ÷ √289 + √( ∛4096 )
2430 × (1 / 9) × (1 / 11) = ?3 ÷ 17 + √16
(270 /11) = ?3 ≈ 17 + 4
24.54 – = (?3 / 17)
20.54 = (?3 / 17)
21 × 17 = ?3
?3 = 357
? ≈ 7
Hence, option D is correct.
| 792
| 1,535
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.40625
| 4
|
CC-MAIN-2021-31
|
latest
|
en
| 0.579471
|
https://convertoctopus.com/157-9-inches-to-meters
| 1,610,963,733,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703514495.52/warc/CC-MAIN-20210118092350-20210118122350-00117.warc.gz
| 290,128,178
| 7,685
|
## Conversion formula
The conversion factor from inches to meters is 0.0254, which means that 1 inch is equal to 0.0254 meters:
1 in = 0.0254 m
To convert 157.9 inches into meters we have to multiply 157.9 by the conversion factor in order to get the length amount from inches to meters. We can also form a simple proportion to calculate the result:
1 in → 0.0254 m
157.9 in → L(m)
Solve the above proportion to obtain the length L in meters:
L(m) = 157.9 in × 0.0254 m
L(m) = 4.01066 m
The final result is:
157.9 in → 4.01066 m
We conclude that 157.9 inches is equivalent to 4.01066 meters:
157.9 inches = 4.01066 meters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 meter is equal to 0.24933552083697 × 157.9 inches.
Another way is saying that 157.9 inches is equal to 1 ÷ 0.24933552083697 meters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred fifty-seven point nine inches is approximately four point zero one one meters:
157.9 in ≅ 4.011 m
An alternative is also that one meter is approximately zero point two four nine times one hundred fifty-seven point nine inches.
## Conversion table
### inches to meters chart
For quick reference purposes, below is the conversion table you can use to convert from inches to meters
inches (in) meters (m)
158.9 inches 4.036 meters
159.9 inches 4.061 meters
160.9 inches 4.087 meters
161.9 inches 4.112 meters
162.9 inches 4.138 meters
163.9 inches 4.163 meters
164.9 inches 4.188 meters
165.9 inches 4.214 meters
166.9 inches 4.239 meters
167.9 inches 4.265 meters
| 478
| 1,694
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875
| 4
|
CC-MAIN-2021-04
|
latest
|
en
| 0.872799
|
http://mathhelpforum.com/advanced-statistics/53053-simple-probability-problem.html
| 1,527,451,237,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794870082.90/warc/CC-MAIN-20180527190420-20180527210420-00346.warc.gz
| 182,918,919
| 10,513
|
# Thread: Simple Probability Problem
1. ## Simple Probability Problem
In a particular town 60% of the population are women. 4% of the men and 1% of the women are taller than 180 cm.
i) What % of the town's population is taller than 180 cm?
ii) If a person is chosen at random and is taller than 180 cm, what is the probability that the person is a woman?
Ummm I'm not that good at probability so could someone please give me some help?
For:
i) I did:
Percentage = 100(0.04*0.4 + 0.01*0.6)
= 0.022*100
= 2.2%
ii) I did:
P = 0.022*0.2
= 11/2500
The answers don't look rite though...
2. You have the first part right, for the second one you need to use Baye's theorem.
$\displaystyle P(woman|taller than 180)= \frac{P(taller than 180|woman)P(woman)}{P(taller than 180|woman)P(woman)+P(taller than 180|man)P(man)}$
3. Originally Posted by xwrathbringerx
In a particular town 60% of the population are women. 4% of the men and 1% of the women are taller than 180 cm.
i) What % of the town's population is taller than 180 cm?
ii) If a person is chosen at random and is taller than 180 cm, what is the probability that the person is a woman?
Ummm I'm not that good at probability so could someone please give me some help?
For:
i) I did:
Percentage = 100(0.04*0.4 + 0.01*0.6)
= 0.022*100
= 2.2% Mr F says: Correct.
ii) I did:
P = 0.022*0.2 Mr F says: Wrong. Where has the 0.2 come from?
= 11/2500
The answers don't look rite though...
ii requires conditional probability.
Here is the magic recipe, the big secret, the great trick, the golden elixir. Read it carefully ........
Set up careful notation and definitions. Carefully define all known probabilities in terms of this notation.
Let T be the event height greater than 180 cm.
You have:
Pr(M) = 2/5, Pr(W) = 3/5,
Pr(T | M) = 1/25, Pr(T | W) = 1/100,
Pr(M and T) = Pr(T| M) Pr(M) = (1/25) (2/5) = 2/125,
Pr(W and T) = Pr(T| W) Pr(W) = (1/100) (3/5) = 3/500,
Pr(T) = Pr(M and T) + Pr(W and T) = 2/125 + 3/500 = 11/500 (= 0.022 as you found).
You require Pr(W | T).
$\displaystyle \Pr(W | T) = \frac{\Pr(W \cap T)}{\Pr(T)} = \frac{3/500}{11/500} = 3/11$.
4. You have the first part right, for the second one you need to use Baye's theorem.
To keep not that messy, Let
W=woman
M=Man
T=taller than 180
$\displaystyle P(W|T)= \frac{P(T|W)P(W)}{P(T|W)P(W)+P(T|M)P(M)}=\frac{(0. 01)(0.60)}{(0.01)(0.60)+(0.04)(0.40)}$
$\displaystyle =\frac{0.006}{0.006+0.016}=\frac{0.006}{0.022}=\fr ac{3}{11}$
| 868
| 2,472
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.28125
| 4
|
CC-MAIN-2018-22
|
latest
|
en
| 0.93007
|
http://nrich.maths.org/public/leg.php?code=-100&cl=2&cldcmpid=1206
| 1,448,540,933,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-48/segments/1448398447266.73/warc/CC-MAIN-20151124205407-00163-ip-10-71-132-137.ec2.internal.warc.gz
| 170,519,764
| 10,446
|
# Search by Topic
#### Resources tagged with Trial and improvement similar to Sliding Game:
Filter by: Content type:
Stage:
Challenge level:
### There are 90 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Trial and improvement
### Sliding Game
##### Stage: 2 Challenge Level:
A shunting puzzle for 1 person. Swop the positions of the counters at the top and bottom of the board.
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Cycling Squares
##### Stage: 2 Challenge Level:
Can you make a cycle of pairs that add to make a square number using all the numbers in the box below, once and once only?
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Strike it Out
##### Stage: 1 and 2 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
### A Mean Tetrahedron
##### Stage: 3 Challenge Level:
Can you number the vertices, edges and faces of a tetrahedron so that the number on each edge is the mean of the numbers on the adjacent vertices and the mean of the numbers on the adjacent faces?
### Strike it Out for Two
##### Stage: 1 and 2 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Junior Frogs
##### Stage: 1 and 2 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### One Million to Seven
##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Pair Sums
##### Stage: 3 Challenge Level:
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?
### 1, 2, 3, 4, 5
##### Stage: 2 Challenge Level:
Using the numbers 1, 2, 3, 4 and 5 once and only once, and the operations x and ÷ once and only once, what is the smallest whole number you can make?
### Number Juggle
##### Stage: 2 Challenge Level:
Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them (in the box).
### Clever Santa
##### Stage: 2 Challenge Level:
All the girls would like a puzzle each for Christmas and all the boys would like a book each. Solve the riddle to find out how many puzzles and books Santa left.
### Magic Triangle
##### Stage: 2 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### Brothers and Sisters
##### Stage: 2 Challenge Level:
Cassandra, David and Lachlan are brothers and sisters. They range in age between 1 year and 14 years. Can you figure out their exact ages from the clues?
### Magic Squares 4x4
##### Stage: 2 Challenge Level:
Fill in the numbers to make the sum of each row, column and diagonal equal to 34. For an extra challenge try the huge American Flag magic square.
### Fifteen Cards
##### Stage: 2 Challenge Level:
Can you use the information to find out which cards I have used?
### Path to the Stars
##### Stage: 2 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
### Lost
##### Stage: 3 Challenge Level:
Can you locate the lost giraffe? Input coordinates to help you search and find the giraffe in the fewest guesses.
### 3388
##### Stage: 3 Challenge Level:
Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.
### In the Bag
##### Stage: 3 Challenge Level:
Can you guess the colours of the 10 marbles in the bag? Can you develop an effective strategy for reaching 1000 points in the least number of rounds?
### A Numbered Route
##### Stage: 2 Challenge Level:
Can you draw a continuous line through 16 numbers on this grid so that the total of the numbers you pass through is as high as possible?
### Paw Prints
##### Stage: 2 Challenge Level:
A dog is looking for a good place to bury his bone. Can you work out where he started and ended in each case? What possible routes could he have taken?
### Tubular Path
##### Stage: 2 Challenge Level:
Can you make the green spot travel through the tube by moving the yellow spot? Could you draw a tube that both spots would follow?
### Inky Cube
##### Stage: 2 and 3 Challenge Level:
This cube has ink on each face which leaves marks on paper as it is rolled. Can you work out what is on each face and the route it has taken?
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### Magic Circles
##### Stage: 2 Challenge Level:
Put the numbers 1, 2, 3, 4, 5, 6 into the squares so that the numbers on each circle add up to the same amount. Can you find the rule for giving another set of six numbers?
### A Shapely Network
##### Stage: 2 Challenge Level:
Your challenge is to find the longest way through the network following this rule. You can start and finish anywhere, and with any shape, as long as you follow the correct order.
### Top Coach
##### Stage: 3 Challenge Level:
Carry out some time trials and gather some data to help you decide on the best training regime for your rowing crew.
### The Clockmaker's Birthday Cake
##### Stage: 2 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### Archery
##### Stage: 3 Challenge Level:
Imagine picking up a bow and some arrows and attempting to hit the target a few times. Can you work out the settings for the sight that give you the best chance of gaining a high score?
### Let's Face It
##### Stage: 2 Challenge Level:
In this problem you have to place four by four magic squares on the faces of a cube so that along each edge of the cube the numbers match.
### Pizza Cut
##### Stage: 2 Challenge Level:
Using only six straight cuts, find a way to make as many pieces of pizza as possible. (The pieces can be different sizes and shapes).
### Augustus' Age
##### Stage: 2 Challenge Level:
In 1871 a mathematician called Augustus De Morgan died. De Morgan made a puzzling statement about his age. Can you discover which year De Morgan was born in?
### Twenty Divided Into Six
##### Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
### Four Colours
##### Stage: 1 and 2 Challenge Level:
Kate has eight multilink cubes. She has two red ones, two yellow, two green and two blue. She wants to fit them together to make a cube so that each colour shows on each face just once.
### One Wasn't Square
##### Stage: 2 Challenge Level:
Mrs Morgan, the class's teacher, pinned numbers onto the backs of three children. Use the information to find out what the three numbers were.
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Cat Food
##### Stage: 2 Challenge Level:
Sam sets up displays of cat food in his shop in triangular stacks. If Felix buys some, then how can Sam arrange the remaining cans in triangular stacks?
### Plenty of Pens
##### Stage: 2 Challenge Level:
Amy's mum had given her £2.50 to spend. She bought four times as many pens as pencils and was given 40p change. How many of each did she buy?
### Magic Matrix
##### Stage: 2 Challenge Level:
Find out why these matrices are magic. Can you work out how they were made? Can you make your own Magic Matrix?
### Fractions in a Box
##### Stage: 2 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each disc. Use the information to find out how many discs of each colour there are in the box.
### Four-digit Targets
##### Stage: 2 Challenge Level:
You have two sets of the digits 0 – 9. Can you arrange these in the five boxes to make four-digit numbers as close to the target numbers as possible?
### Numbered Cars
##### Stage: 2 Challenge Level:
I was looking at the number plate of a car parked outside. Using my special code S208VBJ adds to 65. Can you crack my code and use it to find out what both of these number plates add up to?
### Coded Hundred Square
##### Stage: 2 Challenge Level:
This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up?
### Long Multiplication
##### Stage: 3 Challenge Level:
A 3 digit number is multiplied by a 2 digit number and the calculation is written out as shown with a digit in place of each of the *'s. Complete the whole multiplication sum.
### How Many Eggs?
##### Stage: 2 Challenge Level:
Peter, Melanie, Amil and Jack received a total of 38 chocolate eggs. Use the information to work out how many eggs each person had.
| 2,432
| 10,197
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.03125
| 4
|
CC-MAIN-2015-48
|
longest
|
en
| 0.900382
|
https://codereview.stackexchange.com/questions/162850/calculate-the-sum-at-a-level-of-a-binary-tree-represented-in-a-string/162872
| 1,582,449,972,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875145747.6/warc/CC-MAIN-20200223062700-20200223092700-00222.warc.gz
| 322,943,645
| 36,189
|
# Calculate the sum at a level of a binary tree represented in a String
Fair preface: This is an interview question I would like to improve my knowledge of. I got some rest, and re-solved the problem with a fresh/unpanicked mind.
Given input of the type (10(5(3)(12))(14(11)(17))) which would represent the following tree
n=0 10
n=1 5 14
n=2 3 12 11 17
My task is to find the summation of values at a particular tier, like $5+14=19$ which is the sum for $n=1$, or $3+12+11+17=43$ the sum for $n=2$.
Considering this is a binary tree, a recursive function seems appropriate.
My main utility functions are:
• stripFirstLastParen – strips the first and last paren
• getCurrentVal – retrieves the value of the current node
• getChildren – retrieves the left and right nodes
var input =
"(10(5(3)(12))(14(11)(17)))";
function stripFirstLastParen(input){
if(input[0] !== "(" || input[input.length - 1] !== ")"){
console.error("unbalanced parens")
}
else{
input = input.substr(1);
input = input.substring(0, input.length - 1);
}
return input;
}
function getCurrentVal(input){
var val = "";
while(input[0] !== "(" && input[0] !== ")" && input.length){
val += input[0];
input = input.substr(1);
}
return {
val,
input
}
}
function getChildren(input){
var val = "";
if(input.length == 0){
return {
left: "",
right: ""
}
}
if(input[0] !== "("){
console.error("no open paren at start");
}
val += input[0];
input = input.substr(1);
var parenStack = 1;
while(parenStack > 0){
if(input[0] == ")"){
parenStack--;
}
else if(input[0] == "("){
parenStack++;
}
val += input[0];
input = input.substr(1);
}
return {
left: val,
right: input
}
}
function getValueForLevel(input, k){
var totalValue = 0;
input = stripFirstLastParen(input);
var currentValue = getCurrentVal(input);
var children = getChildren(currentValue.input);
if(k > 0){
if(children.left.length){
totalValue += getValueForLevel(children.left, k-1);
}
if(children.right.length){
totalValue += getValueForLevel(children.right, k-1);
}
}
else if(k == 0){
totalValue += JSON.parse(currentValue.val);
}
}
var test = getValueForLevel(input, 2);
console.log(test);
My main concerns are:
1. String manipulation. How should I properly be passing the altered string throughout the recursive function? Callout for lines like
var currentValue = getCurrentVal(input);
var children = getChildren(currentValue.input);
Where the getChildren function relies on getCurrentVal removing the value from the string itself, that feels dirty.
1. Complexity. I'm having difficulty describing the complexity of my procedure, and would argue that it's somewhere between $O(n)$ and $O(n^2)$, so is it $O(n^2)$? Considering the parens and values are removed as they are parsed, that lends itself to $O(n)$, but I feel it's $O(n^2)$ because of the duplicate parsing in order to create the child nodes. Is that correct? Can it be improved?
I tried to keep the "spirit of the challenge" and re-completed this within the amount of time previously allotted (somewhere around 30-45 minutes). Suggestions and improvements need not consider this time restriction, but providing context is important.
Recursion:
Considering this is a binary tree, a recursive function seems appropriate.
Recursion would be a natural choice if your input were a root tree node with tree nodes as children - a recursive data structure. However, your input is a 'flat' string representation of a tree. A simple iteration might actually be much easier to implement, read and execute.
Passing inputs:
How should I properly be passing the altered string throughout the recursive function?
I think the recursive passing down of substrings representing nodes or node lists (children) is already pretty well done.
However, I would like to suggest a few improvements regarding the parsing functions and their return values:
Style:
Consider the getCurrentVal(input) function:
function getCurrentVal(input){
var val = "";
while(input[0] !== "(" && input[0] !== ")" && input.length){
val += input[0];
input = input.substr(1);
}
return {
val,
input
}
}
The code is not very descriptive: It is not immediately obvious what this function does.
• The function name getCurrentVal is misleading, as you don't just return the node value, but also the children.
• The argument name input is meaningless, as all arguments are inputs.
• The variable names are misleading, as you reuse input to store part of your output.
// Split node 'a(b)(c)' into value 'a' and children '(b)(c)':
function parseNode(node) {
let split = Math.min(node.indexOf('('), node.indexOf(')'));
if (split < 0) split = node.length;
return {
value: node.slice(0, split),
children: node.slice(split)
};
}
• 'Parse' hints at the input being text or some other raw content.
• 'Node' carries more semantics than 'input'
• Math.min and String.indexOf are more descriptive than a while loop.
Destructuring the resulting object improves readability:
var {value, children} = parseNode(node);
Also, when parsing integers, you shouldn't use JSON.parse(string) but the more specialized Number.parseInt(string) or the idiomatic type conversion via +string.
Further naming suggestions:
• getValueForLevel -> getSumForLevel or sumTreeLevel
• Consistency: Choose either val or value when naming currentValue.val, getCurrentVal, getValueForLevel
Complexity:
Considering the parens and values are removed as they are parsed, that lends itself to O(n), but I feel it's O(n^2) because of the duplicate parsing in order to create the child nodes. Is that correct? Can it be improved?
First of all, let's define n as the length of the input string. Then, let's look at the complexity of the helper functions:
• stripFirstLastParen: O(n) because of input = input.substr(1)
• getCurrentVal: O(n²) for worst-case input "1(2)(3)", "11(22)(33)", "111(222)(333)" and so on because of input = input.substr(1) within while (input.length). Yes, we break the loop as soon as we encounter the first parenthesis, but this doesn't happen before parsing roughly 1/3 * n characters for above input. The best-case complexity is however O(n) when the value has constant length.
• getChildren has worst and best-case time complexity of O(n²) due to input = input.substr(1) within a while loop over input.
Now, we can tackle the complexity of getValueForLevel:
• The first four lines are dominated by getChildren(currentValue.input). If the length of currentValue.input - the length of the children - is proportional to n, then it has a runtime complexity of O(n ²).
• Then it calls getValueForLevel(children.left, k-1). It is easy to construct inputs which maximize the length of children.left such as "4(3(2(1))))" without right children. For this input, in each iteration, the length of children.left is reduced by 3.
• The final call to JSON.parse is in O(n).
Adding the complexities together gives us:
n² + (n-3)² + (n-6)² + ... + (n-3k)²
Thus, for k = 0 the runtime complexity is in O(n²) and for maximum k we have a runtime complexity of O(n³):
In general, the runtime complexity is bounded by O(kn²).
Alternative approach:
Following my introductory remarks about a possibly simpler iterative solution, here a possible implementation with linear runtime complexity:
// Sum integer tree nodes at given level:
function sumTreeLevel(tree, level) {
let tokens = tree.match(/($$|[0-9]+|$$)/g);
let sum = 0;
for (let token of tokens) {
if (token === '(') {
level--;
} else if (token === ')') {
level++;
} else if (level === -1) {
sum += +token;
}
}
return sum;
}
// Example:
console.log(sumTreeLevel("(10(5(3)(12))(14(11)(17)))", 2)); // 43
• What tool did you use to generate the complexity chart? – Igor Soloydenko May 8 '17 at 23:01
• @IgorSoloydenko plot.ly - the data doesn't result from a runtime evaluation however, but simply from counting 'primitive' operations according to the model of computation. – le_m May 8 '17 at 23:05
Some notes, in no particular order:
• If there are unbalanced/missing parentheses, you print an error to the console. However, I'd say you should just throw an exception instead. If the parentheses don't make sense, there's something wrong, and probably no way to recover.
• This stood out:
input = input.substr(1);
input = input.substring(0, input.length - 1);
Why use both substr and substring? The two are very subtly different, so don't mix them unless you have good reason. And why do the trimming in two passes? You could just slice from 1 to length-1 (or length-2 in the case of substr, since they're different) in one go.
• This stood out too:
totalValue += JSON.parse(currentValue.val);
Use parseInt, parseFloat, or even * 1 (in a pinch) to convert the string to a number. JSON ain't got nothing to do with this.
• getCurrentVal could be made simpler with a regular expression like /^\d+/ that matches digits only.
Now, if the idea is only to sum the nodes at each level, you can actually skip a lot of stuff.
Namely, there's no reason to build a graph. You can just read the string from beginning to end, and get the sums:
function sumLevels(string) {
var sums = [],
level = -1;
for(let i = 0; i < string.length; i++) {
if(string[i] === '(') {
level += 1; // add a level for each open paren
} else if(string[i] === ')') {
level -= 1; // remove a level for each closing paren
} else {
// match one or more digits
let match = string.slice(i).match(/^\d+/);
// complain if there are no digits
if(!match) throw new Error(Parse error at char \${i});
// fast-forward the loop counter
i += match[0].length - 1;
// add to sum for current level
sums[level] = (sums[level] || 0) + parseInt(match[0], 10);
}
}
return sums;
}
For the test string you had, the function returns [ 10, 19, 43 ]; the sum at each level.
Note, though that the above doesn't do much input checking. It doesn't, for instance, check that parentheses are balanced, meaning level could go negative or skip ahead, which wouldn't make sense. However, checking could be added without too much trouble. I'll leave that as an exercise for the reader though.
Incidentally, another way to write the above, using replace as a regex scan-like function, would be:
function sumLevels(string) {
var sums = [],
level = -1;
string.replace(/($$|$$|\d+)/g, (_, token) => {
switch(token) {
case '(':
level += 1;
break;
case ')':
level -= 1;
break;
default:
sums[level] = (sums[level] || 0) + parseInt(token, 10);
}
});
return sums;
}
This has even less input-checking, though.
• Thank you very much, this is the exact type of feedback I was looking for, it is very much appreciated. I'll be rereading over this in the next few days in order to ensure I understand everything, but thanks again! – Hodrobond May 8 '17 at 22:13
| 2,664
| 10,741
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.828125
| 4
|
CC-MAIN-2020-10
|
latest
|
en
| 0.433927
|
https://byjus.com/question-answer/if-x-c-square-root-of-b-4-find-the-value-of-x-1-x/
| 1,713,538,842,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817438.43/warc/CC-MAIN-20240419141145-20240419171145-00579.warc.gz
| 123,887,271
| 24,129
|
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question
# If x = (c* square root of b) + 4 find the value of x + 1/x
Open in App
Solution
## 1/x = 1/(c√b + 4) = (c√b - 4)/(c√b - 4) x 1/(c√b + 4) = (c√b - 4)/(c2b - 16) So, x + 1/x = (c√b + 4) + (c√b - 4)/(c2b - 16) = (c3b√b -15c√b - 68 + 4c2b)/(c2b - 16)
Suggest Corrections
11
Join BYJU'S Learning Program
Related Videos
Algebraic Identities
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program
| 213
| 485
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.859375
| 4
|
CC-MAIN-2024-18
|
latest
|
en
| 0.664623
|
https://www.futurelearn.com/courses/geohealth/0/steps/19296
| 1,603,909,061,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107900200.97/warc/CC-MAIN-20201028162226-20201028192226-00613.warc.gz
| 739,012,363
| 26,547
|
## Want to keep learning?
This content is taken from the University of Twente's online course, Geohealth: Improving Public Health through Geographic Information. Join the course to learn more.
3.16
# The variogram
You were introduced to the concept of spatial dependence. In this step you will learn about the variogram (also referred to as the semi-variogram). A variogram is used to describe and model spatial dependence. The variogram is used widely in geostatistics. The equation for the variogram is:
$\hat \gamma(h) =\frac{1}{2n(h)} \sum_{i=1}^{n(h)}(y(s)-y(s+h))^2$
Let’s look at what this equation means. $$y$$ is the response variable (e.g., reflectance in an image, air pollution concentration, malaria parasite rate) which is taken at a specific location $$s$$. We calculate the difference in the attribute value between two observations measured at locations separated by a lag distance $$h$$ (hence ($$s$$+$$h$$)). Lag distance refers to the geographic separation between two observations. There are $$n(h)$$ pairs of points separated by $$h$$. We take the average of these squared differences for the $$n(h)$$ pairs. Note the 1/2 on the right-hand side of the equation. We multiply the average squared difference by 1/2 . Hence $$γ ̂(h)$$ is actually the semi-variance for pairs of points separated by $$h$$. For short lags we expect the average squared difference to be small (because measurements separated by short lags are expected to be similar). For large lags we expect the average squared difference to be larger. We can plot the semi-variances for different $$h$$’s against lag. This plot is called the sample semi-variogram or sample variogram (also called the experimental or empirical variogram).
*Figure 1 (top left) An example sample variogram, (top right) Sample variogram with fitted model (curved line). The sill, nugget and range are indicated. (bottom left) Schematic diagram showing a location with no measurement (red diamond) and locations where measurements are taken (black dots). The arrow indicates the distance between two locations. (bottom right) Map of PM10 air pollution concentration across Europe, based on the data in Figure 1 (top right) of Step 3.15.
An example sample variogram is given in Figure 1 (top left). Note that the value of semi-variance increases with increasing lag distance. For short lag distances the attribute values are similar and the semi-variance is small. At large lag distances the attribute values are dissimilar and the semi-variance is large.
We can already gain useful information from the sample semi-variogram. The lag distance where the sample variogram flattens is called the range. This is the maximum spatial separation where we expect two points to be correlated. The range is associated with the variogram sill, which is the maximum variability in the data. Finally, the point where the variogram approaches the y-axis is the nugget. The nugget is the non-spatial variability. These three parameters (sill, nugget, range) can be identified if we fit a model to the sample variogram. The model is a curved line, as illustrated in Figure 1 (top right).
In Figure 1 (top right) the range occurs at approximately 900 m. Two observations separated by less than 900 m would be expected to be correlated whereas two observations separated by more than 900 m would be expected to be uncorrelated. Two observations separated by 300 m are expected to be more correlated than two observations separated by 500 m. The variogram model tells exactly how correlated are two observations that are separated by a given distance.
The sample variogram and variogram model are useful for exploring the spatial dependence in a dataset. They can also be used for mapping. Consider Figure 1 (bottom left). The red diamond indicates a location where we do not have a measurement whereas we do have measurements at the black dots. We know how far the red diamond is from each black dot. Using the variogram we can then say how correlated it is expected to be with that black dot. We can then predict the attribute at the red diamond as a weighted average of the attributes at the black dots – where the weights are based on the correlations. This prediction is also called interpolation. The geostatistical approach to prediction is often called kriging, named after Danie Krige, an early researcher and practitioner in geostatistics.
Figure 1 (bottom left) illustrates prediction at a single location. If we predict at multiple locations on a grid we can create a map. Examples are shown in Figure 1 (bottom right) and Figure 2, which are based on data presented in the step on Spatial Dependence.
Figure 2 Map of malaria parasite rate, based on the data shown in Figure 3 of the Article on Spatial Dependence. See also Hay et al. 2009.
The maps shown in Figure 1 (bottom right) and Figure 2 are the concluding examples in the module on spatial statistics in this online course. These are important examples in the context of geohealth. Consider the air pollution example. We can use such maps to estimate individual exposure as part of an environmental epidemiological study into the health effects of air pollution. The malaria example is different because the disease itself is mapped rather than the possible cause of a disease. Such maps can be used to target interventions aimed at eliminating a disease (e.g., drug treatments, bed nets) and for evaluating the success of interventions.
References
Oliver, M.A., Webster, R. (2014) A tutorial guide to geostatistics: Computing and modelling variograms and kriging. CATENA 113, 56-69.
Hay, S. I., C. A. Guerra, P. W. Gething, A. P. Patil, A. J. Tatem, A. M. Noor, C. W. Kabaria, B. H. Manh, I. R. Elyazar, S. Brooker, D. L. Smith, R. A. Moyeed and R. W. Snow (2009). A world malaria map: Plasmodium falciparum endemicity in 2007. PLoS Medicine 6(3): e1000048. DOI: 10.1371/journal.pmed.1000048.
## Get a taste of this course
Find out what this course is like by previewing some of the course steps before you join:
| 1,390
| 6,047
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.375
| 4
|
CC-MAIN-2020-45
|
latest
|
en
| 0.903135
|
https://www.jiskha.com/display.cgi?id=1389050906
| 1,516,795,340,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084894125.99/warc/CC-MAIN-20180124105939-20180124125939-00110.warc.gz
| 900,431,163
| 3,609
|
# physics
posted by .
A car moving at 50 km/h skids 15 m with locked brakes. Show that a car moving at 150 km/h will skid 135 m with locked brakes. (Hint: The force doing work during the skidding is the same for both speeds. Use the work–energy theorem.)
• physics -
work = force * distance
= F d
F the same for both
work done = kinetic energy lost
= .5 m v^2
same m
V2 = 3 V1
so
V2^2 = 9 V1^2
so work done at 50 = (1/2)m V1^2 = F d1
and at 150 = (1/2) m (9 V1 ^2) = F d2
so
d2/d1 = 9
9 * 15 = 135
## Similar Questions
1. ### PHSC
Hello, i need help. This is Work-Energy Theorum A car moving at 50km/hr skids 15m with locked brakes. How far will the car skid (with locked brakes if the car moves at 100km/hr assume that the mass of the car is 500kg
2. ### Physical science
A car moving at 50km/hr skids 15m with locked brakes. How far will the car skid (with locked brakes if the car moves at 100km/hr assume that the mass of the car is 500kg
3. ### physics
This question is typical on some driver's license exams: A car moving at 50km/hr skids 15m with locked brakes.How far will the car skid with locked brakes at 120 km/h?
4. ### physics
disregard the previous I did not include the first part This question is typical on some driver's license exams: A car moving at 50 km skids 15 m with locked brakes.How far will the car skid with locked brakes at 120 km/h?
5. ### physics
QA: This question is typical on some driver’s license exams: A car moving at 40 km/h skids 12 m with locked brakes. How far will the car skid with locked brakes at 80 km/h?
6. ### physics
This question is typical on some driver’s license exams: A car moving at 40 km/h skids 12 m with locked brakes. How far will the car skid with locked brakes at 80 km/h?
7. ### Physics
A car moving at 43 km/h skids 16 m with locked brakes. How far will the car skid with locked brakes at 129 km/h?
8. ### physics
A car moving at 50 km/h skids 15m with locked brakes. show with locke brakes at 150 km/m the car will skid 135?
9. ### Physict
This question is similar to some on driver's-license exams: A car moving at 50 km/h skids 15 m with locked brakes. Show that with locked brakes at 150 km/h the car will skid 135 m.
10. ### Physics
A car moving at 60 km/h skids 13 m with locked brakes. How far will the car skid with locked brakes at 180 km/h?
More Similar Questions
| 697
| 2,356
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.84375
| 4
|
CC-MAIN-2018-05
|
latest
|
en
| 0.937384
|
https://www.hometheaterforum.com/community/threads/help-with-my-algebra-homework.44906/
| 1,506,017,492,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-39/segments/1505818687834.17/warc/CC-MAIN-20170921172227-20170921192227-00165.warc.gz
| 792,769,004
| 34,996
|
# Help with my algebra homework?
Discussion in 'Archived Threads 2001-2004' started by Andrew S, Feb 24, 2002.
1. ### Andrew S Stunt Coordinator
Joined:
Sep 30, 2001
Messages:
214
0
Trophy Points:
0
Just so you guys know, I've already done like 20 questions... so it's not like you're DOING my homework for me... I just need help with two questions.
We're supposed to solve these word problems through elimination or subsitution.
#1. At a silversmith's shop, they have alloys that contain 40% silver and othets that are 50% silver. A custom order for a breacelet requires 150 grams of 44% silver. How much of each alloy should be melted together to make the bracelet?
My two equations were:
.4a + .5b = .44
a + b = 150
"a" means amount of 40% silver alloy
"b" means amount of 50% silver alloy
The answer in the back of the book is 90 g of 40% and 60 g of 50%, which sounds correct, but doesn't work when I subsititute it into my first equation, so I figure there's something wrong with my first equation.
#2. During a training exercise, a submarine travles 16km/h on the surface, but it only goes 10 km/h underwater. If the submarine traveled a distance of 160 km in 12.5 hours, how long was it underwater.
My equations were:
10a + 16b = 160km
a + b = 12.5 hours
So I changed the second equation to be b = 12.5 - a and then substituted that into the first equation to be:
"a" refers to time spent underwater
"b" refers to time spent above water
10a + 16(12.5 - a) = 160
10a + 200 - 16a = 160
-4a = 160 - 200
-4a = -40
-4a/-4 = -40/-4
a = 10
Then I subsituted the 10 into both equations:
First Equation:
10(10) + 16b = 160
100 + 16b = 160
16b = 160 - 100
16b = 60
16b/16 = 60/16
b = 3.75
Second Equation:
10 + b = 12.5
b = 12.5 - 10
b = 2.5
So as you can see, it doesn't work out. If it helps, the answer in the back of the book is 6 hours and 40 minutes (for how long the submarine was underwater)
I'll count myself lucky if anyone responds to this, so any help would be appreciated..
Thanks for helping out a confused grade 10.
Andrew
2. ### Keith Mickunas Cinematographer
Joined:
Dec 15, 1998
Messages:
2,041
0
Trophy Points:
0
Andrew, on #1 your first equation is forgetting that its 0.44 * c. Where c is the total weight. Try that and see how it works.
I'll get to the second in a moment, I don't want to do to much for you to fast. Gotta keep you honest.
3. ### Keith Mickunas Cinematographer
Joined:
Dec 15, 1998
Messages:
2,041
0
Trophy Points:
0
FWIW, I think you are making a similar mistake on #2.
You have to keep your units consistent on both sides of the equation. In the first, you have a weight of silver and a purity rating on the left, but you only have a total weight on the right. So you aren't balancing the equation. Do you see what I'm getting at?
4. ### Will Pomeroy Stunt Coordinator
Joined:
Feb 9, 2002
Messages:
144
0
Trophy Points:
0
Is that OAC alg/geo, or the new curriculum? (what grade?)
5. ### Andrew S Stunt Coordinator
Joined:
Sep 30, 2001
Messages:
214
0
Trophy Points:
0
This is actually grade 10 math and Keith, I'll see if that helps... that is if I understand you correctly...
Andrew
6. ### MikeAlletto Cinematographer
Joined:
Mar 11, 2000
Messages:
2,369
0
Trophy Points:
0
You got the answers in the back of the book? Hehehe...that reminds me what I would do. I would show all kinds of work and then at the end whatever I came out with if it didn't match the back of the book I would erase the final answer and just write in the back of the book answer. Most of the time the teacher didn't bother to check the work and just looked at the answer even though we were supposed to show all work.
7. ### Andrew S Stunt Coordinator
Joined:
Sep 30, 2001
Messages:
214
0
Trophy Points:
0
You're absolutely correct for the first equation, Keith. I kept coing up with 66, and 44% of 150 is 66.
It still amazes me the help you can get from this forum, even if it has nothing to do with Home Theater.
Thanks agai,
Andrew
8. ### Andrew S Stunt Coordinator
Joined:
Sep 30, 2001
Messages:
214
0
Trophy Points:
0
Mike,
Unfortunately all of my teachers' number one priority is the rough work. There's something like 3-5 marks for each question and if you just write the answer you get 1 mark. Really sucks for the easy questions that everyone can do in their heads because you still have to take up a quarter of the page with the rough work.
Oh well.
Andrew
9. ### Keith Mickunas Cinematographer
Joined:
Dec 15, 1998
Messages:
2,041
0
Trophy Points:
0
Makes me think back to the days of calc in college. On some problems I'd start with a blank sheet of paper, do all my work on there until I got it right, then copy the stuff to the sheet I was handing in. Sometimes you might work only 3 or 4 problems on a single sheet.
Do you see the relation between the two problems yet? They're fundamentally the same. In physics class they often require you to specify your units throughout the problem, which shows you how each side relates to the other. Remember that both sides of #2 are talking about km/h so you should be dealing with a formula where x+y=z and x,y,z are all km/h measurements.
10. ### MichaelPe Screenwriter
Joined:
Feb 22, 1999
Messages:
1,115
0
Trophy Points:
0
11. ### Bill Slack Supporting Actor
Joined:
Mar 16, 1999
Messages:
837
0
Trophy Points:
0
nm
12. ### Bill Catherall Screenwriter
Joined:
Aug 1, 1997
Messages:
1,560
0
Trophy Points:
0
Andrew - You setup the 2nd problem correctly, but you made an error in your math. Michael Perez is showing you the correction.
13. ### Keith Mickunas Cinematographer
Joined:
Dec 15, 1998
Messages:
2,041
0
Trophy Points:
0
Well dammit, now I feel stupid. You just have an arithmetic mistake in #2, like Michael said.
What I was referring too is that you're formula is really this:
a * 10km/h + b * 16km/h = 160km
Since a and b are units of time(h), your equation is balanced because you are multiplying a h value with a km/h value, thus cancelling the time. You can only add and subtract values that have the same unit. Does that make sense? I thought you didn't have all the units accounted for, like in the first problem, but you did.
In your first problem you have:
.4a + .5b = .44
or
(a grams * .4p) + (b grams * 0.5p) = .44p
where p is represent the unit for percantage of silver. So what you were doing was ending up with a formula that has different units on the different sides, so you couldn't work it. What you ended up with was:
(a grams * .4p) + (b grams * 0.5p) = 150 grams * 0.44p
a + b = 150
as your two equations which, as you found, can be solved.
Clear as mud, eh?
Joined:
Sep 30, 2001
Messages:
214
| 1,900
| 6,687
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.21875
| 4
|
CC-MAIN-2017-39
|
latest
|
en
| 0.95045
|
http://www.bardmathcircle.org/2015/08/day-4.html
| 1,519,290,687,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-09/segments/1518891814079.59/warc/CC-MAIN-20180222081525-20180222101525-00088.warc.gz
| 397,524,759
| 12,204
|
## Monday, August 31, 2015
### Day 4
The mathematicians began their day as greedy pirates, swash-buckling, mathematically of course, to get the most gold coins in a famous game theory conundrum.
Here's the problem for those who haven't seen it before:
There are 5 rational pirates A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them.
The pirates have a strict order of seniority: A is superior to B, who is superior to C, who is superior to D, who is superior to E.
The pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins. The pirates, including the proposer, then vote on whether to accept this distribution. In case of a tie vote the proposer has the casting vote. If the distribution is accepted, the coins are disbursed and the game ends. If not, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again.
Pirates base their decisions on three factors. First of all, each pirate wants to survive. Second, given survival, each pirate wants to maximize the number of gold coins each receives. Third, each pirate would prefer to throw another overboard, if all other results would otherwise be equal.The pirates do not trust each other, and will neither make nor honor any promises between pirates apart from a proposed distribution plan that gives a whole number of gold coins to each pirate.
Five pirates vying for 10 coins is an interesting problem but what if there were one hundred pirates each trying to get the most loot out of 100 coins? What if there were more pirates than coins? If there were more pirates than coins would the first pirate be able to get any coins? Is there a strategy where the first pirate would be able to stay onboard? Can you devise a formula for the number of coins any given pirate in such a situation would get where the number of pirates (N) is greater than the number of coins (G)? Use the backwards induction model like we used with the first version of the game!
Speaking of formulating your answer using variables, we continued to derive the pythagorean theorem in Geometry, replicating the work of the indian mathematician Bhakarsa in his famous proof. Walking the same path he did in 1114 CE!
In graph theory we built off of our definitions of days one through three to find isomorphic graphs amidst a maelstrom of interconnected nodes and edges. Mathematicians then used this fresh skill set to draw their own planar graphs.
During lunch I noticed a young mathematician who was playing a game of Cat's Cradle. I wondered, if they were to chop up the string could they make a planar graph with the same number of nodes and edges as we see in the picture below?
If you were to play cat's cradle without chopping up the string, how many different non-planar graphs can you make? How many of these are identical?
In computer science we talked about Boolean logic and the special definitions of and or or in programming.
Here's a joke about the inclusive or:
A logician's wife is in labor. The logician is in the waiting room.
The doctor comes out of the delivery room and says to the logician, "Congratulations, your wife gave birth to a beautiful baby!"
"Is it a boy or a girl?" asks the logician.
"Yes", says the doctor.
In art, mathematicians continued to use their budding architectural acumen to make platonic solids. We're eagerly shedding the templates that we made in the beginning of the week as tetrahedrons, octahedrons, dodecahedrons and more take form.
We're disappointed that C.A.M.P. is discrete and not continuous. The last update (until next summer of course) is tomorrow, stay tuned!
-- Eliana
| 825
| 3,747
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.84375
| 4
|
CC-MAIN-2018-09
|
latest
|
en
| 0.957085
|
http://mathhelpforum.com/calculus/217118-line-integral.html
| 1,532,110,507,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676591719.4/warc/CC-MAIN-20180720174340-20180720194340-00433.warc.gz
| 236,187,963
| 10,346
|
1. ## Line integral
let vector field F be (z^2/x,z^2/y, 2zlnxy). Evaluate the line integral of F
Where C is the path of straight line segments fromP = (1; 2; 1) to Q = (4; 1; 7) to R = (5; 11; 7), and then back to P.
2. ## Re: Line integral
Hey apatite.
Can you show us what you tried? (Hint: Write down the definition of the line integral for one single segment to begin).
3. ## Re: Line integral
i get zero as final answer. I want to double check it. Thank you. Is my answer right? In this particular case i use find the partial function and use that to get the line integral.
4. ## Re: Line integral
Well, yes.
If you calculate the curl $\displaystyle \nabla \times \vec{F} = \left( \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z} \right) \times \vec{F}$ of this vector field, you will find that it equals zero and so the field is conservative. This means that it has path independence and regardless of the path you take, if you end up at the same place (P), the work done on it equals zero.
I do not get what you mean by using the 'partial function' to solve the line integral, do you mean you used the gradient function to solve for the curl?
To solve this line integral you have to parameterise the function and then solve $\displaystyle \int_{t_0}^{t_1} \vec{F}(t) \times \vec{r} (t) dt$
| 385
| 1,347
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5625
| 4
|
CC-MAIN-2018-30
|
latest
|
en
| 0.820693
|
https://www.physicsforums.com/threads/why-not-transmit-electricity-in-4-phase.862397/
| 1,508,775,756,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187826114.69/warc/CC-MAIN-20171023145244-20171023165244-00640.warc.gz
| 949,087,521
| 20,662
|
# Why not transmit electricity in 4-phase?
Tags:
1. Mar 16, 2016
### GregValcourt
Or even better, generate in 4-Phase, convert to two-phase for transmission. Even number phases cancel out nicely (contrary to popular belief, 3-phase does not actually cancel out). 2-phase might be convertible back to four phase for the purpose of industrial motors.
Forgive me for speaking out of ignorance, I'm not an electrician or engineer. Mathematically, an even number of phases seems better for transmission (in terms of loss).
Mathematical graph of 3 sine waves 120 degrees apart, with resultant wave:
#### Attached Files:
• ###### Screen Shot 2016-03-16 at 7.28.25 PM.png
File size:
58 KB
Views:
35
Last edited: Mar 16, 2016
2. Mar 16, 2016
### Averagesupernova
Those 3 sine waves don't look 120 degrees apart to me.
3. Mar 16, 2016
### Baluncore
The sum of the instantaneous currents in all three phases should always be close to zero. That reduces the magnetic radiation from the transmission line.
Given a minimum of three phases and by using transformers, any number of other phase combinations can be generated. For a simple example see; https://en.wikipedia.org/wiki/Scott-T_transformer
The sum of cross sections of conductors in a transmission line is independent of the number of phases, but the cost of insulators is 33% higher for a 4 phase line than for a 3 phase line.
4. Mar 17, 2016
### andrewkirk
Here's why 3-phase cancels:
$$\sin(x-120)+\sin x+\sin(x+120)=\sin x\cos (-120)+\cos x\sin (-120)+\sin x+\sin x\cos 120+\cos x\sin 120$$
$$=-\sin x\cos 60+-\cos x\sin 60+\sin x-\sin x\cos 60+\cos x\sin 60$$
$$=-\sin x\cos 60+\sin x-\sin x\cos 60 =-\tfrac{1}{2}\sin x+\sin x-\tfrac{1}{2}\sin x=0$$
Picking up on the obaservation of @Averagesupernova, the blue and green curves in the graph are 90 degrees out of phase, not 120.
5. Mar 17, 2016
### anorlunda
OP, you really blew it when you said three phases don't add to zero. Ignoring that, high phase order transmission has been studied many times. The short answer is that the added benefits do not outweigh the added costs.
Google the phrase high phase order transmission.
6. Mar 17, 2016
### cabraham
Settled science. Single phase requires 2 wires. Two phase requires 3 wires. Three phase also needs 3 wires, 4 phase needs 4. Computing losses, you will find that to transmit a given amount of power from point A to point B with a given loss, 3 phase requires only 75% the copper area/weight as 1 or 2 phase. Increasing to 4 phase or more does not reduce copper needed, I can look it up later, but more phases may even need more copper than with 3 phases.
Three phase is optimum regarding maximum power ability, for a given amount of loss, with minimum copper consumed. The utility companies have studied this issue since the late 19th century. If a different number of phases is really better, it would have been done decades ago.
Claude
7. Mar 17, 2016
### Baluncore
I believed that the total section of copper needed remained the same for 3PH and above, while the number, mass and cost of insulators increased.
If anything I would expect total copper mass to fall slightly for more phases due to the skin effect not fully utilising the centre of thick conductors needed for 3PH.
Can you give a reference to the weight of copper needed for polyphase lines of three and above ?
8. Mar 17, 2016
### The Electrician
From the first paragraph here: https://en.wikipedia.org/wiki/Polyphase_system
"A major advantage of three phase power transmission (using three conductors, as opposed to a single phase power transmission, which uses two conductors), is that, since the remaining conductors act as the return path for any single conductor, the power transmitted by a balanced three phase system is three times that of a single phase transmission but only one extra conductor is used. Thus, a 50% / 1.5x increase in the transmission costs achieves a 200% / 3.0x increase in the power transmitted."
9. Mar 17, 2016
### cabraham
Wikipedia is incorrect here. I will find or recompute, then post. Here is the conclusion, I know this is right. For transmitting power from A to B, same amount using 1-phase & 3-phase, with equal power lost in transmission, a 3-ph system uses 75% the copper of 1-ph, as well as 2-ph.
10. Mar 18, 2016
### anorlunda
Can't you both be right?
• The Electrician holds wire diameter constant and increases the power transmitted, holding watts/(mile*phase) constant.
• Cabraham holds power and losses constant and decreases wire diameter.
Edit: By the way, high phase order transmission is interesting not only because of losses, but also because of series reactance.
11. Mar 18, 2016
### cabraham
Here is a 2-page computation for copper needed re 1-phase, 3-ph, and 4-ph transmission. Here is the synopsis. The generated power is 1.000 watt per unit, the power loss in cables in 0.010 watts per unit, and the load power is 0.990 watts per unit. These values were held constant for comparison purposes. Also, maximum line to line voltages were fixed at 1.0 volts per unit.
The result is what I seem to recall. The copper requirement for 1-phase is set at 100% as the reference. A 3-phase system requires just 75% the copper of 1-phase, a considerable savings. But, a 4-phase system did not improve on the 75% figure. Actually it came in at 100%, identical to 1 phase. I remember computing the general relation for any number of phases, i.e. "n-phase". For any value of "n" other than 3, the result is the same, the same amount of copper, i.e. 100%, is needed. If n=3, then 75% of the copper weight is needed.
I'm not surprised, because if 4 phases, or 5, 6, 10, whatever, saved on copper usage, the power companies & others would have quickly done it. I've heard of 6, 12, 24, & even 36 phase power employed. But it's advantage is usually very low ac ripple after rectification negating the need for large smoothing capacitors. But for long distance power transmission, 3 phase is pretty much the only game in town. Comments/feedback/questions welcome. Best regards.
Claude
#### Attached Files:
• ###### c abraham 1-3-4 phase cu needs0.pdf
File size:
109.8 KB
Views:
31
12. Mar 18, 2016
### mheslep
Or HVDC.
13. Mar 18, 2016
### GregValcourt
Your right, I must have did something wrong when building the spreadsheet. The second time I did it, 3-phase does cancel out.
| 1,677
| 6,394
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.90625
| 4
|
CC-MAIN-2017-43
|
longest
|
en
| 0.835147
|
http://www.jpetrie.net/2016/07/22/if-a-divides-b-and-a-divides-c-then-a-divides-b-c/
| 1,518,916,270,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-09/segments/1518891811243.29/warc/CC-MAIN-20180218003946-20180218023946-00241.warc.gz
| 483,870,240
| 13,962
|
# If a divides b and a divides c, then a divides (b-c)
In reading about Euclid’s proof of the infinitude of prime numbers, the only part that wasn’t completely clear to me was this:
If $$p$$ divides $$P$$ and $$q$$, then $$p$$ would have to divide the difference of the two numbers, which is $$(P + 1) − P$$ or just $$1$$.
Well, I don’t know…why is that true? Why does a number have to divide the difference of two numbers if it divides each of those numbers separately?
It turns out that my textbook from my introduction to proofs class, Mathematical Proofs by Chartrand, Polimeni, and Zhang (a class that was taught by Dr. Zhang herself at Western Michigan University), contains the proof of a more general statement. (Note: For this theorem, the vertical bar $$\vert$$, which looks identical to the absolute value symbol, is used to mean “divides”; that is, $$a ~\vert~ b$$ i.f.f. $$a$$ is a factor of $$b$$, that is, $$a$$ “guzzinta” $$b$$ an integer number of times, that is, $$b \div a$$ equals an integer.)
Theorem: If $$a~\vert~b$$ and $$a ~\vert~c$$, then $$a ~\vert~ (bx + cy)$$ for all integers $$x$$ and $$y$$.
Proof: Let $$a ~\vert~ b$$ and $$a ~\vert~ c$$. Then there exist integers $$q_{\scriptscriptstyle 1}$$ and $$q_{\scriptscriptstyle 2}$$ such that $$b=aq_{\scriptscriptstyle 1}$$ and $$c=aq_{\scriptscriptstyle 2}$$. Hence, for integers $$x$$ and $$y$$,
$$\begin{eqnarray} bx + cy = aq_{\scriptscriptstyle 1}x + aq_{\scriptscriptstyle 2}y = a(q_{\scriptscriptstyle 1}x + q_{\scriptscriptstyle 2}y). \end{eqnarray}$$
Since $$q_{\scriptscriptstyle 1} x + q_{\scriptscriptstyle 2} y$$ is an integer, $$a ~\vert~ (bx + cy)$$. $$\tag*{\blacksquare}$$
The specific case where $$x=1$$ and $$y=-1$$ is used in Euclid’s proof of the infinitude of primes.
Aside from this abstract proof, it’s easy to see why the theorem would be true with a simple example. Consider $$a = 5$$ and $$b$$ and $$c$$ any multiples of $$5$$, say $$25$$ and $$100$$. Since these are both multiples of $$5$$, they are some multiple of $$5$$ apart from each other, so their difference is also obviously a multiple of $$5$$. It’s just counting by fives, or by whatever the factor is in the example you choose. It’s hard to think of a theorem that could be more obvious and intuitive, even to an elementary-schooler.
As a side note, Mathematical Proofs is an extremely good introduction to mathematical proofs. It is one of my favorite textbooks. It is detailed, thorough, and extremely long, with great explanations that they call “proof strategy” before many proofs and summaries that they call “proof analysis” after many proofs. It has chapters on sets, logic, direct proofs, proof by contradiction, mathematical induction, equivalence relations, functions, cardinalities of sets, number theory, calculus, linear algebra, topology, group theory, and ring theory. I love this book and I highly recommend it for anyone who might want to learn or review how to do a lot of basic and important proofs from many fields of mathematics.
This entry was posted in Math, Theorems. Bookmark the permalink.
| 837
| 3,095
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.53125
| 5
|
CC-MAIN-2018-09
|
latest
|
en
| 0.797534
|
https://ambitiousbaba.com/reasoning-quiz-for-ibps-po-pre-sbi-clerk-mains-2019-7-august-2019/
| 1,675,630,341,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00467.warc.gz
| 111,829,892
| 53,705
|
# Reasoning Quiz for IBPS PO PRE & SBI CLERK MAINS 2019 | 7 August 2019
Table of Contents
## Reasoning Quiz for IBPS PO PRE & SBI CLERK MAINS
Reasoning Quiz to improve your Reasoning for SBI Po & SBI clerk exam Reasoning, IBPS PO Reasoning , IBPS Clerk Reasoning , IBPS RRB Reasoning, LIC AAO , LIC ADO, and other competitive exams.
Directions (Q1-Q5): Given below are five statements out of which one can be logically concluded from other four, you have to mark that statement as your answer.
Q1. Statements:
(a) Some D are A
(b) Some C are E
(c) All B are D
(d) Some D are E
(e) Some B are A
Answer & Explanation
Q1. Ans(a)
Q2. Statements:
(a) All bill are nut
(b) All bill are pot
(c) Some nut are rat
(d) Some rat are bat
(e) All nut are pot
Answer & Explanation
Q2. Ans(b)
Q3. Statements:
(a) No din are gin
(b) No rum are bum
(c) All din are rum
(d) No bum is gun
(e) No gin are rum
Answer & Explanation
Q3. Ans(a)
Q4. Statements:
(a) All toy are boy
(b) All boy are strong
(c) All strong are hard
(d) Some loud are boy
(e) All hard are loud
Answer & Explanation
Q4. Ans(d)
Q5. Statements:
(a) Some rum are num
(b) All fun are bun
(c) Some bun are rum
(d) All gun are rum
(e) Some fun are rum
Answer & Explanation
Q5. Ans(c)
Q6. Among J, K, L, M and N, who is the shortest? J is shorter than only N. K is taller than M and L. If they are arranged in ascending order of their heights from left to right then M is second from the left?
A) J
B) K
C) L
D) M
E) Can’t be determined.
Answer & Explanation
Q6. Ans(C)
Q7. Each of the five friends Kamal, Rahul, Suri, Kalia and Bhoopendra has a different age. Kamal is youger than only Rahul. Suri is older than Kalia. Who is not the youngest. Who among the following is/are older than Suri?
A) Only Kamal and Bhoopendra
B) Only Kalia and Kamal
C) Only Bhoopendra
D) Only Bhoopendra and Kalia
E) None of these
Answer & Explanation
Q7. Ans(E)
Q8. On which day was Shilpi definitely born? Shilpi’s mother correctly remembers that Shilpi was born before Friday but after Monday. Shilpi’s brother correctly remembers that his sister was born before Saturday but after Wednesday.
A) Monday
B) Tuesday
C) Thursday
D) Friday
E) Can’t be determined.
Answer & Explanation
Q8. Ans(C)
Q9. Five people A, B, C, D, E lives on five different floors from bottom to top. Two people live between B and C. A lives immediately above C. D lives on one of the floor above E. Then who among the following lives on fourth floor?
A) B
B) A
C) Can’t be determine
D) E
E) D
Answer & Explanation
Q9. Ans(C)
Q10. E is heavier than G and K but not as much as F. L is only lighter than M. The one who is second heaviest is 56kg. The weight of third lightest is 47 kg. K is 10kg lighter than E. G is of 18kg. What can be the weight of F?
A) 58 kg
B) 70 kg
C) 50 kg
D) 45 kg
E) None of these
Answer & Explanation
Q10. Ans(C)
### You Can Read This Also:
Best E-books for LIC AAO 2019 : Get PDF here
ambitiousbaba.com need your support to Grow
I challenge you will get Best Content in Our PDFs with Detail solutions and Latest Pattern
Memory Based Puzzle E-book | 2016-19 Exams Covered
Get PDF here
Caselet Data Interpretation 200 Questions
Get PDF here
Puzzle & Seating Arrangement E-Book for BANK PO MAINS (Vol-1)
Get PDF here
### ARITHMETIC DATA INTERPRETATION 2019 E-book
Get PDF here
The Banking Awareness 500 MCQs E-book| Bilingual (Hindi + English)
Get PDF here
High Level DATA INTERPRETATION Practice E-BOOK
Get PDF here
### How to Access on App:-
1. Go to Playstore search Ambitious Baba or Click here to Install App
2. After Install Login with Google Account or Facebook Account
3
| 1,090
| 3,700
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.734375
| 4
|
CC-MAIN-2023-06
|
latest
|
en
| 0.853838
|
https://edurev.in/course/quiz/attempt/16247_Test-Sequences-And-Series-1/c653c0df-4e9f-457f-af1e-9434bd7208c4
| 1,660,094,590,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882571097.39/warc/CC-MAIN-20220810010059-20220810040059-00771.warc.gz
| 236,106,093
| 42,797
|
# Test: Sequences And Series - 1
## 15 Questions MCQ Test Practice Questions for GMAT | Test: Sequences And Series - 1
Description
Attempt Test: Sequences And Series - 1 | 15 questions in 30 minutes | Mock test for GMAT preparation | Free important questions MCQ to study Practice Questions for GMAT for GMAT Exam | Download free PDF with solutions
QUESTION: 1
### If integer k is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of k?
Solution:
First, let us simplify the problem by rephrasing the question. Since any even number must be divisible by 2, any even multiple of 15 must be divisible by 2 and by 15, or in other words, must be divisible by 30. As a result, finding the sum of even multiples of 15 is equivalent to finding the sum of multiples of 30. By observation, the first multiple of 30 greater than 295 will be equal to 300 and the last multiple of 30 smaller than 615 will be equal to 600.
Thus, since there are no multiples of 30 between 295 and 299 and between 601 and 615, finding the sum of all multiples of 30 between 295 and 615, inclusive, is equivalent to finding the sum of all multiples of 30 between 300 and 600, inclusive. Therefore, we can rephrase the question: “What is the greatest prime factor of the sum of all multiples of 30 between 300 and 600, inclusive?”
The sum of a set = (the mean of the set) × (the number of terms in the set)
Since 300 is the 10th multiple of 30, and 600 is the 20th multiple of 30, we need to count all multiples of 30 between the 10th and the 20th multiples of 30, inclusive.
There are 11 terms in the set: 20th – 10th + 1 = 10 + 1 = 11
The mean of the set = (the first term + the last term) divided by 2: (300 + 600) / 2 = 450
k = the sum of this set = 450 × 11
Note, that since we need to find the greatest prime factor of k, we do not need to compute the actual value of k, but can simply break the product of 450 and 11 into its prime factors:
k = 450 × 11 = 2 × 3 × 3 × 5 × 5 × 11
Therefore, the largest prime factor of k is 11.
QUESTION: 2
### If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn-1 + 6,..., what is the sum of all terms in the set {S13, S14, ..., S28}?
Solution:
For sequence S, any value Sn equals 6n. Therefore, the problem can be restated as determining the sum of all multiples of 6 between 78 (S13) and 168 (S28), inclusive. The direct but time-consuming approach would be to manually add the terms: 78 + 84 = 162; 162 + 90 = 252; and so forth.
The solution can be found more efficiently by identifying the median of the set and multiplying by the number of terms. Because this set includes an even number of terms, the median equals the average of the two ‘middle’ terms, S20 and S21, or (120 + 126)/2 = 123. Given that there are 16 terms in the set, the answer is 16(123) = 1,968.
QUESTION: 3
### In an increasing sequence of 5 consecutive even integers, the sum of the second, third, and fourth integer is 132. What is the sum of the first and last integers?
Solution:
Let the five consecutive even integers be represented by x, x + 2, x + 4, x + 6, and x + 8. Thus, the second, third, and fourth integers are x + 2, x + 4, and x + 6. Since the sum of these three integers is 132, it follows that
3x + 12 = 132, so
3x = 120, and
x = 40.
The first integer in the sequence is 40 and the last integer in the sequence is x + 8, or 48.
The sum of 40 and 48 is 88.
QUESTION: 4
What is the sum of the multiples of 7 from 84 to 140, inclusive?
Solution:
84 is the 12th multiple of 7. (12 x 7 = 84)
140 is the 20th multiple of 7.
The question is asking us to sum the 12th through the 20th multiples of 7.
The sum of a set = (the mean of the set) x (the number of terms in the set)
There are 9 terms in the set: 20th - 12th + 1 = 8 + 1 = 9
The mean of the set = (the first term + the last term) divided by 2: (84 + 140)/2 = 112
The sum of this set = 112 x 9 = 1008
Alternatively, one could list all nine terms in this set (84, 91, 98 ... 140) and add them.
When adding a number of terms, try to combine terms in a way that makes the addition easier
(i.e. 98 + 112 = 210, 119 + 91 = 210, etc).
QUESTION: 5
If each term in the sum a1 + a2 + a3 + ... +an is either 7 or 77 and the sum is equal to 350, which of the
following could equal to n?
Solution:
QUESTION: 6
For any integer k from 1 to 10, inclusive, the kth of a certain sequence is given by [(-1)(k+1)] × (1 / 2k). If T is the sum of the first 10 terms of the sequence, then T is:
Solution:
T= 1/2-1/22+1/23-...-1/210
= 1/4+1/42+1/43+1/44+1/45
Notice that 1/42+1/43+1/44+1/45 < 1/4, we can say that 1/4<T<1/2.
QUESTION: 7
S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk–1 + 2(10k–1). If p is the sum of the first 30 terms of S, what is the eleventh digit of p, counting right to left from the units digit?
Solution:
The first few terms of the sequence are 2, 22, and 222 and each subsequent term has an additional 2 added on. The 30th term then is a string of 30 2’s. If we line up the first 30 terms of the sequence to add them up, we will get rows in the following pattern:
To find p, the sum of the first 30 terms of S, we would simply be adding columns of 2’s. The key here is to see a pattern in the addition process. Starting with the units digit column, all 30 of the terms have a 2 in that position so the sum of the units column would be 30 x 2 = 60. A zero would be written as the units digit of the sum and a six would be carried over to the tens column.
In the tens column, 29 of the 30 terms would have a 2 because the first term has no tens digit. The sum of the tens digits would be 29 x 2 = 58, to which we must add the 6 for a total of 64. The 4 gets written down as the second digit of p and the 6 is carried over to the hundreds column.
In the hundreds column, 28 of the 30 terms would have a 2, the sum of the hundreds digits would be 28 x 2 = 56, to which we must add the 6 again for a total of 62. The 2 gets written down as the third digit of p and the 6 is carried over to the thousands column.
There are two ways to finish this problem. We can do out the remaining 8 columns and find that the 11th digit (i.e. the 10 billions column) will have a sum of 2(20) + 4 = 44 (where the 4 was carried over from the 10th column). 4 then will be the 11th digit (from the right) of p (and a 4 will be carried over into the 12th column).
We could also have seen that each column has one less 2 than the previous, so if we started out with 30 2’s in the first column, the 11th column must have 11 – 1 = 10 less 2’s, for a total of 20 2’s. The amount that is carried over from the previous column could be calculated by realizing that the 10th column had 21 2’s for a total of 42. Since there is no way that the 10th column inherited more than 8 from the 9th column, the total must be forty-something and the amount that is carried over to the 11th column MUST BE 4. This makes the total for the 11th column 40 + 4 = 44 and the 11th digit of p4.
QUESTION: 8
Sequence S is defined as Sn = 2Sn-1 – 2. If S1 = 3, then S10 – S9 =
Solution:
We can use the formula to calculate the first 10 values of S:
S1 = 3
S2 = 2(3) – 2 = 4
S3 = 2(4) – 2 = 6
S4 = 2(6) – 2 = 10
S5 = 2(10) – 2 = 18
S6 = 2(18) – 2 = 34
S7 = 2(34) – 2 = 66
S8 = 2(66) – 2 = 130
S9 = 2(130) – 2 = 258
S10 = 2(258) – 2 = 514
S10 – S9 = 514 – 258 = 256.
Alternatively, we could solve this problem by noticing the following pattern in the sequence:
S2 – S1 = 1 or (20)
S3 – S2 = 2 or (21)
S4 – S3 = 4 or (22)
S5 – S4 = 8 or (23)
We could extrapolate this pattern to see that S10 S9 = 28 = 256.
QUESTION: 9
Sn = 2Sn-1 + 4 and Qn = 4Qn-1 + 8 for all n > 1. If S5 = Q4 and S7 = 316, what is the first value of n for whichQn is an integer?
Solution:
If S7 = 316, then 316 = 2S6 + 4, which means that S6=156.
We can then solve for S5
156 = 2S5 + 4, so S5 = 76
Since S5 = Q4, we know that Q4 = 76 and we can now solve for previous Qn’s to find the first n value for which Qn is an integer.
To find Q3: 76 = 4Q3 + 8, so Q3 = 17
To find Q2: 17 = 4Q2 + 8, so Q2 = 9/2
It is clear that Q1 will also not be an integer so there is no need to continue.
Q3 (n = 3) is the first integer value.
QUESTION: 10
What is the sixtieth term in the following sequence? 1, 2, 4, 7, 11, 16, 22...
Solution:
Noting that a1 = 1, each subsequent term can be calculated as follows:
a1 = 1
a2 = a1 + 1
a3 = a1 + 1 + 2
a4 = a1 + 1 + 2 + 3
ai = a1 + 1 + 2 + 3 + ... + i-1
In other words, ai = a1 plus the sum of the first i - 1 positive integers. In order to determine the sum of the first i - 1 positive integers, find the sum of the first and last terms, which would be 1 and i - 1 respectively, plus the sum of the second and penultimate terms, and so on, while working towards the median of the set. Note that the sum of each pair is always equal to i:
1 + (i - 1) = i
2 + (i – 2) = i
3 + (i – 3) = i
…
Because there are (i - 1)/2 such pairs in a set of i - 1 consecutive integers, this operation can be summarized by the formula i(i - 1)/2. For this problem, substituting a1 = 1 and using this formula for the sum of the first (i-1) integers yields:
ai = 1 + (i)(i - 1)/2
The sixtieth term can be calculated as:
a60 = 1 + (59)(60)/2
a60 = 1,771
QUESTION: 11
Sequence S is defined as Sn = X + (1/X), where X = Sn – 1 + 1, for all n > 1.
If S1= 201, then which of the following must be true of Q, the sum of the first 50 terms of S?
Solution:
To find each successive term in S, we add 1 to the previous term and add this to the reciprocal of the previous term plus 1.
S1= 201
The question asks to estimate (Q), the sum of the first 50 terms of S. If we look at the endpoints of the intervals in the answer choices, we see have quite a bit of leeway as far as our estimation is concerned. In fact, we can simply ignore the fractional portion of each term. Let’s use S2 ≈ 202, S3 ≈ 203. In this way, the sum of the first 50 terms of S will be approximately equal to the sum of the fifty consecutive integers 201, 202 … 250.
To find the sum of the 50 consecutive integers, we can multiply the mean of the integers by the number of integers since average = sum / (number of terms).
The mean of these 50 integers = (201 + 250) / 2 = 225.5
Therefore, the sum of these 50 integers = 50 x 225.5 = 11,275, which falls between 11,000 and 12,000. The correct answer is C.
QUESTION: 12
In a certain sequence, every term after the first is determined by multiplying the previous term by an integerconstant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number ofnonnegative integer values possible for the first term?
Solution:
The equation of the sequence can be written as follows: an= (an-1 )(x) , where x is the integer constant. So for every term after the first, multiply the previous term by x. Essentially, then, all we are doing is multiplying the first term by x over and over again. For example, (a2)= (a1)(x) and (a3)=(a2)(x) or a3 = ((a1)(x))(x), which is the same as (a3) = (a1)(x2)
If we keep going, we’ll see that a3 = (a1)(x2) and so on for the rest of the sequence. We can thus rewrite the equation of the sequence as an = (a1)(xn-1) , for all n >1.
We also know from the question that a5 < 1000, which means that (a1)(x4) < 1000
We are asked for the maximum number of possible nonnegative integer values for ; we can get this by minimizing the value of the integer constant, x. Since x is an integer constant greater than 1, the smallest possible value for x is 2. When x = 2, then x4= 16
We can solve for a1 as follows:
Thus all the integers from 1 to 62, inclusive, are permissible for a1. So far we have 62 permissible values. If a1=0 then it doesn’t matter what x is, since every term in the sequence will always be 0. So 0 is one more permissible value for a1
There is a maximum of 62 +1 (or 63) nonnegative integer values for ain which a5 < 1000.
QUESTION: 13
The sum of the squares of the first 15 positive integers (12 + 22 + 32 + . . . + 152) is equal to 1240. What is the sum of the squares of the second 15 positive integers (162 + 172 + 182 + . . . + 302) ?
Solution:
The key to solving this problem is to represent the sum of the squares of the second 15 integers as follows: (15 + 1)2 + (15 + 2)2 + (15 + 3)2 + . . . + (15 + 15)2.
Recall the popular quadratic form, (a + b)2 = a2 + 2ab + b2. Construct a table that uses this expansion to calculate each component of each term in the series as follows:
In order to calculate the desired sum, we can find the sum of each of the last 3 columns and then add these three subtotals together. Note that since each column follows a simple pattern, we do not have to fill in the whole table, but instead only need to calculate a few terms in order to determine the sums.
The column labeled a2 simply repeats 225 fifteen times; therefore, its sum is 15(225) = 3375.
The column labeled 2ab is an equally spaced series of positive numbers. Recall that the average of such a series is equal to the average of its highest and lowest values; thus, the average term in this series is (30 + 450) / 2 = 240. Since the sum of n numbers in an equally spaced series is simply n times the average of the series, the sum of this series is 15(240) = 3600.
The last column labeled b2 is the sum of the squares of the first 15 integers. This was given to us in the problem as 1240.
Finally, we sum the 3 column totals together to find the sum of the squares of the second 15 integers: 3375 + 3600 + 1240 = 8215. The correct answer choice is (D).
QUESTION: 14
The infinite sequence Sk is defined as Sk = 10 Sk – 1 + k, for all k > 1. The infinite sequence An is defined as An = 10 An – 1 + (A1 – (n - 1)), for all n > 1. q is the sum of Sk and An. If S1 = 1 and A1 = 9, and if An is positive, what is the maximum value of k + n when the sum of the digits of q is equal to 9?
Solution:
In complex sequence questions, the best strategy usually is to look for a pattern in the sequence of terms that will allow you to avoid having to compute every term in the sequence.
In this case, we know that the first term of Sk is 1 and the first term of An is 9. So when n = 1 and k = 1, q = 9 + 1 = 10 and the sum of the digits of q is 1 + 0 = 1.
since S1= 1, S2= (10)(1) +(2) = 10+2 = 12. since A1 =9,
A2= (10) (9)+(9-(2-1)) = 90+(9-1) = 90+8 = 98. so When K=2 and n = 2, q = 12 + 98 = 110 and the sum of the digits of q is 1 + 1 + 0 = 2.
Since S2= 12, S3 =(10)(12)+3= 120+3= 123. Since A2= 98, It is true that
A3= (10)(98) + (9-(3-1)) = 980+7 = 987. So when n = 3 and k = 3, q = 123 + 987 = 1110 and the sum of the digits of q is 1 + 1 + 1 + 0 = 3.
At this point, we can see a pattern: sproceeds as 1, 12, 123, 1234..., and An proceeds as 9, 98, 987, 9876.... The sum q therefore proceeds as 10, 110, 1110, 11110... The sum of the digits of q, therefore, will equal 9 when q consists of nine ones and one zero. Since the number of ones in q is equal to the value of n and k (when n and k are equal to each other), the sum of the digits of q will equal 9 when n = 9 and k = 9:
S9 = 123456789 and A9 = 987654321. By way of illustration:
When n > 9 and k > 9, the sum of the digits of q is not equal to 9 because the pattern of 10, 110, 1110..., does not hold past this point and the additional digits in q will cause the sum of the digits of q to exceed 9.
Therefore, the maximum value of k + n (such that the sum of the digits of q is equal to 9) is 9 + 9 = 18.
QUESTION: 15
A certain club has exactly 5 new members at the end of its first week. Every subsequent week, each
of the previous week's new members (and only these members) brings exactly x new members into the club. If y is the number of new members brought into the club during the twelfth week, which of the following could be y?
Solution:
At the end of the first week, there are 5 members. During the second week, 5x new members are brought in (x new members for every existing member). During the third week, the previous week's new members (5x) each bring in x new members:
(5x)x= 5xnew members. If we continue this pattern to the twelfth week, we will see that 5x11 new members join the club that week. Since y is the number of new members joining during week 12, y=5x11.
if y=5x11, we can set each of the answer choices equal to 5x11 and see which one yields an integer value (since y is a specific number of people, it must be an integer value). The only choice to yield an integer value is (D)
5x11= 311512
x11 = 311512
x=(3)(5)
Therefore x = 15.
Since choice (D) is the only one to yield an integer value, it is the correct answer.
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code
| 5,258
| 16,729
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.6875
| 5
|
CC-MAIN-2022-33
|
latest
|
en
| 0.935509
|
https://mathematica.stackexchange.com/questions/58032/how-to-find-the-integer-solution-number-of-a-linear-system-with-inequalities
| 1,716,074,408,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971057516.1/warc/CC-MAIN-20240518214304-20240519004304-00350.warc.gz
| 342,331,492
| 40,180
|
# How to find the integer solution number of a linear system with inequalities?
I have the following system with mixed equality and inequalities defined as:
n = 100;
eqn = (n1 + n3 + n5 == n2 + n4 + n6) && (n1 >=
n2) && (n1 - n2 + n3 - n4 >= 0) && (n1 <= n) && (n1 - n2 + n3 <=
n) && (n1 - n2 + n3 - n4 + n5 <= n) && (2 n <=
n1 + n2 + n3 + n4 + n5 + n6 <= 6 n) && (n1 > 0) && (n2 > 0) && (n3 > 0) && (n4 > 0) && (n5 > 0) && (n6 > 0);
How can I find all the integer solutions and especially the number of all integer solutions?
It seems Reduce or Solve does not work:
sol = Reduce[eqn, {n1, n2, n3, n4, n5, n6}, Integers]
produces:
(n1 | n2 | n3 | n4 | n5 | n6) \[Element]
Integers && ((1 <= n1 <= 99 &&
0 < n2 <=
n1 && ((0 < n3 < 100 - n1 && 0 < n4 <= n1 - n2 + n3 &&
100 - n1 - n3 <= n5 <= 100 - n1 + n2 - n3 + n4 &&
n6 == n1 - n2 + n3 - n4 + n5) || (100 - n1 <= n3 <=
100 - n1 + n2 && 0 < n4 <= n1 - n2 + n3 &&
0 < n5 <= 100 - n1 + n2 - n3 + n4 &&
n6 == n1 - n2 + n3 - n4 + n5))) || (n1 == 100 &&
1 <= n2 <= 100 && 0 < n3 <= n2 && 0 < n4 <= 100 - n2 + n3 &&
0 < n5 <= n2 - n3 + n4 && n6 == 100 - n2 + n3 - n4 + n5))
• FindInstance[eqn, {n1, n2, n3, n4, n5, n6}, Integers, 10] will give you the first ten sols, but there are more ... Aug 24, 2014 at 4:14
• I tried FIndInstance by setting the number 10 as 10,000 instead; it has more than 10,000 solutions! The computation is very slow and has a very high demand on memory. Aug 24, 2014 at 9:28
SetSystemOptions[
"ReduceOptions" -> {"DiscreteSolutionBound" -> 10000000}];
f[n_] := Module[{eqn},
eqn = (n1 + n3 + n5 == n2 + n4 + n6) && (n1 >=
n2) && (n1 - n2 + n3 - n4 >= 0) && (n1 <= n) && (n1 - n2 + n3 <=
n) && (n1 - n2 + n3 - n4 + n5 <= n) && (2 n <=
n1 + n2 + n3 + n4 + n5 + n6 <= 6 n) && (n1 > 0) && (n2 >
0) && (n3 > 0) && (n4 > 0) && (n5 > 0) && (n6 > 0);
Length@Reduce[eqn, {n1, n2, n3, n4, n5, n6}, Integers]]
Flatten[{#, Timing[f@#]}] & /@ Range[5, 15] //
TableForm[#, TableHeadings -> {None, {"n", "timing", "numbers"}}] &
Edit: First
Reduce[n1 <= n, n1, Reals]
n1 <= n
Reduce[n1 >= n2, n2, Reals]
n2 <= n1
Reduce[n1 - n2 + n3 <= n, n3, Reals]
n3 <= n - n1 + n2
Reduce[n1 - n2 + n3 - n4 >= 0, n4, Reals]
n4 <= n1 - n2 + n3
Reduce[n1 - n2 + n3 - n4 + n5 <= n, n5, Reals]
n5 <= n - n1 + n2 - n3 + n4
sol = Solve[n1 + n3 + n5 == n2 + n4 + n6, n6]
(2 n <= n1 + n2 + n3 + n4 + n5 + n6 <= 6 n) /. sol
{{n6 -> n1 - n2 + n3 - n4 + n5}}
{2 n <= 2 n1 + 2 n3 + 2 n5 <= 6 n}
Reduce[2 n <= 2 n1 + 2 n3 + 2 n5 <= 6 n && n > 0, n5, Reals]
n > 0 && n - n1 - n3 <= n5 <= 3 n - n1 - n3
Reduce[(n6 /. sol) > 0, n5, Reals]
n5 > -n1 + n2 - n3 + n4
So the range of n5 is:
Max[-n1 + n2 - n3 + n4 + 1, n - n1 - n3, 1]< n5 < Min[3 n - n1 - n3, n - n1 + n2 - n3 + n4]
g1 = Compile[{{n, _Integer}},
Sum[1, {n1, n}, {n2, n1}, {n3, n - n1 + n2}, {n4,
n1 - n2 + n3}, {n5, Max[-n1 + n2 - n3 + n4 + 1, n - n1 - n3, 1],
Min[3 n - n1 - n3, n - n1 + n2 - n3 + n4]}], CompilationTarget -> "C"];
Flatten[{#, Timing[g1@#]}] & /@ Range[5, 15] //
TableForm[#, TableHeadings -> {None, {"n", "timing", "numbers"}}] &
g1[50] // AbsoluteTiming
{0.999057, 39324265}
g2 = Compile[{{n, _Integer}},
Sum[Max[0, Min[3 n - n1 - n3, n - n1 + n2 - n3 + n4] -
Max[-n1 + n2 - n3 + n4 + 1, n - n1 - n3, 1] + 1], {n1, n}, {n2,
n1}, {n3, n - n1 + n2}, {n4, n1 - n2 + n3}], CompilationTarget -> "C"];
g2[50] // AbsoluteTiming
g2[100] // AbsoluteTiming
{0.248014, 39324265}
{3.890223, 1212505405}
• Thank you! Where can I 学到 learn such insights in using Mathematica this way? It seems the online documents of Wolfram have no such contents. Aug 25, 2014 at 5:00
• One possible issue is when $n$ increases, the memory demand becomes high quickly. If only solution number is required, can it be improved to be more efficient and less memory demanding? Aug 25, 2014 at 5:24
• +1 for a great answer and your first Enlightened badge. :-) Aug 29, 2014 at 12:06
| 1,759
| 3,907
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625
| 4
|
CC-MAIN-2024-22
|
latest
|
en
| 0.555705
|
http://mathhelpforum.com/calculus/71993-vector-parametric-equation.html
| 1,480,960,257,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698541773.2/warc/CC-MAIN-20161202170901-00118-ip-10-31-129-80.ec2.internal.warc.gz
| 171,027,757
| 11,896
|
1. ## Vector Parametric Equation
How do I find the equation of a plane with this information:
Its through the point (-3, 9, 10)
and x = 5+t y = 4t and z = 3-2t
2. Originally Posted by acg716
How do I find the equation of a plane with this information:
Its through the point (-3, 9, 10)
and x = 5+t y = 4t and z = 3-2t
I'm guessing that the plane contains the line you mention. If so, then the vector
$\bold{u} = < 1, 4, -2>$
is on the plane. We can also find another vector on the plane, a point on the line to the point given
$\bold{v} = <5-(-3),0-9 , 3 -10> = <8,-9,-7>$
The normal to the plane is then obatin from
$\bold{n} = \bold{u} \times \bold{v}$ (*)
If $\bold{n} = $
then the plane is given by
$n_1(x+3) + n_2(x-9) + n_3(x-10) = 0$
So all you need to do is find n *
3. When I did that I got -46(x+3)-9(y-9)-41(z-10)=0
but that isn't the correct answer...
4. Originally Posted by acg716
When I did that I got -46(x+3)-9(y-9)-41(z-10)=0
but that isn't the correct answer...
Well, when you substitute the point $(-3,9,10)$ into this plane we get 0, right? And when we substitute $x = 5+t,\;\;y = 4t,\;\;z = 3-2t$ we get
$-46(5+t+3)-9(4t-9)-41(3-2t-10)=$
$-46(8+t)-9(4t-9)-41(-7-2t)=$
$-368 - 46 t +81-36t + 287 + 82t = 0$
So both the point and line lie on the plane. Did the question say that the line is perpendicular to the plane?
5. yes it is perpendicular, so sorry that i forgot to include that!
6. Originally Posted by acg716
yes it is perpendicular, so sorry that i forgot to include that!
Makes a huge difference (and easier). The normal to the plane will be parallel so the line and so
$\bold{n} = <1,4,-2>$
so the line is
$(x+3) + 4(x-9) - 2(x-10)=0$
See, way easier.
7. ahh it still says it wrong so maybe I'm leaving something out from the directions so here is what the directions say exactly:
Find an equation of the plane through the point and perpendicular to the given line.
(-3, 9, 10)
x= 5 + t
y= 4t
z= 3-2t
8. Originally Posted by acg716
ahh it still says it wrong so maybe I'm leaving something out from the directions so here is what the directions say exactly:
Find an equation of the plane through the point and perpendicular to the given line.
(-3, 9, 10)
x= 5 + t
y= 4t
z= 3-2t
What answer are you given? Are you given $x + 4y - 2z = 13$?
9. I'm not given an answer, its on my webassign homework so when I submit it it tells me if I'm right or not.
10. Originally Posted by danny arrigo
Makes a huge difference (and easier). The normal to the plane will be parallel so the line and so
$\bold{n} = <1,4,-2>$
so the line is
$(x+3) + 4(x-9) - 2(x-10)=0$
See, way easier.
Originally Posted by danny arrigo
What answer are you given? Are you given $x + 4y - 2z = 13$?
Did you try both answers? If so, I don't know what to suggest. Can some of the other regulars chime in here.
11. The second one was correct, thank you so much!!!
| 975
| 2,890
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.3125
| 4
|
CC-MAIN-2016-50
|
longest
|
en
| 0.903276
|
https://www.jobtestprep.co.uk/free-abstract-reasoning-test?idev_username=jtp-dk
| 1,568,651,977,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-39/segments/1568514572879.28/warc/CC-MAIN-20190916155946-20190916181946-00118.warc.gz
| 883,880,190
| 20,320
|
You are logged in as customer LOG OUT
4.9
7.5 Minute Time Limit
9,477 Have Taken This Test
15 Questions included
XXX
XXX
XXX
XXX
XXX
XXX
## What Is an Abstract Reasoning?
Abstract reasoning tests, also known as conceptual reasoning, are non-verbal in nature and consist of questions including shapes and images. They are widely used as part of pre-employment aptitude or psychometric assessments.
The exam measures your lateral thinking and fluid intelligence with the sole purpose of determining how quickly you can identify patterns, logical rules and data trends.
The test is meant to inform your future employer how capable you are at learning new skills, thinking strategically about problems and analysing new information.
For this challenging assessment, your current skills may not be enough.
## Free Abstract Reasoning Test Practice
Test Time 7:30 min
Questions 15
Pass Score 8
## Abstract Example Questions
There are many different types of abstract questions, however, most are similar to the common core.
Below we have grouped together 4 most frequently used question types to help you distinguish and answer them quickly.
### 1) Matrice
Matrices are a group of separate images that follow a particular pattern. Your task is to understand the pattern and cypher which image should appear next. Below is an example:
In each row two shapes are united to create the third with the exception of the outline of the outermost shape which remains constant (the square in the first row, circle in the second row and diamond in the third row). In the first row, the united figure is on the left; in the second figure it is on the right and in the bottom row it should be in the centre. Therefore, the missing figure should have a diamond outline and have only a black dot in the centre so when combined with the figure on the left will create the figure in the centre.
### 2) Next in Series
Here you are presented with several images in a row. It is your job to comprehend the pattern and find the missing image.
Note: The missing image will not necessarily be at the end of the row. Frequently it can be found in the middle of the line.
There are two characteristics in this question: the number of chords (lines) drawn, and shading. First, look at the chords. In each frame, an additional chord is drawn from one corner (vertex) of the pentagon to another corner directly across from it. If you label each corner of the pentagon with a number, it is easier to see.
The first chord is drawn from vertex one (to vertex three).
The next chord is drawn from vertex two (to vertex four).
The next from vertex three (to vertex five).
The final one is drawn from vertex four (to vertex one).
Therefore, the missing shape is the one with the chord drawn from vertex five, creating the star shape inside the pentagon.
This eliminates answer choices one and four as they do not contain the star shape. Next, look at the shading in the shape. Each time a new chord is drawn, it reduces the amount of grey shading. The shading always lies on the inside of the chord boundaries. When drawing the final chord from vertex five, the remaining space inside that chord is just in the centre of the star. Therefore, it can only be answer choice two.
### 3) Odd One Out
Unlike matrices and next in the series, odd one out questions do not follow a particular pattern. Rather you must determine which shape is different from the rest in the series. Below is an example:
Choose the figure that does not share a common pattern with the other four:
Every shape in the sequence, aside from this one, is symmetrical on a vertical axis, while the parallelogram is symmetrical on a diagonal axis.
### 4) Analogies
In this series of questions, you are presented with two pairs of images. One pair will be complete, the second incomplete. While the two pairs are similar in pattern, there will be differences.
So, you will have to identify what is similar in the first set and organically implement it to second set of shapes.
The relationship between figure X and figure Y is as follows:
Figure Y represents figure X with:
1) The dotted section becoming white, and
2) The white section becoming shaded
The correct answer must have the same relationship with figure Z.
Answer 2 can be eliminated as the blank section remains blank, and the dotted section becomes shaded instead of becoming blank.
Answer 3 can be eliminated as the dotted section remains dotted.
We are left with answer 1, which is the correct answer, as it portrays figure Z while the blank section becomes shaded and the dotted section becomes blank.
We Help Over 40,000 Job-Seekers Each Year Get A better Score
Continue The Practice You Started to See Results
within Less than a Week
## How to Improve Your Abstract Reasoning Results
Improving your abstract reasoning results can prove difficult if you do not know where to begin. Thus, when given a new question, remember that the first step is identifying the question type.
To do this, you must pay close attention to the patterns and colours within the image. For example, what are the differences or similarities?
Once you feel confident in knowing the question types, your mental ability of breaking down each puzzle will greatly improve, allowing you to score higher on your test.
## Abstract Reasoning Test Tips
Oftentimes when given a new abstract question, candidates become overwhelmed with the sequence of pictures. With the right information however, you can build your confidence and answer correctly.
The first step is to take into account the shapes, colour, location, direction, rotation, movement and order of each image.
If you are having difficulty solving the question, try beginning at the end of the sequence. Next, use the information you have learned from the sample questions.
Remember:
1) Matrices include a group of figures in a single panel. Each panel follows a certain rule or pattern. You will determine what this pattern is and choice the answer that fits best.
2) Next in the Series are questions that have a row of images that follow a pattern. Identify the pattern to determine the next in the sequence.
3) Odd one Out present you with several images in an order that follow a particular pattern. Find the images that do not fit the pattern.
4) Analogies begin with two pairs of shapes that share a relationship. Your task is to find the locate the relationship and implement it into the one of the given answers.
Hopefully now, you will feel more familiar with the common question types. If you still are having trouble visualising the questions, reread the sample questions show above.
Start Practising to See Results Within Just a Few Days
## Prepare for a Specific Test Provider
Optimise Your Efforts with our Specialised PrepPacks™ - We Help Over 40,000 Job-Seekers Each Year Get A better Score
All major test provers have at least one abstract reasoning test, which is why practising any generic version of abstract practice is crucial during your preparation.
However, it is best to find out which test provider whose test you will take. Each assessment provider differs from the rest.
For example, while the general information will remain the same, you will find the format, style and questions represented differently. So, to give yourself the best chance of passing try to find out which test provider the employer uses.
Our exclusive PrepPacks™ have been created by experts in the field. Furthermore, We collect feedback from our customers to gain insight into the efficacy of our practise material and to ensure they are up to date.
Need Help
Need Help
Please fill out the form below and we will contact you soon.
Your message was sent. We will contact you shortly.
There was a problem sending your message. Please try again in a few minutes.
| 1,625
| 7,846
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.671875
| 4
|
CC-MAIN-2019-39
|
latest
|
en
| 0.93261
|
https://govtvacancy.net/sarkari-job/what-is-the-time-complexity-of-matrix-chain-multiplication-2
| 1,718,521,646,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861643.92/warc/CC-MAIN-20240616043719-20240616073719-00823.warc.gz
| 239,951,996
| 8,635
|
# What is the time complexity of matrix chain multiplication?
## What is the time complexity of matrix chain multiplication?
Matrix chain multiplication is a classic optimization problem in computer science and is often used in algorithm analysis and dynamic programming. It involves finding the most efficient way to multiply a chain of matrices to minimize the total number of scalar multiplications. In this essay, we will explore the concept of matrix chain multiplication, its time complexity, the dynamic programming approach to solving it, and the underlying principles that make it an essential problem in algorithm design.
1. Introduction to Matrix Chain Multiplication:
Matrix chain multiplication is a problem where we are given a chain of matrices, and we need to determine the most efficient way to multiply them to minimize the total number of scalar multiplications. The order in which we multiply the matrices can significantly affect the overall computational cost.
2. Problem Formulation:
Let's consider a sequence of matrices A_1, A_2, ..., A_n, where the dimensions of the matrices are given by d_i x d_{i+1} for 1 ≤ i < n. The goal is to find the optimal parenthesization of the matrices that minimizes the total number of scalar multiplications required.
For example, suppose we have four matrices with dimensions:
A_1: 10x30 A_2: 30x5 A_3: 5x60 A_4: 60x10
There are several possible ways to parenthesize the matrices, resulting in different numbers of scalar multiplications. The optimal parenthesization is the one that requires the fewest scalar multiplications.
3. Naive Approach:
A naive approach to solving the matrix chain multiplication problem is to consider all possible parenthesizations and compute the number of scalar multiplications for each. However, this approach is inefficient and has an exponential time complexity, as the number of possible parenthesizations grows exponentially with the number of matrices.
4. Dynamic Programming Approach:
To overcome the inefficiency of the naive approach, we can use dynamic programming to find the optimal solution efficiently. The dynamic programming approach breaks the problem into smaller subproblems and stores the results of these subproblems to avoid redundant computations.
The dynamic programming solution to the matrix chain multiplication problem involves the following steps:
Step 1: Define Subproblems:
We define the subproblem as finding the optimal parenthesization for a subchain of matrices within the given chain.
For a given chain of matrices A_i, A_{i+1}, ..., A_j, where 1 ≤ i < j ≤ n, we want to find the optimal k such that the matrices are parenthesized as (A_i)(A_{i+1})...(A_k) and (A_{k+1})(A_{k+2})...(A_j). The objective is to minimize the total number of scalar multiplications for the entire chain.
Step 2: Recurrence Relation:
To solve the subproblem efficiently, we need to find a recurrence relation that relates the optimal solution for the subproblem to the optimal solutions for its smaller subproblems.
Let m[i][j] represent the minimum number of scalar multiplications needed to multiply the matrices A_i to A_j inclusively. Then, we can express the recurrence relation as follows:
m[i][j] = min(m[i][k] + m[k+1][j] + d_i x d_{k+1} x d_{j+1}) for all i ≤ k < j
The recurrence relation states that the minimum number of scalar multiplications needed to multiply the matrices A_i to A_j can be obtained by considering all possible values of k, where i ≤ k < j. The term m[i][k] represents the minimum number of scalar multiplications needed for the subchain A_i to A_k, and m[k+1][j] represents the minimum number of scalar multiplications needed for the subchain A_{k+1} to A_j. The term d_i x d_{k+1} x d_{j+1} represents the number of scalar multiplications needed to compute the resulting matrix of multiplying A_i to A_k with A_{k+1} to A_j.
Step 3: Base Case:
The base case of the dynamic programming solution is when the subchain contains only one matrix (i = j). In this case, no scalar multiplications are needed, so m[i][j] = 0.
Step 4: Filling the Table:
To find the optimal solution for the entire chain of matrices (A_1, A_2, ..., A_n), we need to fill in the dynamic programming table m[][] in a bottom-up manner.
We start by filling in the base cases (i = j), where m[i][j] = 0. Then, we proceed to fill in the table for increasing subchain lengths (j - i). For each subchain length, we compute the values of m[i][j] using the recurrence relation described earlier.
Finally, the value m[1][n] represents the minimum number of scalar multiplications needed to multiply the entire chain of matrices A_1 to A_n.
5. Time Complexity of Dynamic Programming Solution:
The time complexity of the dynamic programming solution to the matrix chain multiplication problem is O(n^3), where n is the number of matrices in the chain.
Explanation of Time Complexity:
The dynamic programming table m[][] has dimensions n x n, representing all possible subchains of the given chain of matrices. Filling in the table requires computing the optimal solution for each subchain, which involves considering all possible values of k for each subchain.
For each subchain length (j - i), there are at most n - 1 possible values of k to consider (since i ≤ k < j). Hence, the number of computations required to fill in each entry m[i][j] is proportional to n - 1. As there are n^2 entries in the dynamic programming table, the overall time complexity is O(n^3).
6. Conclusion:
Matrix chain multiplication is a classic optimization problem in computer science, where the goal is to find the most efficient way to multiply a chain of matrices to minimize the total number of scalar multiplications. The dynamic programming approach offers an efficient solution to this problem, with a time complexity of O(n^3), where n is the number of matrices in the chain.
By breaking the problem into smaller subproblems and using a recurrence relation, dynamic programming allows us to compute the optimal solution for the entire chain in a systematic and efficient manner. The matrix chain multiplication problem is not only important in algorithm design and analysis but also finds practical applications in various fields, such as computer graphics, robotics, and numerical computation.
Thank You
| 1,372
| 6,329
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375
| 4
|
CC-MAIN-2024-26
|
latest
|
en
| 0.903084
|
http://www.slideserve.com/Gabriel/class-26-one-way-analysis-of-variance-with-tukey-hsd-post-test
| 1,475,186,912,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-40/segments/1474738661953.95/warc/CC-MAIN-20160924173741-00026-ip-10-143-35-109.ec2.internal.warc.gz
| 721,265,428
| 20,369
|
This presentation is the property of its rightful owner.
1 / 57
# Oneway ANOVA PowerPoint PPT Presentation
Oneway ANOVA Analysis of variance is used to test for differences among more than two populations. It can be viewed as an extension of the t-test we used for testing two population means.
Oneway ANOVA
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
### Oneway ANOVA
• Analysis of variance is used to test for differences among more than two populations. It can be viewed as an extension of the t-test we used for testing two population means.
• The specific analysis of variance test that we will study is often referred to as the oneway ANOVA. ANOVA is an acronym for ANalysis Of VAriance. The adjective oneway means that there is a single variable that defines group membership (called a factor). Comparisons of means using more than one variable is possible with other kinds of ANOVA analysis.
### Why Not Use Multiple T-tests
• It might seem logical to use multiple t-tests if we wanted to compare a variable for more than two groups. For example, if we had three groups, we might do three t-tests: group 1 versus group 2, group 1 versus group 3, and group 2 versus group3.
• However, doing three hypothesis tests to compare groups changes the probability that we are making an error (the alpha error rate). When conducting multiple tests of significance, the chance of making at least one alpha error over the series of tests is greater than the selected alpha level for each individual test. Thus, if we do multiple t-tests on the same variables with an alpha level of 0.05, the chances that we are making a mistake in applying our findings to the population is actually greater than 0.05.
### Logic of Analysis of Variance
• The logic of the analysis of variance test is the same as the logic for the test of two population means.
• In both tests, we are comparing the differences among group means to a measure of dispersion for the sampling distribution.
• In ANOVA, differences of group means is computed as the difference for each group mean from the mean for all subjects regardless of group. The measure of dispersion for the sampling distribution is a combination of the dispersion within each of the groups.
### Step 1. Assumptions for the Test
• Level of measurement of the group variable can be any level of variable that identifies groups.
• Level of measurement of the test variable is interval.
• The test variable is normally distributed in the population:
• skewness and kurtosis between –1.0 and +1.0, or
• number is each group is greater than 10 (central limit theorem)
• The variances (dispersion) of the groups are equal. The Levene test of equality of population variances is used to test this assumption.
### Levene Test of Homogeneity of Variances
• The Levene test of equality of population variances tests whether or not the variances for the groups are equal. It is a test of the research hypothesis that the variance (variability) of one or more groups is different from the others. The null hypothesis states that the variances of all groups are equal.
• If the probability of the test statistic is greater than 0.05, we do not reject the null hypothesis and conclude that none of the variances are different. This is the desired outcome.
• If the probability of the test statistic is less than or equal to 0.05, we conclude the variances are different. There is no alternative formula adjusting for unequal variances for an ANOVA test, like there is for a t-test. Though we violate the assumption, we will add a caution to any true answers instead of deciding that it is an incorrect application of a statistic because the analysis of variable is robust to violations of the assumption.
### Step 2. Hypotheses and alpha
• The research hypothesis is that the mean of at least one of the population groups is different from the means of the other groups.
• The null hypothesis is that the means of all of the population groups are equal.
• If we don’t have a specific reason for setting the level of significance to a specific probability, we can use the traditional social science benchmark of 0.05. This means that we are willing to risk making a mistake in our decision to reject the null hypothesis if it only happens once in every 20 decisions, or our decision would be correct 19 out of 20 times. The alpha level to use will be stated in the problems.
### Step 3. Sampling distribution and test statistic
• In the ANOVA test, the probability is obtained from the “F” distribution instead of the normal curve distribution.
• The test statistic is also referred to as the F-ratio or F-test because it follows the f-distribution.
### Step 4. Computing the Test Statistic
• Conceptually the test statistic is computed in a way similar to the independent samples t-test. Both are computed by dividing the differences in means by the measure of variability among the groups.
• We identify the probability of the test statistic from the SPSS statistical output.
### Step 5. Decision and Interpretation
• If the probability of the test statistic is less than or equal to the probability of the level of significance (alpha error rate), we reject the null hypothesis and conclude that our data supports the research hypothesis.
• If the probability of the test statistic is greater than the probability of the level of significance (alpha error rate), we fail to reject the null hypothesis and conclude that our data does not support the research hypothesis.
### Interpreting Differences in Population Means
• If we fail to reject the null hypothesis, we can state that we found no differences among the means for the population groups for this characteristic. We do not say they are equal.
• If we reject the null hypothesis, we can conclude that the mean for at least one population group is different from the others.
• The ANOVA test itself does NOT tell us which group means are different. To determine this, we use a Post Hoc test, such as the Tukey HSD (honestly significant differences) Post Hoc Test.
### Post Hoc Test for Difference in Means
• Just as we used a post hoc test to identify which cells in a frequency table were responsible for the statistically significant result, we use a post hoc test to identify the differences in pairs of means that produce a statistically significant result in an ANOVA table.
• We only look at the post hoc test when the probability of the ANOVA statistic causes us to reject the null hypothesis, i.e. the probability of the test statistic is less than the level of significance.
• The Post Hoc Test may NOT reveal differences among group means even when we reject the null hypothesis in the ANOVA test.
### Inflation of Type I Error (Alpha)
• Type I Error: Probability of falsely rejecting null hypothesis when it is true.
• The only time you need to worry about inflation of Type I error rate is when you look for a lot of effects in your data.
• The more effects you look for, the more likely it is that you will turn up an effect that doesn't really exist (Type I error!).
• Doing all possible pair-wise comparisons (t-test) on a one-way ANOVA would increase the overall Type I error rate.
### The Tukey HSD Post Hoc Test
• The Tukey HSD Post Hoc Test compares all possible pairs of group means to determine which differences in group means are statistically different.
• The post hoc test finds the numeric difference between means that is statistically different at the 0.05 and 0.01 levels of significance. It does this without inflating the alpha error rate, which would happen if we used a series of t-tests to identify differences.
• We use the same level of significance of the ANOVA test and the post hoc test.
### ANOVA post hoc Test Practice Problem – 1
This question asks you to use ANOVA to answer whether there is a relationship between country [lcouncu] and age [age] and, if there is, to do a Tukey HSD post hoc test to see if respondents from US were older than respondents from the UK.
A one-way analysis of variance requires that the independent variable specify groups or categories and the dependent variable be interval level. The independent variable [lcouncu] is nominal and the dependent variable [age] is interval, satisfying the requirement for the independent and dependent variables.
### ANOVA post hoc Test in SPSS (1)
Next step is to examine the distribution of the dependent variable. You can check whether the dependent variable is normally distributed or not in:
Analyze > Descriptive Statistics > Descriptives…
### ANOVA post hoc Test in SPSS (2)
After moving [age] into “Variable(s):” box, click “Options…” button to select the distribution statistics.
### ANOVA post hoc Test in SPSS (3)
Select “Kurtosis” and “Skewness” to examine whether [age] is normally distributed or not.
Then, click “Continue” and “OK” buttons.
### ANOVA post hoc Test in SPSS (4)
[Age] satisfied the criteria for a normal distribution. The skewness of the distribution (.590) was between -1.0 and +1.0 and the kurtosis of the distribution (-.150) was between -1.0 and +1.0.
### ANOVA post hoc Test in SPSS (5)
You can conduct ANOVA by clicking:
Analyze > Compare Means > One-Way ANOVA…
### ANOVA post hoc Test in SPSS (6)
Dependent variable [age] goes to “Dependent List:” box and the independent variable [lcouncu] goes to “Factor:” box.
Then, click “Options…” button to select statistics options.
### ANOVA post hoc Test in SPSS (7)
Select “Descriptive” and “Homogeneity of variance test” in the “Statistics” section of “One-Way ANOVA: Options” window.
Then, click “Continue”.
### ANOVA post hoc Test in SPSS (8)
Now, click “Post Hoc…” button to select post hoc test option.
### ANOVA post hoc Test in SPSS (9)
Select “Tukey” in “Equal Variances Assumed” panel.
Enter alpha in the “Significance level:” textbox. It is same as the alpha level (.01) in the problem.
Then, click “Continue” and “OK” buttons.
### ANOVA post hoc Test in SPSS (10)
First of all, you have to check the equal variance assumption. The probability associated with Levene's Test for Equality of Variances (p<0.001) is less than or equal to the level of significance (0.01). The assumption of equal variances is not satisfied.
However, since analysis of variance is robust to violations of this assumption, we will add a caution to any true findings rather than conclude that this is an incorrect application of a statistic.
### ANOVA post hoc Test in SPSS (11)
The probability of the F test statistic (F=32.638) was p<0.001, less than or equal to the alpha level of significance of 0.01. The null hypothesis that the mean "age" [age] is the same for all groups defined by the variable "current country of residence" [lcouncu] is rejected.
The research hypothesis that the mean "age" [age] for groups defined by the variable "current country of residence" [lcouncu] is not the same for all groups is supported by this analysis.
### ANOVA post hoc Test in SPSS (12)
Based on the Tukey post hoc test, the difference in the mean for survey respondents from the United States (40.47) and the mean for survey respondents from the United Kingdom (33.95) was 6.53, a statistically significant difference at the 0.01 level of significance.
Survey respondents from the United States were older than survey respondents from the United Kingdom.
### ANOVA post hoc Test in SPSS (13)
If you identify a statistically significant difference, but aren’t sure which group is older than the other, look back at the table of descriptive statistics.
In this table, it is clear that the mean age for respondents in the United states (40.47) is higher than the mean age for respondents in the United Kingdom (33.95).
Survey respondents from the United States were older than survey respondents from the United Kingdom.
The answer to the question is true with caution, with the caution added for the violation of the assumption of equality of variances.
### ANOVA post hoc Test Practice Problem – 2
This question asks you to use ANOVA to answer whether there is a relationship between country [lcouncu] and web surfing [intera06] and, if there is, to do a Tukey HSD post hoc test to see if respondents from Canada said they surf the web for recreational purposes less frequently than respondents from the UK.
A one-way analysis of variance requires that the independent variable specify groups or categories and the dependent variable be interval level. The independent variable [lcouncu] is nominal and the dependent variable [intera06] is ordinal, satisfying the requirement for the independent and dependent variables.
### ANOVA post hoc Test in SPSS (13)
Next step is to examine the distribution of the dependent variable. You can check whether the dependent variable is normally distributed or not in:
Analyze > Descriptive Statistics > Descriptives…
### ANOVA post hoc Test in SPSS (14)
After moving [intera06] into “Variable(s):” box, click “Options…” button to select the distribution statistics.
### ANOVA post hoc Test in SPSS (15)
Select “Kurtosis” and “Skewness” to examine whether [age] is normally distributed or not.
Then, click “Continue” and “OK” buttons.
### ANOVA post hoc Test in SPSS (16)
“Surf Web for recreational purposes” [intera06] satisfied the criteria for a normal distribution. The skewness of the distribution (-0.687) was between -1.0 and +1.0 and the kurtosis of the distribution (-0.296) was between -1.0 and +1.0.
### ANOVA post hoc Test in SPSS (17)
You can conduct ANOVA by clicking:
Analyze > Compare Means > One-Way ANOVA…
### ANOVA post hoc Test in SPSS (18)
Dependent variable [intera06] goes to “Dependent List:” box and the independent variable [lcouncu] goes to “Factor:” box.
Then, click “Options…” button to select statistics options.
### ANOVA post hoc Test in SPSS (19)
Select “Descriptive” and “Homogeneity of variance test” in the “Statistics” section of “One-Way ANOVA: Options” window.
Then, click “Continue”.
### ANOVA post hoc Test in SPSS (20)
Now, click “Post Hoc…” button to select post hoc test option.
### ANOVA post hoc Test in SPSS (21)
Select “Tukey” in “Equal Variances Assumed” section and make sure the “Significance level:” is same as the alpha level (.01) in the problem.
Then, click “Continue” and “OK” buttons.
### ANOVA post hoc Test in SPSS (22)
First of all, you have to check the equal variance assumption. The probability associated with Levene's Test for Equality of Variances (p = 0.003) is less than or equal to the level of significance (0.01). The assumption of equal variances is not satisfied.
However, since analysis of variance is robust to violations of this assumption, we will add a caution to any true findings rather than conclude that this is an incorrect application of a statistic.
### ANOVA post hoc Test in SPSS (23)
The probability of the F test statistic (F=0.086) was p=.918, larger than the alpha level of significance of 0.01. The null hypothesis that the mean “surf the web for recreational purpose" [intera06] is the same for all groups defined by the variable "current country of residence" [lcouncu] is not rejected.
The research hypothesis that the mean [intera06] for groups defined by the variable [lcouncu] is not the same for all groups is not supported by this analysis.
### ANOVA post hoc Test in SPSS (24)
When the F test statistic is not significant, the results of the post hoc tests are not interpreted, even if statistically significant differences between pairs of groups are found.
The answer to the question is false.
### ANOVA post hoc Test Practice Problem – 3
This question asks you to use ANOVA to answer whether there is a relationship between country [lcouncu] and closeness to community[valatt01] and, if there is, to do a Tukey HSD post hoc test to see if respondents from Canada agreed more strongly that they feel close to other people in their community than respondents from the US.
A one-way analysis of variance requires that the independent variable specify groups or categories and the dependent variable be interval level. The independent variable [lcouncu] is nominal and the dependent variable [valatt01] is ordinal, satisfying the requirement for the independent and dependent variables.
### ANOVA post hoc Test in SPSS (25)
Next step is to examine the distribution of the dependent variable. You can check whether the dependent variable is normally distributed or not in:
Analyze > Descriptive Statistics > Descriptives…
### ANOVA post hoc Test in SPSS (26)
After moving [valatt01] into “Variable(s):” box, click “Options…” button to select the distribution statistics.
### ANOVA post hoc Test in SPSS (27)
Select “Kurtosis” and “Skewness” to examine whether [age] is normally distributed or not.
Then, click “Continue” and “OK” buttons.
### ANOVA post hoc Test in SPSS (28)
“I feel close to other people in my community” [valatt01] satisfied the criteria for a normal distribution. The skewness of the distribution (-0.655) was between -1.0 and +1.0 and the kurtosis of the distribution (-0.835) was between -1.0 and +1.0.
### ANOVA post hoc Test in SPSS (29)
You can conduct ANOVA by clicking:
Analyze > Compare Means > One-Way ANOVA…
### ANOVA post hoc Test in SPSS (30)
Dependent variable [valatt01] goes to “Dependent List:” box and the independent variable [lcouncu] goes to “Factor:” box.
Then, click “Options…” button to select statistics options.
### ANOVA post hoc Test in SPSS (31)
Select “Descriptive” and “Homogeneity of variance test” in the “Statistics” section of “One-Way ANOVA: Options” window.
Then, click “Continue”.
### ANOVA post hoc Test in SPSS (32)
Now, click “Post Hoc…” button to select post hoc test option.
### ANOVA post hoc Test in SPSS (33)
Select “Tukey” in “Equal Variances Assumed” section and make sure the “Significance level:” is same as the alpha level (.01) in the problem.
Then, click “Continue” and “OK” buttons.
### ANOVA post hoc Test in SPSS (34)
First of all, you have to check the equal variance assumption. The probability associated with Levene's Test for Equality of Variances (p<0.001) is less than or equal to the level of significance (0.01). The assumption of equal variances is not satisfied.
However, since analysis of variance is robust to violations of this assumption, we will add a caution to any true findings rather than conclude that this is an incorrect application of a statistic.
### ANOVA post hoc Test in SPSS (35)
The probability of the F test statistic (F=8.754) was p<0.001, less than or equal to the alpha level of significance of 0.01. The null hypothesis that the mean [valatt01] is the same for all groups defined by the variable [lcouncu] is rejected.
The research hypothesis that the mean [valatt01] for groups defined by the variable [lcouncu] is not the same for all groups is supported by this analysis.
### ANOVA post hoc Test in SPSS (36)
Based on the Tukey post hoc test, the difference in the mean for survey respondents from Canada (4.85) and the mean for survey respondents from the United States (4.61) was 0.24. This difference was not statistically significant (p = .094) at the 0.01 level of significance .
The mean "agreement that respondent feels close to other people in their community" for survey respondents from Canada is not different from the mean for survey respondents from the United States.
The answer to the question is false.
### Steps in solving One-Way ANOVA post hoc Test Problems - 1
The following is a guide to the decision process for answering
homework problems about one-way ANOVA post hoc test problems:
Is the dependent variable ordinal or interval level and does independent variable define groups?
Incorrect application of a statistic
No
Yes
Compute the skewness, and kurtosis for the variable to test assumption of normality.
### Steps in solving One-Way ANOVA post hoc Test Problems - 2
Assumption of normality satisfied? (skew, kurtosis between -1.0 and + 1.0)
Yes
No
Sample size 10+ in each group to apply Central Limit Theorem?
No
Incorrect application of a statistic
Yes
Compute the one-way ANOVA with Tukey HSD post hoc option selected
### Steps in solving One-Way ANOVA post hoc Test Problems - 3
Is the p-value for the Levene’s test for equality of variances <= alpha?
No
Yes
Add a caution to any true findings rather than conclude that this is an incorrect application of a statistic.
Is the p-value for the F ratio test <= alpha?
No
False
Yes
### Steps in solving One-Way ANOVA post hoc Test Problems - 4
Examine Tukey HSD post hoc test result
Is the p-value for the Tukey HSD post hoc test <= alpha?
No
False
Yes
Is the dependent variable ordinal level or assumption of equality of variances violated?
No
True
Yes
True with caution
| 4,927
| 21,263
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.84375
| 4
|
CC-MAIN-2016-40
|
latest
|
en
| 0.918669
|
https://plainmath.net/15853/find-value-which-series-converges-sum_-equal-infty-series-those-values
| 1,642,949,823,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320304287.0/warc/CC-MAIN-20220123141754-20220123171754-00233.warc.gz
| 517,252,009
| 8,081
|
# Find the value of x for which the series converges \sum_{n=1}^\infty(x+2)^n Find the sum of the series for those values of x.
tinfoQ 2021-06-02 Answered
Find the value of x for which the series converges
$$\sum_{n=1}^\infty(x+2)^n$$ Find the sum of the series for those values of x.
### Expert Community at Your Service
• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers
### Solve your problem for the price of one coffee
• Available 24/7
• Math expert for every subject
• Pay only if we can solve it
## Expert Answer
d2saint0
Answered 2021-06-03 Author has 28286 answers
Consider the series $$\sum_{n=1}^\infty(x+2)^n$$
Let $$a_n=(x+2)^n,\ a_{n+1}=(x+2)^{n+1}$$
$$\frac{a_{n+1}}{a_n}=\frac{(x+2)^{n+1}}{(x+2)^n}$$
$$=x+2$$
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}(|x+2|)=|x+2|$$
Using ratio test, this will converge, when
$$|x+2|<1$$
$$-1<(x+2)<1$$
$$-3$$
So, the series is converges at the values lies in the interval, (-3,-1)
And $$\sum_{n=1}^\infty(x+2)^n=(x+2)^1+(x+2)^2+(x+2)^3+...+(x+2)^\infty$$
$$=\frac{x+2}{1-(x+2)}$$
$$=\frac{x+2}{-1-x}$$
### Expert Community at Your Service
• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers
### Solve your problem for the price of one coffee
• Available 24/7
• Math expert for every subject
• Pay only if we can solve it
...
| 526
| 1,417
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4
| 4
|
CC-MAIN-2022-05
|
latest
|
en
| 0.692224
|
https://www.jiskha.com/display.cgi?id=1266961262
| 1,516,512,066,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084890187.52/warc/CC-MAIN-20180121040927-20180121060927-00082.warc.gz
| 954,946,789
| 4,471
|
# College Physics
posted by .
a 1200kg car coasts from rest down a driveway that is inclined 20degrees. to the horizontal and is 15m long. how fast is the car going at the end of the delivery if....
(a) friction is negligible
(b) a friction force of 3000n opposes the motion?
dont know where to begin to solve these 2questions. I tried differant formulas, but nothing I tried came out to the answers in the book.
help
• College Physics -
see this very related problem:
http://www.jiskha.com/display.cgi?id=1266960070
• College Physics -
Many if not most problems in physics are not solved by plugging directly into a single equation. It is better to prceed in logical steps using formulas you are familiar with.
In this case, the place to start is with conservation of energy. The vertical drop of the runway is
H = 15 sin 20 = 5.13
The Potential energy loss MgH equals the Kinetic Energy gain, so
MgH = (1/2) M V^2
V = sqrt(2 g H) = 100.3 m/s
(I used 9.81 m/s^2 for g)
Note that you did not need to use the mass for this one.
To do the second case, subtract friction work done
3000N*15 m = 45,000 J from the potential energy decrease, and solve for V again.
M g H - 45,000 = (1/2) M V^2
You will need to use the mass this time.
• College Physics -
Two long, parallel wires separated by 30 cm each carry currents of 4.0 A in a horizontal direction.
B)Find if they are in opposite directions.
## Similar Questions
1. ### Physics
a 3262 kg car starts from rest at the top of a 5m long driveway that is sloped at 20 degrees with the horizontal. If an average friction force of 612N impedes the motion, fing the speed of the car at the bottom of the driveway.
2. ### physics
A 2.10 103 kg car starts from rest at the top of a 5.8 m long driveway that is sloped at 25° with the horizontal. If an average friction force of 4.0 103 N impedes the motion, find the speed of the car at the bottom of the driveway.
3. ### physics
A 2.1 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an average friction force of 4.0 103 N impedes the motion, find the speed of the car at the bottom of the driveway. …
4. ### physics
A 2.1 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an average friction force of 4.0 103 N impedes the motion, find the speed of the car at the bottom of the driveway.
5. ### physics
A 2100 car starts from rest at the top of a 5.0m long driveway that is sloped at 20 degrees with the horizontal. If an average friction force of 4000N impedes the motion find the speed of the car at the bottom of the driveway.
6. ### Physics
A 4.2 × 10^3 kg car accelerates from rest at the top of a driveway that is sloped at an angle of 18.4◦ with the horizontal. An average frictional force of 4.1×10^3 N impedes the car’s motion so that the car’s speed at the …
7. ### physics
A 1,600 kg car coasts down a hill starting from rest and decreases its height by 22 m over a distance of 90 m of travel. How fast will it be going at the end of the hill, if friction has no effect in slowing its motion?
8. ### physics
A 2.1 multiplied by 103-kg car starts from rest at the top of a 5.8-m-long driveway that is inclined at 22° with the horizontal. If an average friction force of 4.0 multiplied by 103 N impedes the motion, find the speed of the car …
9. ### physics
A 1500 kg car at rest rolls down a hill inclined at 10 degrees with the horizontal. A 400 N friction force opposes its motion on the inclined plane and a 600 N on the horizontal surface a) how fast will it be moving at point B b) how …
10. ### trig
A car is on a driveway that is inclined 12 degrees to the horizontal. A force of 470lb is required to keep the car from rolling down the driveway. a) Find the weight of the car b) Find the force the car exerts against the driveway
More Similar Questions
| 1,059
| 3,912
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.640625
| 4
|
CC-MAIN-2018-05
|
latest
|
en
| 0.921053
|
https://www.coursehero.com/file/29759245/ConstMathdoc/
| 1,547,601,206,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-04/segments/1547583656577.40/warc/CC-MAIN-20190116011131-20190116033131-00421.warc.gz
| 752,403,465
| 129,152
|
ENC
ConstMath.doc
# ConstMath.doc - Career Technical Mathematics Academic...
• Notes
• 16
This preview shows pages 1–3. Sign up to view the full content.
Career Technical Mathematics Academic Alignment Construction Trades Grading on performance-based assessments is based on the quality and quantity of the work students do in class. Often students are required to do the task until they are able to complete it accurately (mastery learning). If a standard is covered partially, then the part that is covered is underlined. Teacher: Jack Decker Date: Academic Year 2009 – 2010 Construction Trades - HSCE Unit of Study: ON SITE Time Allocated: (180 days) CTE Mathematics Application Mathematics Standards Addressed Math Assessment 1. Students convert inches to feet, feet to inches, feet to yards, yards to feet, cubic feet to cubic yards. 2. Students round or truncate units in order to buy appropriate quantities. 3. Students carry units through inches to feet and the through feet to yards. L2.3.1 Convert units of measurement within and between systems: explain how arithmetic operations on measurements affect units, and carry units through calculations correctly. Performance based: Students demonstrate proficiency on-site in the following ways: 1. students calculate quantity of concrete in pours, use tape measures to accurately measure for needed building material. 2. students buy appropriate amounts of material based on calculations 1. Students measure framing accuracy to an 1/8”, finish trim accuracy to 1/16” and drywall accuracy to ¼” 2. When marking measurements for cutting L2.4.1 Determine what degree of accuracy is reasonable for measurements in a given situation ; express accuracy through use of significant digits, error tolerance, or percent of error; describe how errors in measurements are magnified by computation; recognize accumulated error in applied situations. Students demonstrate proficiency on-site. Daily in-class and on-site. Clare-Gladwin RESD March 31, 2018 1
This preview has intentionally blurred sections. Sign up to view the full version.
Career Technical Mathematics Academic Alignment Construction Trades - HSCE CTE Mathematics Application Mathematics Standards Addressed Math Assessment Students measure and cut corresponding angles on a knee- wall or interior gable to floor wall. Students know that a plumb wall is perpendicular to both the floor and ceiling and that all angel types are congruent. G1.1.2 Solve multi –step problems and construct multi- step problems and construct proofs involving corresponding angles, alternate interior angles, alternate exterior angels and same side (consecutive) interior angels. Students demonstrate competency with framing squares and speed squares Various aspects of this concept are applied everyday for 100 days. 1. Student use complimentary angle relationships for cutting siding, cedar trim and OSB. 2. Students use area and perimeter for estimating siding quantities. 3. Students use the Pythagorean Theorem to calculate missing side lengths of gable ends.
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern
| 825
| 4,095
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.65625
| 4
|
CC-MAIN-2019-04
|
latest
|
en
| 0.912819
|
http://math.stackexchange.com/questions/53150/i-heard-that-0-0-is-undetermined-why-is-infty-infty-also-undetermined
| 1,467,113,528,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-26/segments/1466783396875.58/warc/CC-MAIN-20160624154956-00156-ip-10-164-35-72.ec2.internal.warc.gz
| 197,436,367
| 22,813
|
# I heard that $0/0$ is undetermined, why is $\infty/\infty$ also undetermined?
I heard that $0/0$ is undetermined, but why is $\infty/\infty$ also undetermined? Also what is $\lim\limits_{x \to \infty} \frac{x}{x}$ and what is $\lim\limits_{x \to \infty} \frac{2^x}{2^x}$?
-
Well, contrast the limiting behaviors of $\exp(x)/x$ and $x/\exp(x)$ for instance... – J. M. Jul 22 '11 at 18:42
@J.M.-what's so special about the contrasting? – abcde Jul 22 '11 at 18:45
Graph both and observe what happens to them as you go rightward... – J. M. Jul 22 '11 at 18:49
Hmm. Didn't you learn just yesterday how to prove that one of them is zero? The same argument shows that the other one is infinity. Since you seem to like l'Hôpital, use that one! – t.b. Jul 22 '11 at 18:51
@abcde: Nothing wrong with using a calculator to get some numerical insight. – André Nicolas Jul 23 '11 at 1:49
A short and non rigorous explanation: in a fraction $\dfrac{P(x)}{Q(x)}$ (assume that both $P(x)$ and $Q(x)$ are positive) it may happen that $P(x)$ goes to infinity "faster", "slower" or "as fast as" $Q(x)$. Depending on these speeds the fraction tends to infinity, zero or a constant, or it may happen that the limit does not exist, as commented by Samuel. To find the limit one can apply L'Hôpital's rule, if $P(x)$ and $Q(x)$ are differentiable.
Both $\lim\limits_{x \to \infty} \frac{x}{x}$ and $\lim\limits_{x \to \infty} \frac{2^x}{2^x}$ is $1$.
-
To add to this: It may also happen that the limit does not exist. – Samuel Jul 23 '11 at 21:22
@Samuel: That's right. I will add this possibility to the answer. – Américo Tavares Jul 23 '11 at 21:42
Your question is completely reasonable, given the unfortunate survival of inappropriate notation that has confused many generations of students.
Here I am referring to phrases of the kind "indeterminate form of the type $\infty/\infty$," which one sees even in very good calculus books.
This notation encourages the notion that there is such a thing as the "number" infinity, that it makes sense to divide this "number" by things, including "$\infty$," but that somehow the result of this so-called division can be various things. Sadly, we also meet "$\cdot \infty$," and "$\infty-\infty$."
After a while, most students either find out what's really going on, or else at least understand that they must, for some unknown reason, use special rules, and manage to get by.
The reality is really quite simple. We have two functions, $f(x)$ and $g(x)$ such that when $x$ gets very large, each of $f(x)$ and $g(x)$ both get very large. The question is: What happens to $$\frac{f(x)}{g(x)}$$ as $x$ gets very large?
The answer is: It depends. Here are some examples.
1. Let $f(x)=x^2+17x$, and $g(x)=x^3+1$. Each gets very large when $x$ gets large. But it is fairly easy to see that after a while, $x^3+1$ is much larger than $x^2+17x$, so after a while the ratio is close to $0$.
2. Let $f(x)=x^2+17x$, and $g(x)=5x^2 -11$. When $x$ is large, the behaviour of $f(x)$ is dominated by the $x^2$ term, and the behaviour of $g(x)$ is dominated by the $5x^2$ term. So the ratio $f(x)/g(x)$ should be more or less $x^2/5x^2$ when $x$ is large. One can informally verify with a calculator that it indeed looks as if $f(x)/g(x)$ is close to $1/5$ when $x$ is large.
3. Let $f(x)=e^x$, and $g(x)=x^{10}$. Here things are not obvious. By remembering that, in the long run, $e^x$ grows faster than any polynomial, we can expect that $f(x)/g(x)$ gets very large as $x$ gets large. But maybe in this case we should use some reliable machinery to find out what happens.
4. There can be more complicated behaviour, with $f(x)$ being way ahead for a while, then falling way behind, then racing ahead, and so on. It takes some work to come up with examples of this kind of behaviour, but it can happen.
No "$\infty/\infty$" anywhere! All we have been looking at is questions about the long-term behaviour of certain ratios of functions. Fairly often, we need information about such ratios. In many cases, intuition is not sufficiently accurate to provide an answer, so we need special tools. In the case of your two questions, we need no tools, since in each case the ratio is $1$ for all $x$.
The term $\infty/\infty$ should be viewed merely as a label that calculus books to questions of the types discussed above. It is a terribly unfortunate label, since it invites misunderstanding.
One can hope that the label $\infty/\infty$, and its brethren, such as $0/0$, will someday disappear from calculus books. But we should not hold our breath: they have been around since the early eighteenth century!
-
HINT $\rm\ \ f/g\$ of form $\:0/0$ $\rm\ \Rightarrow\ (f/g)^2 =\: (1/g^2)/(1/f^{\:2})\$ of form $\:\infty/\infty$
-
Actually, $\frac{\infty}{\infty } = \frac{5}{2}$. Don't believe me? Look. Take these three facts:
(1) $\lim_{x\to\infty} 5x+2 = \infty$
(2) $\lim_{x\to\infty} 2x = \infty$
(3) $\lim_{x\to\infty} \frac{5x+2}{2x} = \frac{5}{2}$
Now, divide (1) by (2) and compare with (3) to conclude $\frac{\infty}{\infty } = \frac{5}{2}$.
-
Your question has some issues, since it depends on what you mean by $\frac 0 0$ and $\frac \infty \infty$.
Either you're asking why $\frac 0 0$ is not defined and why $\frac \infty \infty$ is not defined. The latter one would be a meaningless statement in $\mathbf R$ since $\infty$ does not denote a real number. Generally it is either an indicator that a limit diverges (in which case an equality statement is something of an abuse of notation) or an element of an extension of $\mathbf R$ where $\infty$ is an "added" element greater than any other. There are some issues with doing calculus in this structure though (and I would hazard a guess that this is not what you are trying to do).
The former expression being meaningless is due to $0$ being identity of addition of what is known as a field. Quickly the reason for this is that division $\frac a b$ is then defined as $a \cdot b^{-1}$, where $b^{-1}$ is the (unique) element ssuch that $b \cdot b^{-1} = 1$, but clearly there cannot exist such a $0^{-1}$ since $0 \cdot a = 0 \not = 1$ (this is generally true for fields, and especially for $\mathbf R$) for any $a$.
My other interpretation is why there is no general value for $\lim \limits_{x \to a} \frac {f (x)} {g (x)}$ for $f, g$ such that either (corresponding to the first statement) $\lim \limits_{x \to a} \; f(x) = 0$ and $\lim \limits_{x \to a} \; g(x) = 0$ or (corresponding to the second statment) $\lim \limits_{x \to a} \; f(x) = \infty$ and $\lim \limits_{x \to a} \; g(x) = \infty$. In this case it's just as the other answers have suggested, that different functions can tend differently "fast" to their limits (or their lack thereof).
As for your example functions $\frac x x$ and $\frac {2^x} {2^x}$ both are equal to the constant function $f(x) = 1$, so their limits (when x tends to any value) is just 1.
-
| 2,058
| 6,925
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.859375
| 4
|
CC-MAIN-2016-26
|
longest
|
en
| 0.911433
|
https://www.atulocal1309.com/faq/often-asked-how-far-can-you-see-from-the-beach.html
| 1,652,860,910,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662521152.22/warc/CC-MAIN-20220518052503-20220518082503-00519.warc.gz
| 755,596,967
| 11,802
|
## Can you see 100 miles away?
The Earth is round and you cannot see 100 miles out to sea from sea level. If you stand at the water’s edge and your eyes are 180 cm from the ground the horizon is about 4.8km or nearly 3 miles away. For the horizon to be 100 miles away your eyes would have to be 1878 metres above the ground.
## How far away can you see a ship at sea?
Assuming a clear day, you can actually see another ship breach the horizon about 12 miles away (19.3 kilometres or 10.4 nautical miles).
## How far can you see before the curve of the earth?
Your visual curvature is limited to about 5KM (3.3miles) before the object you see on Earth will start appearing below the horizon.
You might be interested: FAQ: How many devices can you connect to netflix?
## How far can you see from 1 mile up?
Earth’s curvature
The Earth curves about 8 inches per mile. As a result, on a flat surface with your eyes 5 feet or so off the ground, the farthest edge that you can see is about 3 miles away.
## How far can a human walk in a day?
While your body is made for walking, the distance you can achieve at an average walking pace of 3.1 miles per hour depends on whether you have trained for it or not. A trained walker can walk a 26.2-mile marathon in eight hours or less, or walk 20 to 30 miles in a day.
## How far can the human eye see a light?
Considering the absolute threshold, the brightness of a candle flame, and the way a glowing object dims according to the square of the distance away from it, vision scientists conclude that one could make out the faint glimmer of a candle flame up to 30 miles away.
## How far out to sea can you see land?
For an observer standing on the ground with h = 2 metres (6 ft 7 in), the horizon is at a distance of 5 kilometres (3.1 mi). For an observer standing on a hill or tower 30 metres (98 ft) above sea level, the horizon is at a distance of 19.6 kilometres (12.2 mi).
## Is the Ocean Flat?
Most people are surprised to learn that, just as the surface of the Earth is not flat, the surface of the ocean is not flat, and that the surface of the sea changes at different rates around the globe. For instance, the absolute water level height is higher along the West Coast of the United States than the East Coast.
You might be interested: How can i raise my hemoglobin level?
## How far can the human eye see on a mountain?
Often, the curvature of the Earth gets in the way first. For example, at sea level, the horizon is only 4.8 kilometres (2.9 miles) away. On the top of Mt Everest, you could theoretically see for 339 kilometres (211 miles), but in practice clouds get in the way. For a truly unobstructed view, though, look up.
## Can you see earth curvature from Everest?
Even while on top of Mount Everest, it is impossible to see the curvature. Studies place the threshold to see the curvature at 35,000 feet, but even at this height, one must have at least a 60° angle of vision.
## Where in the world can you see the farthest?
The furthest photographed sightline in the world is 443 km, from Pic de Finestrelles in the Spanish Pyrenees to Pic Gaspard in the French Alps, almost 100x further than what can be seen driving along the prairies and staring at the horizon.
## Is it possible to see the curvature of the earth?
It is possible to see the effect of the curvature of the earth from sea level (ships disappearing over the horizon). To actually see the curvature directly (ie to be able to see the horizon as an arc, rather than a straight line) requires an altitude of over 50,000 feet.
## How far can a 6 foot man see at sea level?
For a six-foot tall person, the horizon is a little more than 3 miles (5 km) away. Geometry tells us that the distance of the horizon – i.e. the farthest point the eye can see before Earth curves out beneath our view – depends simply on the height of the observer.
You might be interested: Readers ask: How old are kittens when they can leave their mother?
## How far away can you see a plane?
So, how far can you see from a plane? Here is the theoretical distance you can see from a plane at different altitudes, given perfect weather conditions: at 1000 feet: 38.7 Miles (62km) at 5,000 feet: 86.6 Miles (140 km)
## How far can you see at 400 feet?
If you are on a light house at 200 feet above the sea level the horizon is at a distance of 17.3 statute miles.
Tabel: Distance to Horizon (height in feet and distance in statute miles)
Height [feet] Distance Horizon [miles]
200 17,3
300 21,2
400 24,5
500 27,4
| 1,118
| 4,545
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.90625
| 4
|
CC-MAIN-2022-21
|
latest
|
en
| 0.920294
|
http://perplexus.info/show.php?pid=15&op=sol
| 1,553,182,992,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-13/segments/1552912202526.24/warc/CC-MAIN-20190321152638-20190321174638-00408.warc.gz
| 151,026,733
| 4,838
|
All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Move the 2 - double the number (Posted on 2002-04-19)
A certain number ends with the digit 2. Moving the 2 from the end of the number to its front doubles it. Can you find this number?
(Hint: it's quite large)
Submitted by levik Rating: 4.1111 (9 votes) Solution: (Hide) The number is 105263157894736842. Hey, I said it was quite large, didn't I? Start by realizing that the second to last digit of the original number is 4. (This is because the new number's last digit has to be double the old last digit which was two. And the new last digit is the original next to last digit - everything gets shifted when you move the 2 to the front). Next write down:``` ..42 + ..42 = ...4``` (This is because you are doubling the first number to get the new one.) It will become obvious that the second to last digit of the new number (and third to last of the new one) is 8: ``` ...842 + ...842 = ....84``` Note that for the next step, 8+8 will be 16, and so we will use the 6 as the third to last digit for the new number, but remember to carry over the 10 for the next time, when 6+6+1 will yield 13: ``` ...6842 + ...6842 = ...3684``` If you keep this up long enough (don't get discouraged, but be careful in your arithmetics) you will eventually get a valid equasion: ``` 105263157894736842 + 105263157894736842 = 210526315789473684```
Subject Author Date The number Math Man 2015-11-08 19:30:18 re: The continued fraction way TheKPAXian 2009-10-06 19:48:57 answer K Sengupta 2007-06-28 23:03:30 Infinitely many Michael Cottle 2004-12-31 23:13:01 The continued fraction way Federico Kereki 2003-12-04 09:49:37 Another method DJ 2003-06-25 16:07:43 another method Charlie 2003-06-04 04:01:16 another approach steve 2003-05-22 03:09:30 My Answer Tom 2002-05-09 12:15:01 Typo in your solution FingLao 2002-05-09 09:43:32
Search: Search body:
Forums (1)
| 593
| 1,955
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.03125
| 4
|
CC-MAIN-2019-13
|
latest
|
en
| 0.889927
|
https://socratic.org/questions/what-s-the-ratio-of-16-cm-over-3-6-cm
| 1,597,266,903,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439738944.95/warc/CC-MAIN-20200812200445-20200812230445-00257.warc.gz
| 498,769,251
| 5,844
|
# What's the ratio of 16 cm over 3.6 cm?
Jun 26, 2015
The ratio of 16 cm over 3.6 cm is $40 : 9$ or $\frac{40}{9}$
#### Explanation:
$\frac{16}{3.6} = \frac{160}{36} = \frac{40}{9}$
Ratios can be expressed in the form:
$\textcolor{w h i t e}{\text{XXXX}}$$a : b$
or equivalently as a fraction:
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{a}{b}$
| 142
| 348
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875
| 4
|
CC-MAIN-2020-34
|
latest
|
en
| 0.623689
|
https://www.studypool.com/discuss/234992/math-problem-distance-rate-time?free
| 1,508,806,796,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187827662.87/warc/CC-MAIN-20171023235958-20171024015958-00337.warc.gz
| 992,305,258
| 14,562
|
Time remaining:
##### Math Problem distance rate time
label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Timothy gets a phone call from his friends. They say they will be at the library in 25 minutes and ask him to meet them there. The library is 5/8 of a mile away. Timothy walks at a steady pace of 2 1/2 miles per hour. Will he get there first.
Oct 23rd, 2017
2.5 miles per hour = 1/24 miles per minute
Speed x Time = Distance; thus, Time = Distance / Speed
Therefore: Time = (5/8 miles) / (1/24 miles per minute) = 15 minutes
Timothy will be at the library in 15 minutes, so yes, he will get there first (10 minutes before his friends, to be exact)
Sep 17th, 2014
...
Oct 23rd, 2017
...
Oct 23rd, 2017
Oct 24th, 2017
check_circle
| 232
| 777
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.703125
| 4
|
CC-MAIN-2017-43
|
latest
|
en
| 0.893901
|
https://blog.xkcd.com/2009/09/02/urinal-protocol-vulnerability/?like_comment=21530&_wpnonce=fce5d9edd8
| 1,621,324,340,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00182.warc.gz
| 169,220,951
| 21,264
|
# Urinal protocol vulnerability
When a guy goes into the bathroom, which urinal does he pick? Most guys are familiar with the International Choice of Urinal Protocol. It’s discussed at length elsewhere, but the basic premise is that the first guy picks an end urinal, and every subsequent guy chooses the urinal which puts him furthest from anyone else peeing. At least one buffer urinal is required between any two guys or Awkwardness ensues.
Let’s take a look at the efficiency of this protocol at slotting everyone into acceptable urinals. For some numbers of urinals, this protocol leads to efficient placement. If there are five urinals, they fill up like this:
The first two guys take the end and the third guy takes the middle one. At this point, the urinals are jammed — no further guys can pee without Awkwardness. But it’s pretty efficient; over 50% of the urinals are used.
On the other hand, if there are seven urinals, they don’t fill up so efficiently:
There should be room for four guys to pee without Awkwardness, but because the third guy followed the protocol and chose the middle urinal, there are no options left for the fourth guy (he presumably pees in a stall or the sink).
For eight urinals, the protocol works better:
So a row of eight urinals has a better packing efficiency than a row of seven, and a row of five is better than either.
This leads us to a question: what is the general formula for the number of guys who will fill in N urinals if they all come in one at a time and follow the urinal protocol? One could write a simple recursive program to solve it, placing one guy at a time, but there’s also a closed-form expression. If f(n) is the number of guys who can use n urinals, f(n) for n>2 is given by:
The protocol is vulnerable to producing inefficient results for some urinal counts. Some numbers of urinals encourage efficient packing, and others encourage sparse packing. If you graph the packing efficiency (f(n)/n), you get this:
This means that some large numbers of urinals will pack efficiently (50%) and some inefficiently (33%). The ‘best’ number of urinals, corresponding to the peaks of the graph, are of the form:
The worst, on the other hand, are given by:
So, if you want people to pack efficiently into your urinals, there should be 3, 5, 9, 17, or 33 of them, and if you want to take advantage of the protocol to maximize awkwardness, there should be 4, 7, 13, or 25 of them.
These calculations suggest a few other hacks. Guys: if you enter a bathroom with an awkward number of vacant urinals in a row, rather than taking one of the end ones, you can take one a third of the way down the line. This will break the awkward row into two optimal rows, turning a worst-case scenario into a best-case one. On the other hand, say you want to create awkwardness. If the bathroom has an unawkward number of urinals, you can pick one a third of the way in, transforming an optimal row into two awkward rows.
And, of course, if you want to make things really awkward, I suggest printing out this article and trying to explain it to the guy peeing next to you.
Discussion question: This is obviously a male-specific issue. Can you think of any female-specific experiences that could benefit from some mathematical analysis, experiences which — being a dude — I might be unfamiliar with? Alignments of periods with sequences of holidays? The patterns to those playground clapping rhymes? Whatever it is that goes on at slumber parties? Post your suggestions in the comments!
Edit: The protocol may not be international, but I’m calling it that anyway for acronym reasons.
## 1,135 replies on “Urinal protocol vulnerability”
1. will talk says:
Now that people have raised the question, I’m not even sure that I interpreted the scene this way when I was watching it. chin
Like
It is our intention to drive the first Electric Van for approximately 90 days and discover all of the nuances that might come with an EV or if there are any design changes we would like to have included with future vans,” said Robichaud. Closet Door Ideas
Like
I love it! Maybe you can do one about periods that are supposed to come at regular intervals, but instead always fall on the worst possible dates. Job interview? Enjoy trying to sit comfortably with that going on. Pool party? Hope you like tanning only. Camping trip? There are no sanitary disposal units in the wild, enjoy packing that shit out with you.
Like
4. are some says:
Surely agree with simply what you said. Your explanation was certainly the simplest to comprehend. I Know Seo
Like
5. Curently says:
However taxation are not measured that way. The first 25K of earnings would have been 2013 tax supports at the reduced 15% segment first, thus producing a much reduced determine than what you display.I am not disagreeing that Shrub does not have reduced taxation. He certainly does. tweeting from Russia
Like
6. ares someassssw says:
The oyster mixture (\$15) contains eight completely fresh-shucked oysters diving in a dish of hot red mixture marinade. great info
Like
7. Curently says:
From the experiences of his career encounter, I think this is probably one of these non-compliant discarded steel traders that FBI is making reference to in its evaluation. in order to follow Stephen
Like
8. ares someassssw says:
According to the quite a number of HVACR experts that be current at the various market conventions every year, it would seem so. which makes his service unique
Like
9. ares someassssw says:
I was wanting to know your situation; many of us have developed some nice methods and we are looking to trade methods with others, why not shoot me an e-mail if interested. for another blog
Like
10. Curently says:
Exhibitors must be more inspired to advertise the Show to their focus on viewers and offer them with rewards for coming to their unit. The display organizer’s job is to get individuals to the Show, not to a particular company’s unit. taking the advice
Like
11. ares someassssw says:
Property components are a naturally and considerably utilized claim on the value of property solutions because some of the payments for those solutions must be used for devaluation, investing on advanced goods. Susan McGalla
Like
12. Curently says:
Often it is the amazing deficiency of way of lifestyle of someone near that provides this home; and then the regularity of medical center trips and memorials gradually begins to select up amount, like a drumbeat in the forests. real estate photos
Like
13. ares someassssw says:
Property components are a naturally and considerably utilized claim on the value of property solutions because some of the payments for those solutions must be used for devaluation, investing on advanced goods. from another one of these sites
Like
14. ares someassssw says:
It’s discussed at length elsewhere, but the basic premise is that the first guy picks an end urinal, and every subsequent guy chooses the urinal which puts him furthest from anyone else peeing. Nonton Film Gratis
Like
15. ares someassssw says:
I also have GeoIP information; if you’d like to do geocorrelation of some kind, I’ll be providing a version of the data with basic region-level lat/long information (limited to protect privacy) sometime in the next few days. Note: The ColorDB data is the main survey. park office
Like
16. ares someassssw says:
So they get to foul our stuff not the other way around. It’s sort of a version of the George Carlin line: “how come other people’s stuff is shit, and your shit is stuff. try this web-site
Like
17. ares someassssw says:
Now I am found which I actually want. I check your blog everyday and try to learn something from your blog. Thank you and waiting for your new post. one of those members
Like
18. ares someassssw says:
The website loading speed is amazing. It kind of feels that you’re doing any distinctive trick. Moreover, The contents are masterpiece. Stansberry & Associates
Like
19. Curently says:
We also discovered about Viega’s MegaPress System and each one of us pushed black tube. We also used a tube threads machine, so we could comprehend the different processes available, and how long engaged when an experienced is on a job, threads tube. We were helped by Jerr McKinnon, manager of coaching and tech assistance group, and Andrew Richards, item instructor, when using the tools and operating on set ups. after reading that report
Like
20. Curently says:
So while we are assured the highly effective company activities that offer actual value are here to stay, the second part of the question is “Why are some reveals like AHR Expo doing so well while others fight.” Here are some of our results based on conversations with a wide range of display managers, members and members. info here
Like
21. williama says:
Automobiles can be ballasted with front side and rear loads to distribute 60 percent of the body weight on the returning axle for a 60:40 rate, or the other way around for improved taking power. mary kom movie
Like
22. ares someassssw says:
Whether you might be building a fresh pool and also want any cooler as compared to concrete, non-slip surface area or remodeling a preexisting pool to be able to freshen the particular image, the Sundek Vintage Texture Overlay System is a superb solution. important link
Like
23. ares someassssw says:
Compared to concrete, non-slip surface area or remodeling a preexisting pool to be able to freshen the particular image, the Sundek Vintage Texture Overlay System is a superb solution. https://savingandbuycoupons.wordpress.com/
Like
24. technology says:
Di dalam artikel kali ini kita akan mengupas tuntas tentang kanker. Penyebab terkena sakit kanker, gejala-gejala kanker paru-paru serta tahap pengobatan yang paling aman dalam mengobati penyakit yang paling mematikan ini. try this web-site
Like
25. jani says:
Now, what outstanding is it if you use a conventional image and individuals don’t see your encounter next to your comments. corporate media training nyc
Like
26. Marcos Chicote says:
Hi
I recently come across a counterexample for this. The problem relays on the International Choice of Urinal Protocol not taking into consideration the distance from the urinals to the toilets (where you poop).
If the urinals and the toilets are facing each other, people pooping might alter your choice for the urinal. Even more, if someone there’s just one person in the urinal (and he follows the ICUP) and one person pooping (on the oposite corner), the next person that goes into the bathroom to might not choose the most far urinal as he would be closer to the guy pooping.
I have a fix for this bug but it is too large to fit on this textarea.
Like
27. I remember a situation… there were 4 urinals. I come in, there’s already a guy. So I take the last possible urinal. Then another guy comes in and goes directly between us. Nobody spoke. sometimes I happen to need to pee at the same time as another male family member. We go to the toilet separately but if there is an awkward situation already, then it is OK to use two urinals next to each other. I wonder what this does to the math.
Like
28. You have great blogging skills i must say, i am quite inspired from your blogging skills
Like
29. robert says:
And then from any delivery provide shop you can pick-up dual on the edges awkward tape to position around the edges of the mattress. Yet one more way to mistake and observe bed bug visitors. her explanation
Like
30. sultankhan1212 says:
On the other hand, say you want to create awkwardness. If the bathroom has an unawkward number of urinals, you can pick one a third of the way in, transforming an optimal row into two awkward rows. Alpha Fuel XT Trial
Like
31. robert says:
You might discover several experts on the internet that would help you to complete the access and display fix in a simple manner. These experts would assess your house and figure out the most essential access and display to begin the improvement or fix process with. overview
Like
32. sultankhan1212 says:
It’s discussed at length elsewhere, but the basic premise is that the first guy picks an end urinal, and every subsequent guy chooses the urinal which puts him furthest from anyone else peeing. 4 piece grinder
Like
| 2,796
| 12,361
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.09375
| 4
|
CC-MAIN-2021-21
|
latest
|
en
| 0.905448
|
https://www.askiitians.com/forums/Magical-Mathematics%5BInteresting-Approach%5D/29/39741/logarithm.htm
| 1,712,976,753,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816535.76/warc/CC-MAIN-20240413021024-20240413051024-00776.warc.gz
| 601,938,799
| 42,845
|
# Log1/8cosec2 π/8 sin2 3π/8 =?
APURV GOEL
39 Points
12 years ago
Log1/8cosec2 π/8 sin2 3π/8 = Log 1/(2√2 sinΠ/8) sin3Π/8
= Log 1/(2√2 sinΠ/8) cosΠ/8
we have sinΠ/4 = 2.sinΠ/8.cosΠ/8
so sinΠ/8 = (sinΠ/4)/2.cosΠ/8
so 1/(2√2 sinΠ/8) = cosΠ/8
hence Log 1/(2√2 sinΠ/8) cosΠ/8 = LogcosΠ/8cosΠ/8 = 1
| 177
| 302
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.78125
| 4
|
CC-MAIN-2024-18
|
latest
|
en
| 0.40078
|
https://www.physicsforums.com/threads/derivitives-help.534725/
| 1,527,123,980,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794865863.76/warc/CC-MAIN-20180523235059-20180524015059-00535.warc.gz
| 808,536,757
| 15,089
|
# Homework Help: Derivitives Help!
1. Sep 28, 2011
### Gabriel1234
1. The problem statement, all variables and given/known data
A projectile is fired with initial speed Vo at an angle Θ above the horizontal over flat ground.
a. show that the distance d that the projectile goes and height h reached by the projectile are given by:
d = (Vo^2 sin2Θ)/g
h = (Vo^2 sin^2Θ)/2g
2. Relevant equations
X = VT
Vx = V(cosΘ)
Vy = V(sinΘ)
T = d / Vo(sinΘ)
3. The attempt at a solution
d = (Vo^2 sin2Θ)/g
Vt = (Vo^2 sin2Θ)/g
(Vocos(Θ))(d / Vosin(Θ)) = (Vo^2 sin2Θ)/g
dcot(Θ) = (Vo^2 sin(2Θ)/g
dcot(Θ) = (Vo^2 * 2sin(Θ)cos(Θ))/g
Then this is where I get lost.
If I do divide cotangent of theta then would distance be proved.
I haven't started on height yet.
2. Sep 29, 2011
### grzz
The last equation is not correct.
3. Sep 29, 2011
### HallsofIvy
The only acceleration is -g in the y direction, the initial position is (0,0), and the initial velocity is $v_x= V_0 cos(\theta)$, $v_y= V_0 sin(\theta)$:
$a_x= 0$, $a_y= -g$
$v_x= V_0 cos(\theta)$, $v_y= -gt+ V_0 sin(\theta)$
$x= V_0 cos(\theta)t+ 0$, $-(g/2)t^2+ V_0 sin(\theta)+ 0$.
Since the problem asks for the x and y distances separately, I see no reason to combine x and y and so no reason to look at $tan(\theta)$ or $cot(\theta)$. The projectile is NOT moving in a straight line so there is no need to look for a "slope".
(y will be largest when $v_y= 0$. x will be largest when y= 0 again.)
4. Sep 29, 2011
### GrantB
Yes. Think about it like this: What is the slope of the parabola at its maximum?
It appears as though you are just trying to throw equations together, and work backwards, to get the answer. Take a minute and think about the problem.
| 572
| 1,725
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.9375
| 4
|
CC-MAIN-2018-22
|
longest
|
en
| 0.830214
|
https://community.boredofstudies.org/threads/vector-space-pls-help.366858/
| 1,624,051,064,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623487641593.43/warc/CC-MAIN-20210618200114-20210618230114-00067.warc.gz
| 176,359,251
| 17,113
|
# Vector space pls help (1 Viewer)
#### hayabusaboston
##### Well-Known Member
https://imgur.com/a/J5ZRs
2 c) and all of three have me completely lost pls advise what to do
Also is 2 b) a subspace? I said it was after some working
#### hayabusaboston
##### Well-Known Member
Imgur.com/a/4za4z
#### Drongoski
##### Well-Known Member
If you have even heard of Calabi-Yau Manifold, then vector spaces should be a piece of cake.
By the way - can't see the question, so unable to help.
#### hayabusaboston
##### Well-Known Member
help pls someone :'(
Heeeeeeelep
aw
#### He-Mann
##### Vexed?
2 b) is a subspace.
What do you need to prove for 2 c) to be a subspace?
#### hayabusaboston
##### Well-Known Member
2 b) is a subspace.
What do you need to prove for 2 c) to be a subspace?
I got 2) but I cant get 3) now :/
#### Drongoski
##### Well-Known Member
My knowledge of vector space is rusty. For 3(a), I think you can use: (x-1) and (x-1)2 as a basis of W; dimension = 2
#### He-Mann
##### Vexed?
My knowledge of vector space is rusty. For 3(a), I think you can use: (x-1) and (x-1)2 as a basis of W; dimension = 2
Still astonishing that you remember this stuff.
Unfortunately, your basis cannot produce x^2 - x - 2.
_____________________________________________
\bg_white \begin{align*}p(1) = 0 &\iff (x-1)(ax + b) = 0 \quad [\text{where }a,b \in \mathbb{C}]\\ &\iff ax(x-1)-b(x-1) = 0 \\ &\iff \mathrm{span}\{x-1, x(x-1)\}\end{align*} \\Thus, a basis for W is \{x-1, x(x-1)\} which means \mathrm{dim}(W) = 2.
_____________________________________________
In linear algebra, a basis (for a vector space) is a set of vectors that are linearly independent and span the vector space.
If you unpack this, then you can interpret a basis as a set of the most fundamental building blocks for some world.
Last edited:
#### seanieg89
##### Well-Known Member
Unfortunately, your basis cannot produce x^2 - x - 2
That polynomial does not vanish at 1 and so is not in the span of either of your bases (nor does it need to be).
Your basis is equivalent to his since $\bg_white (x-1)^2=x(x-1)-(x-1)$.
(Also you probably shouldn't use the double implication symbol as informally as in the above post, you should only use it between two propositions that imply each other. In one line, the RHS is an algebraic expression rather than a mathematical statement, and in the others you refer to a polynomial with previously undefined/unquantified constants a and b in it. These early linear algebra courses aren't hard mathematically, but they are meant to drill you in making (semi)formal logical arguments and chasing definitions.)
Last edited:
#### He-Mann
##### Vexed?
That polynomial does not vanish at 1 and so is not in the span of either of your bases (nor does it need to be).
Your basis is equivalent to his since $\bg_white (x-1)^2=x(x-1)-(x-1)$.
(Also you probably shouldn't use the double implication symbol as informally as in the above post, you should only use it between two propositions that imply each other. In one line, the RHS is an algebraic expression rather than a mathematical statement, and in the others you refer to a polynomial with previously undefined/unquantified constants a and b in it. These early linear algebra courses aren't hard mathematically, but they are meant to drill you in making (semi)formal logical arguments and chasing definitions.)
Woops, arithmetic error.
Thanks, I will take note of that in the future.
#### hayabusaboston
##### Well-Known Member
Ohh thankyou goyss! <3
Im trying to figure out 3c atm heh
#### hayabusaboston
##### Well-Known Member
pls help 3C I have no idea lol
#### sida1049
##### Well-Known Member
pls help 3C I have no idea lol
What you need to do is to find the general form of a matrix that satisfies A=-A^T. That is, make each of the 9 entries a variable, and use that equation to make as many substitutions as you can to simplify A.
Once you've done that, you should have found the minimum number of variables you can use to describe a matrix satisfying such an equation. That number is the dimension of the vector space W.
Then, you need to decompose the matrix A as a linear combination of linearly independent 3x3 matrices. This shouldn't be too hard, as all you need to do is to look at entries with the same variable, and make it so A is equal to a sum of 3x3 matrices, with a variable as a coefficient in front of each matrix. Then each of those 3x3 matrices are your basis vectors.
I hope that makes sense.
For example, the matrix {{a,2b},{b,-a}} (first row is a and b, second row is b and a) has dimension 2, because the variables a and b can freely vary. I can decompose the matrix as a{{1,0},{0,-1}} + b{{0,2},{1,0}}. And so my basis vectors are the matrices {{1,0},{0,-1}} and {{0,2},{1,0}}.
Edit: corrected something and used a slightly better example.
Last edited:
#### hayabusaboston
##### Well-Known Member
What you need to do is to find the general form of a matrix that satisfies A=-A^T. That is, make each of the 9 entries a variable, and use that equation to make as many substitutions as you can to simplify A.
Once you've done that, you should have found the minimum number of variables you can use to describe a matrix satisfying such an equation. That number is the dimension of the vector space W.
Then, you need to decompose the matrix A as a linear combination of linearly independent 3x3 matrices. This shouldn't be too hard, as all you need to do is to look at entries with the same variable, and make it so A is equal to a sum of 3x3 matrices, with a variable as a coefficient in front of each matrix. Then each of those 3x3 matrices are your basis vectors.
I hope that makes sense.
For example, the matrix {{a,2b},{b,-a}} (first row is a and b, second row is b and a) has dimension 2, because the variables a and b can freely vary. I can decompose the matrix as a{{1,0},{0,-1}} + b{{0,2},{1,0}}. And so my basis vectors are the matrices {{1,0},{0,-1}} and {{0,2},{1,0}}.
Edit: corrected something and used a slightly better example.
So I wrote out a 3x3 matrix abdefghi equals minus abcdefghi, so b+d=0, h+f=0 and g+c=0, so taking that into account I got basis as a long thing with dimension 6 right?
https://imgur.com/ai837hu
also is this right for b)
https://imgur.com/h1WnGLK
#### hayabusaboston
##### Well-Known Member
Yep, and yep! You got this.
Oh snap yay! lol I was blank for a while because totally forgot how to approach these things.
Linear algebra 80% final exam is in like 5 weeks, Imma start the heavy prep 2 weeks before i think.
#### sida1049
##### Well-Known Member
Oh snap yay! lol I was blank for a while because totally forgot how to approach these things.
Linear algebra 80% final exam is in like 5 weeks, Imma start the heavy prep 2 weeks before i think.
You got this. If you haven't already, I recommend 3blue1brown's YouTube series on linear algebra to get some intuition for it.
| 1,871
| 6,957
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.90625
| 4
|
CC-MAIN-2021-25
|
latest
|
en
| 0.885784
|
https://trafficsteed.com/qa/quick-answer-what-is-the-fastest-way-to-find-a-prime-number.html
| 1,606,509,264,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-50/segments/1606141194171.48/warc/CC-MAIN-20201127191451-20201127221451-00311.warc.gz
| 515,972,406
| 7,883
|
# Quick Answer: What Is The Fastest Way To Find A Prime Number?
## How do you find the square root of a prime number?
Divide the number you are testing, one by one, by each number between 2 and the square root of the tested number.
One of the traits of numbers is that, if they have a factor pair, one of the factors must be equal to or less than the square root..
## What is the fastest way to figure out prime numbers?
To prove whether a number is a prime number, first try dividing it by 2, and see if you get a whole number. If you do, it can’t be a prime number. If you don’t get a whole number, next try dividing it by prime numbers: 3, 5, 7, 11 (9 is divisible by 3) and so on, always dividing by a prime number (see table below).
## Why is 11 not a prime number?
For 11, the answer is: yes, 11 is a prime number because it has only two distinct divisors: 1 and itself (11). As a consequence, 11 is only a multiple of 1 and 11.
## Why isn’t 1 considered a prime number?
Proof: The definition of a prime number is a positive integer that has exactly two positive divisors. However, 1 only has one positive divisor (1 itself), so it is not prime.
## What’s the opposite of a prime number?
composite numbersThe opposite of prime numbers are composite numbers. A composite number is a positive nutural number that has at least one positive divisor other than one or itself.
## What is the smallest prime number?
The smallest prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23. The number 2 is the only even prime number. The number 7 has only two factors: 1 and itself.
## How do you find prime numbers that are efficient?
The Sieve of Eratosthenes is one of the most efficient ways to find the prime numbers smaller than n when n is smaller than around 10 million. A program that demonstrates the Sieve of Eratosthenes is given as follows.
## What is the formula to find prime numbers?
A Formula for PrimesConsider a polynomial F(x) = x^{2} + x + 41. … Let’s check a couple more values: F(10) = 151\, is a prime; F(11) = 173\, and F(12) = 197\, are both prime. … G(x) = x^{2} – x + 41\, is prime for x\, from 0\, through 40,\, and H(x) = x^{2} – 79x + 1601\, is prime for x\, from 1\, through 80.
## Are all odd numbers prime numbers?
Explanation: By definition a prime number has only 2 factors – itself and 1. Hence the smallest natural prime number is 2, and the only on that is even. All other prime numbers are odd, and there are infinitely many prime numbers.
## Why 0 and 1 is not a prime number?
0 is not prime, because you can not create any new number by having it as a factor. 1 was considered prime at some time, but was dropped because many rules for primes would need to make a special case exemption for 1 – which was impractical.
| 739
| 2,771
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375
| 4
|
CC-MAIN-2020-50
|
latest
|
en
| 0.921938
|
https://metanumbers.com/18870
| 1,623,904,240,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623487629209.28/warc/CC-MAIN-20210617041347-20210617071347-00291.warc.gz
| 354,069,980
| 7,660
|
## 18870
18,870 (eighteen thousand eight hundred seventy) is an even five-digits composite number following 18869 and preceding 18871. In scientific notation, it is written as 1.887 × 104. The sum of its digits is 24. It has a total of 5 prime factors and 32 positive divisors. There are 4,608 positive integers (up to 18870) that are relatively prime to 18870.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 24
• Digital Root 6
## Name
Short name 18 thousand 870 eighteen thousand eight hundred seventy
## Notation
Scientific notation 1.887 × 104 18.87 × 103
## Prime Factorization of 18870
Prime Factorization 2 × 3 × 5 × 17 × 37
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 18870 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 18,870 is 2 × 3 × 5 × 17 × 37. Since it has a total of 5 prime factors, 18,870 is a composite number.
## Divisors of 18870
1, 2, 3, 5, 6, 10, 15, 17, 30, 34, 37, 51, 74, 85, 102, 111, 170, 185, 222, 255, 370, 510, 555, 629, 1110, 1258, 1887, 3145, 3774, 6290, 9435, 18870
32 divisors
Even divisors 16 16 8 8
Total Divisors Sum of Divisors Aliquot Sum τ(n) 32 Total number of the positive divisors of n σ(n) 49248 Sum of all the positive divisors of n s(n) 30378 Sum of the proper positive divisors of n A(n) 1539 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 137.368 Returns the nth root of the product of n divisors H(n) 12.2612 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 18,870 can be divided by 32 positive divisors (out of which 16 are even, and 16 are odd). The sum of these divisors (counting 18,870) is 49,248, the average is 1,539.
## Other Arithmetic Functions (n = 18870)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 4608 Total number of positive integers not greater than n that are coprime to n λ(n) 144 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2152 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 4,608 positive integers (less than 18,870) that are coprime with 18,870. And there are approximately 2,152 prime numbers less than or equal to 18,870.
## Divisibility of 18870
m n mod m 2 3 4 5 6 7 8 9 0 0 2 0 0 5 6 6
The number 18,870 is divisible by 2, 3, 5 and 6.
• Arithmetic
• Abundant
• Polite
• Practical
• Square Free
## Base conversion (18870)
Base System Value
2 Binary 100100110110110
3 Ternary 221212220
4 Quaternary 10212312
5 Quinary 1100440
6 Senary 223210
8 Octal 44666
10 Decimal 18870
12 Duodecimal ab06
20 Vigesimal 273a
36 Base36 ek6
## Basic calculations (n = 18870)
### Multiplication
n×i
n×2 37740 56610 75480 94350
### Division
ni
n⁄2 9435 6290 4717.5 3774
### Exponentiation
ni
n2 356076900 6719171103000 126790758713610000 2392541616925820700000
### Nth Root
i√n
2√n 137.368 26.623 11.7204 7.16398
## 18870 as geometric shapes
### Circle
Diameter 37740 118564 1.11865e+09
### Sphere
Volume 2.81452e+13 4.47459e+09 118564
### Square
Length = n
Perimeter 75480 3.56077e+08 26686.2
### Cube
Length = n
Surface area 2.13646e+09 6.71917e+12 32683.8
### Equilateral Triangle
Length = n
Perimeter 56610 1.54186e+08 16341.9
### Triangular Pyramid
Length = n
Surface area 6.16743e+08 7.91862e+11 15407.3
## Cryptographic Hash Functions
md5 7780dee418096d1e5cc1cdde8de01679 7e526554fdce1e34b32edc3de615309ca8137e4d 26fab3e83143619829145200d057e95f0f373bfe72b936d2d1eddfc1d8465d16 cc2e9845db120a6f6a524dd0b261d4d537d50f643e5055631e33ac3d75d42e66349c9534f5aed42ba8451927b3c48a99bdc380982d0e7a296947c315aa43b435 f60aef4ff5205d6500ac2d9410939b01f8f113be
| 1,528
| 4,203
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.59375
| 4
|
CC-MAIN-2021-25
|
latest
|
en
| 0.780929
|
https://www.david-cook.org/what-is-the-square-root-of-1-to-25/
| 1,695,512,260,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233506539.13/warc/CC-MAIN-20230923231031-20230924021031-00538.warc.gz
| 817,041,922
| 10,386
|
# What is the square root of 1 to 25?
## What is the square root of 1 to 25?
List of Square Root 1 to 25
Number (x) Square root of the Number (√X) (Rounded to 3 Decimal Places)
22 4.690
23 4.796
24 4.899
25 5.000
What are the perfect squares from 1 to 30?
Therefore, the value of squares of numbers 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, and 30 will be even.
### What’s the perfect square for 25?
The correct answer is 625 as is written above. Alan P.
What is the square root of 1 25 simplified?
Therefore the square root of the fraction 125 is 0.2 in short.
## Is the square root of 1 25 A rational number?
In square roots 1 to 25, the numbers 1, 4, 9, 16, and 25 are perfect squares and the remaining numbers are non-perfect squares i.e. their square root will be irrational.
What are the first 25 square numbers?
Informally: When you multiply an integer (a “whole” number, positive, negative or zero) times itself, the resulting product is called a square number, or a perfect square or simply “a square.” So, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, and so on, are all square numbers.
### What are the squares of 25?
List of Perfect Squares
NUMBER SQUARE SQUARE ROOT
23 529 4.796
24 576 4.899
25 625 5.000
26 676 5.099
What are the perfect squares from 1 to 20?
Square Root of Perfect Squares – 1 to 20 Hence, there are four perfect squares between 1 to 20, that are rational numbers. Apart from these four numbers, i.e.,1, 4, 9 and 16, the square root of numbers (2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20) will all be decimal values.
## What are all the square root of 25?
The square roots of 25 are √25=5 and −√25=−5 since 52=25 and (−5)2=25 . The principal square root of 25 is √25=5 .
What is 1/25th as a fraction?
5/4
Answer: 1.25 as a fraction is expressed as 5/4 In order to express 1.25 as a fraction, we will first try to remove the decimal.
### What is the principal root of square root of 1 25?
In square roots 1 to 25, the numbers 1, 4, 9, 16, and 25 are perfect squares and the remaining numbers are non-perfect squares i.e. their square root will be irrational. The square root 1 to 25 in radical form is expressed as √x and in the exponential form, it is expressed as (x)½.
What are the 25 square numbers?
Square Numbers List
Number Square
22 484 =22 X 22
23 529 =23 X 23
24 576 =24 X 24
25 625 =25 X 25
## How do you find perfect squares?
To check the perfectness of your square, you can simply calculate the square root of a given number. If the square root is an integer, your number is the perfect square. Let’s calculate the squares of the following numbers: 49 and 53 . √49 = 7 – 7 is an integer → number 49 is a perfect square.
How many perfect squares are there between 1 and 25?
They are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100.
| 921
| 2,826
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.4375
| 4
|
CC-MAIN-2023-40
|
latest
|
en
| 0.906125
|
https://www.coursehero.com/file/5460543/A-Units-of-Measurements/
| 1,490,426,531,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-13/segments/1490218188824.36/warc/CC-MAIN-20170322212948-00060-ip-10-233-31-227.ec2.internal.warc.gz
| 875,839,193
| 22,583
|
A_-_Units_of_Measurements
# A_-_Units_of_Measurements - Units of Measurement LEARNING...
This preview shows page 1. Sign up to view the full content.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: Units of Measurement LEARNING OBJECTIVES Be able to identify the basic quantities. Be able to differentiate between the SI and FPS systems of unit. Be able to convert quantities from one unit system to another. Be able to define dimensional homogeneity LEARNING RESOURCES Text book Lecture notes Classroom discussion Concept questions Homework assignment Practice problems (text book) INTERNATIONAL SYSTEM OF UNITS (SI) Basic Quantities Length is measured in meter (m) Time is measured in second (s) Mass is measured in kilogram (kg) Force is measured in Newton (N) SI UNITS Unit Force In the SI system, the unit force is measured in Newton (N), which is defined as: F = m a If a = 1 m/s2 and m = 1 kg then F = 1 N Hence; 1 N = 1 kg . m/s2 SI UNITS Prefix Giga Mega Kilo Milli Micro Nano Multiple 1,000,000,000 1,000,000 1,000 0.001 0.000 000 1 0.000 000 000 1 Exponential Symbol 109 106 103 10-3 10-6 10-9 G M k m n In addition, multiple of ten terms (kilo, hecto, deca, deci, centi and milli) are often used. U.S. CUSTOMARY SYSTEM OF UNITS (FPS) Basic Quantities Length is measured in foot (ft) Time is measured in second (s) Force is measured in pound (lb) Mass is measured in slug CONVERSION FACTORS FPS Units 1 mile = 5280 ft; 1 ft = 12 in; 1 in = 1000 mil; 1 kip = 1000 lb; 1 ton = 2 kip = 2000 lb, FPS and SI Units 1 lb = 4.4482 N 1 slug = 14.5938 kg 1 ft = 0.3048 m Force Mass Length FPS UNITS Unit Mass In the FPS system, the unit mass is measured in slug and is defined as: F = m.a If a = 32.2 ft/s2 and m = 1 slug then F = 32.2 lb Hence; 1 slug = 1 lbs /ft SIGNIFICANT FIGURS A significant figure is any digit from 0 to 9 of a given number. The zero is counted as a significant figure if it does not specify the location of the decimal point. For example each of the numbers 4072 and 0.04072 has 4 significant figures, whereas the number 0.0025 has only 2 significant figures. The accuracy of a number is specified by the number of significant figures it contains. Exponential notation (in multiple of 3) is typically used to facilitate conversion of SI units. SIGNIFICANT FIGURS The numbers 450245, 256, 31207, and 0.003467 can be expressed in three significant figures as follows: 450245 = 450(10)3 256 = 256 31207 = 31.2(10)3 0.003467 = 3.47(10)-3 ROUNDING OFF NUMBERS The number of significant figures in an answer should be equal to or less than the number of significant figures of the data. The answer should be rounded off. Rounding to 3 significant figures 0.005689 = 5.69(10)-3 2657.243 = 2.66 (10)3 1.345557 = 1.35 244.4399 = 244 DIMENSIONAL HOMOGENEITY In a given equation, all terms must be Expressed in the same unit system, for example, in the equation S = vt + 0.5 at S = distance in m; v = velocity in m/s; t = time in s; and a = acceleration in m/s DIMENSIONAL HOMOGENEITY In the equation S = vt + 0.5 at whether All terms must maintain their dimensional homogeneity ...
View Full Document
## This note was uploaded on 07/26/2009 for the course CE 221 taught by Professor Buch during the Spring '08 term at Michigan State University.
Ask a homework question - tutors are online
| 967
| 3,364
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.8125
| 4
|
CC-MAIN-2017-13
|
longest
|
en
| 0.862858
|
https://www.myeduwaves.com/2022/10/arithmetic-operators-in-python-hackerrank-solution.html
| 1,695,660,300,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233509023.57/warc/CC-MAIN-20230925151539-20230925181539-00523.warc.gz
| 1,006,072,467
| 41,340
|
# Arithmetic Operators in Python - HackerRank Solution
## Arithmetic Operators - Python HackerRank Solution
Hello Friends, How are you? Today I am going to solve the HackerRank Arithmetic Operators Problem in Python with a very easy explanation. In this article, you will get one or more approaches to solving this problem. So let's start-
{tocify} \$title={Table of Contents}
### HackerRank Python Arithmetic Operators Solution - Problem Statement
Task
The provided code stub reads two integers from STDIN, a and b. Add code to print three lines where:
The first line contains the sum of the two numbers.
The second line contains the difference between the two numbers (first - second).
The third line contains the product of the two numbers.
Example
a = 3
b = 5
Print the following:
8 -2 15 {codeBox}
Input Format
The first line contains the first integer, a.
The second line contains the second integer, b.
Constraints
1 <= a <= 10^10
1 <= b <= 10^10
Output Format:
Print the three lines as explained above.
Sample Input 0
3 2 {codeBox}
Sample Output 0
5 1 6 {codeBox}
Explanation:
3 + 2 => 5
3 - 2 => 1
3 * 2 => 6
### Python Arithmetic Operators - Hacker Rank Solution
Approach I: Arithmetic Operators HackerRank Python Solution
``````# ========================
# Information
# ========================
# Name: Arithmetic Operators in Python HackerRank
# Direct Link: https://www.hackerrank.com/challenges/python-arithmetic-operators/problem
# Difficulty: Easy
# Max Score: 10
# Language: Pypy 3
# ========================
# Solution Start
# ========================
#Arithmetic Operators in Python - Hacker Rank Solution
if __name__ == '__main__':
a = int(input())
b = int(input())
print(a+b);
print(a-b);
print(a*b);
#Arithmetic Operators in Python - Hacker Rank Solution END
# MyEduWaves``````
Disclaimer: The above Problem ( Python Arithmetic Operators ) is generated by Hackerrank but the Solution is Provided by MyEduWaves. This tutorial is only for Educational and Learning purposes. Authority if any queries regarding this post or website fill out the contact form.
I hope you have understood the solution to this HackerRank Problem. All these three solutions will pass all the test cases. Now visit Python Arithmetic Operators Hackerrank Problem and try to solve it again.
All the Best!
| 557
| 2,346
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625
| 4
|
CC-MAIN-2023-40
|
latest
|
en
| 0.720531
|
https://www.jiskha.com/questions/1352285/the-decibel-level-of-a-sound-is-given-by-db-10log-ee0-where-e0-10-12wattsm2-and-e-is
| 1,618,993,092,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618039526421.82/warc/CC-MAIN-20210421065303-20210421095303-00033.warc.gz
| 874,365,151
| 5,224
|
# LOGS
The decibel level of a sound is given by dB=10log(EE0), where E0=10−12wattsm2 and E is the intensity of the sound.
(a) Find the intensity of a 80-decibel sound.
E= watts per square meter.
(b) How many times more intense is the sound from (a) than a 50-decibel sound?
A sound at 80 decibels is ____ times more intense than a sound at 50 decibels.
(c) The loudness (perceived volume by the human ear) of a sound of d decibels doubles each time the intensity of a sound increases by a factor of ten. To the human ear, how many times louder does a 80-decibel sound seem than a 50-decibel sound?
A sound at 80 decibels sounds______ times louder than a sound at 50 decibels.
1. 👍
2. 👎
3. 👁
And what about that E0 = 10-12wattsm2 ??
1. 👍
2. 👎
## Similar Questions
1. ### physics
A person standing 1.00 m from a portable speaker hears its sound at an intensity of 7.50 103 W/m2. (a) Find the corresponding decibel level. (b) Find the sound intensity at a distance of 35.0 m, assuming the sound
2. ### Algebra 2
Can someone help me? I'm really confused. The loudness L of a sound in decibels is given by L= 10log(base10)R, where R is the sound's relative intensity. If the intensity of a certain sound is tripled, by how many decibels does
3. ### alg
An amplifier has an input power of 5 mW. The output power is 100 mW. What is the decibel gain to the nearest decibel?
4. ### Physics
In the afternoon, the decibel level of a busy freeway is 80 dB with 100 cars passing a given point every minute. Late at night, the traffic flow is only 5 cars per minute. What is the late-night decibel level?
1. ### Algebra
Find the number of decibels for the power of the sound. Round to the nearest decibel. A rock concert, 5.3 10-6 watts/cm2
2. ### Math
On th decibel scale, the loudness of a sound, in decibels, is given by D=10log(I/I0), where I is the intensity of a sound barely audible to the human ear. If the intensity of a sound is 10^11I0, what is the loudness in decibels?
3. ### Physics
Part One: The decibel level of the noise from a jet aircraft is 130 dB when measured 19.3 m from the aircraft. How much sound power does the jet aircraft emit? Answer in units of W - 46673.5W I need help with Part Two: Part Two:
4. ### Algebra 2
The loudness L of sound in decibels is given by L= 10log (l/lo) where l is the intensity of sound and lo is the intensity of the least audible sound. If lo= 10^-12 w/m^2, about how many times more intense is a 108 decibel sound
1. ### Math
How would I solve this Equation? DB = 10log(I/Io) DB = 10log (140 Io/Io) DB = 10 log (140) DB = ????? This is as far I have gotten on this equation. Can someone please help me finish? I am trying to determine how many times louder
2. ### Physics
The intensity due to a number of indipendent sound sources is the sum of the individual intensities. A) When four quadruplets cry simultaneously, how many decibels greater is the sound intensity level than when a single one cries?
3. ### PreCalc
The decibel (dB) is defined as dB= 10log (P2/P1), where P2 is the power of a particular signal. P1 is the power of some reference signal. In the case of sounds, the reference signal is a sound level that is just barely audible.How
4. ### physics
a rock group is playing in a club. Sound emerging outodoors from an open door spreads uniformly in all directions. If the decibel level is 70 dB at a distance of 1.0m from the door, at whiat distance is the music just barely
| 982
| 3,464
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.65625
| 4
|
CC-MAIN-2021-17
|
latest
|
en
| 0.82974
|
https://scallope.com/qa/quick-answer-what-is-a-complete-vector-space.html
| 1,600,804,310,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600400206763.24/warc/CC-MAIN-20200922192512-20200922222512-00536.warc.gz
| 592,075,788
| 7,899
|
# Quick Answer: What Is A Complete Vector Space?
## How do you show a subspace is closed?
A subspace is closed under the operations of the vector space it is in.
In this case, if you add two vectors in the space, it’s sum must be in it.
So if you take any vector in the space, and add it’s negative, it’s sum is the zero vector, which is then by definition in the subspace..
## What does it mean for a space to be complete?
In mathematical analysis, a metric space M is called complete (or a Cauchy space) if every Cauchy sequence of points in M has a limit that is also in M or, alternatively, if every Cauchy sequence in M converges in M. Intuitively, a space is complete if there are no “points missing” from it (inside or at the boundary).
## How do you show a space is complete?
A metric space (X, d) is called complete if every Cauchy sequence (xn) in X converges to some point of X. A subset A of X is called complete if A as a metric subspace of (X, d) is complete, that is, if every Cauchy sequence (xn) in A converges to a point in A.
## What is a real vector space?
A real vector space is a vector space whose field of scalars is the field of reals. A linear transformation between real vector spaces is given by a matrix with real entries (i.e., a real matrix).
## Are the rationals complete?
The real numbers are complete in the sense that every set of reals which is bounded above has a least upper bound and every set bounded below has a greatest lower bound. The rationals do not have this property because there is a “gap” at every irrational number.
## Do all Cauchy sequences converge?
In a complete metric space, every Cauchy sequence is convergent. This is because it is the definition of Complete metric space . … This shows that every Cauchy sequence in converges to a point in , so is a complete metric space.
## Is a complete metric space closed?
A metric space (X, d) is said to be complete if every Cauchy sequence in X converges (to a point in X). Theorem 4. A closed subset of a complete metric space is a complete sub- space. … A complete subspace of a metric space is a closed subset.
## Is the set of integers complete?
The integers, for the metric defined by d(x,y) = |x-y|, are a discrete metric space and thus they are complete since a Cauchy sequence of integers is eventually constant. … Every discrete metric space is complete.
## What is a Cauchy?
In mathematics, a Cauchy sequence (French pronunciation: [koʃi]; English: /ˈkoʊʃiː/ KOH-shee), named after Augustin-Louis Cauchy, is a sequence whose elements become arbitrarily close to each other as the sequence progresses.
## Is a metric space open?
Given any metric space, , is both open and closed. Let , then is open if and only if for every there exist such that . This definition is always satisfied for the empty set, , and the entire space, . … In the case of any metric space, the entire space and the empty set fall under “…they [are] both…”
| 720
| 2,963
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4
| 4
|
CC-MAIN-2020-40
|
latest
|
en
| 0.936776
|
http://mathhelpforum.com/differential-geometry/107544-convergence-proof.html
| 1,527,006,265,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794864798.12/warc/CC-MAIN-20180522151159-20180522171159-00033.warc.gz
| 189,562,070
| 9,858
|
1. ## Convergence Proof
Suppose that A > 0 is a given real number. Define the sequence {$\displaystyle a_n$}$\displaystyle n \in N$ recursively by:
$\displaystyle a_1$ is a positive real number; $\displaystyle a_{n+1} = \frac{1}{2} (a_n + \frac{A}{a_n})$.
Prove that {$\displaystyle a_n$}$\displaystyle n \in N$ converges to $\displaystyle \sqrt{A}$
No clue where to start...
2. Originally Posted by thaopanda
Suppose that A > 0 is a given real number. Define the sequence {$\displaystyle a_n$}$\displaystyle n \in N$ recursively by:
$\displaystyle a_1$ is a positive real number; $\displaystyle a_{n+1} = \frac{1}{2} (a_n + \frac{A}{a_n})$.
Prove that {$\displaystyle a_n$}$\displaystyle n \in N$ converges to $\displaystyle \sqrt{A}$
Use induction to prove the sequence becomes decreasing bounded below.
From that you know that the sequence converges to say $\displaystyle L$.
Then solve $\displaystyle L=\frac{1}{2}\left(L+\frac{A}{L}\right)$ for $\displaystyle L$
3. thanks! that was a lot easier than I thought
| 313
| 1,021
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.09375
| 4
|
CC-MAIN-2018-22
|
latest
|
en
| 0.785329
|
https://cs.stackexchange.com/questions/13055/time-complexity-and-space-complexity-in-recursive-algorithm
| 1,716,983,993,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971059239.72/warc/CC-MAIN-20240529103929-20240529133929-00877.warc.gz
| 149,983,486
| 41,057
|
Time complexity and space complexity in recursive algorithm
"The designer of an algorithm needs to balance between space complexity and time
complexity." - Comment on the validity of the statement in the context of recursive
algorithms.
This is a question from my university's previous paper. But i couldn't find a decent answer. Actually i am confused about how can a developer minimize the time-complexity for any recursive function. I get that if there is a tail-recursion then space complexity can be minimized. But can't get the idea of time-complexity.
• I don't think you quite understand the question. The space-time tradeoff is about decreasing space or time at the cost of the other ("balance between" points to this being what the question is talking about). So I don't think tail-recursion classifies - I believe it decreases both. Jul 3, 2013 at 11:55
One thing comes in mind is memoization. Simple well studied problem for this is Fibonacci numbers, simple recursion is as follow:
fib(int n)
{
if (n < 3)
return 1;
return fib(n-1) + fib(n-2);
}
But with memoization, we can use an auxiliary array to get rid of extra calls:
f[1]=f[2] = 1;
fib(int n)
{
if (n < 3)
return 1;
if (f[n] == 0)
f[n] = fib(n-1) + fib(n-2);
return f[n];
}
This simple change, reduces the time from $\Theta(\phi^n)$ to $\Theta(n)$.
The memoization technique sometimes uses more memory, but very faster in time, and one of a tradeoffs that software developer should be care about it is this.
Explanation of the memoization of Fibonacci numbers:
First we create an array $f$, to save the values that already computed. This is the main part of all memoization algorithms. Instead of many repeated recursive calls we can save the results, already obtained by previous steps of algorithm. As shown in the algorithm we set the $f[1],f[2]$ to $1$.
In the first if we actually check whether we are in the start or not. But we can remove this if statement. But for make it simpler to read I left it.
In the second if, we check if the value of fib(n) is already computed or not. This prevents us from multiple call for the same number, for example suppose we want to compute f(6), then in normal recursion we have the first recursion tree as shown in the following figure and in the memoization version we have the second tree. The reason is, in memoization we just compute the green vertices one time and then we save them into the memory (array $f$) and if needed we fetch them later.
In the following figure, green nodes are parts which are necessary to be computed (in this way), yellow nodes are precomputed ones, and red nodes are the nodes that are repeatedly computed in the first recursion.
As is clear from the image, in the normal case we have just precomputed f(1) and f(2), but in the memoization case, all functions for less than $n-1$ are precomputed, and this causes exponentially smaller recursion tree. (In memoization number of red nodes is zero, which is exponential in the normal recursion).
• I'm not sure this is correct. The recursive solution requires the stack to be stored, which likely uses more memory than the memoization's array. Jul 3, 2013 at 12:09
• @Dukeling, If you see my answer, I didn't say that in this case, I said sometimes. On the other hand you are right, memoization is very good even in memory, in most times. But I mentioned this to notify OP about the usage of this technique to reduce the time complexity (not memory).
– user742
Jul 3, 2013 at 12:14
• Ok, the memoization code you provided does actually seem to use slightly more memory. But IMO it's a non-issue, since you can just use a simple for-loop to populate the array, which is faster and uses less memory than either. And this can further be optimized (assuming you only do a single call) to only store the last 2 values, thus constant space. Jul 3, 2013 at 12:21
• But a for-loop wouldn't be recursive any more, so I suppose this answer answers the question - I'm just not too fond of giving examples of things that can easily be done a lot better, even though Fibonacci numbers is a popular example for recursive functions. Jul 3, 2013 at 12:28
• @SaeedAmiri yes actually i'm having difficulty understanding the code.I will be very glad if you can please explain with proper code and comments. Jul 3, 2013 at 16:53
| 1,033
| 4,325
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.796875
| 4
|
CC-MAIN-2024-22
|
latest
|
en
| 0.897956
|
https://archivemore.com/archives/528
| 1,659,896,934,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00722.warc.gz
| 153,925,284
| 8,908
|
999
## What is the largest number you can write by using only 3 digits it’s not 999?
Answer: Biggest numbers that can be formed using 3 digits are as follows: If repetition of all the 3 digits is allowed, then the required number is 999. If repetition of only 2 digits is allowed, then it is 998. If no repetition is permitted, the number is 987.
900 numbers
## What is the greatest number of 3 digits which when divided by 6 9 12 leaves a remainder 3 in each case?
Answer: Required Number is 975. Step-by-step explanation: Given: Required Number when divided by 3 , 9 and 12 leaves 3 as remainder in each case.
## When the largest 3-digit number is divided by 3 we get?
We know the divisibility rule for 3: If the sum of the digits of a number is divisible by 3, then the number is also divisible by 3. Here, 9+9+6=24 is multiple of 3. So, 996 is the largest number divisible by 3.
## What is the greatest number of 4 digits that when divided by any of the number 6 9 12 17?
The Largest number of four digits is 9999.by dividing 9999 to 612 get 207 as remainder. So, the greatest number of 4 digits divisible by any numbers from 6, 9, 12 or 17 = (9999-207) = 9792.
## What is the least number when divided by 36 24 and 16?
Explanation: For 16, 24 and 36 the smallest number which would be perfectly divisible by them is their LCM which is 144.
9996
## What 4 digit numbers are divisible by 7?
The first 4-digit number divisible by 7 is 1001. This is sometimes also referred to as the smallest four digit number divisible by 7 or the lowest 4-digit number divisible by 7. What is the last four digit number divisible by 7? The last 4-digit number divisible by 7 is 9996.
## What is the greatest 4 digit number divisible by 3?
The last 4-digit number divisible by 3 is 9999. This is sometimes also referred to as the largest four digit number divisible by 3 or the greatest 4-digit number divisible by 3.
9940
| 515
| 1,925
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.65625
| 5
|
CC-MAIN-2022-33
|
latest
|
en
| 0.920928
|
http://alex.state.al.us/lesson_view.php?id=33080
| 1,498,634,237,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128322873.10/warc/CC-MAIN-20170628065139-20170628085139-00185.warc.gz
| 11,968,229
| 8,061
|
Home | Add Bookmark | Print Friendly | Rate This Lesson Plan | Suggest a Variation
## Lesson Plan
You may save this lesson plan to your hard drive as an html file by selecting "File", then "Save As" from your browser's pull down menu. The file name extension must be .html.
This lesson provided by:
Author: Tim McKenzie Organization:
General Lesson Information
Lesson Plan ID: 33080 Title: How Far Can You Leap? Overview/Annotation: This lesson will allow students to become familiar with the concept of unit rate. Through an open investigation students will develop methods to find unit rate with a table, equivalent ratios, or an equation. This is a lesson to be used as part of a unit with "Painter Problems" and "How Big Should It Be?"This is a College- and Career-Ready Standards showcase lesson plan.
Associated Standards and Objectives
Content Standard(s):
MA2015 (6) 1. Understand the concept of a ratio, and use ratio language to describe a ratio relationship between two quantities. [6-RP1] Examples: "The ratio of wings to beaks in the bird house at the zoo was 2:1 because for every 2 wings there was 1 beak." "For every vote candidate A received, candidate C received nearly three votes." MA2015 (6) 2. Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. [6-RP2] Examples: "This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar." "We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger." (Expectations for unit rates in this grade are limited to non-complex fractions.) MA2015 (6) 3. Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations. [6-RP3] a. Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios. [6-RP3a] b. Solve unit rate problems including those involving unit pricing and constant speed. [6-RP3b] Example: If it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours' At what rate were lawns being mowed' c. Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent. [6-RP3c] d. Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities. [6-RP3d] MA2015 (7) 1. Compute unit rates associated with ratios of fractions, including ratios of lengths, areas, and other quantities measured in like or different units. [7-RP1] MA2201007070000102.jpg
Local/National Standards:
Math Practice Standards:
1. Make sense of problems and persevere in solving them.
2. Reason abstractly and quantitatively.
3. Construct viable arguments and critique the reasoning of others.
4. Model with mathematics.
5. Use appropriate tools strategically.
6. Attend to precision.
7. Look for and make use of structure.
8. Look for and express regularity in repeated reasoning.
Primary Learning Objective(s):
I CAN identify and develop ratios in real world situations.
I CAN identify equivalent ratios.
I CAN compare ratios in real world situations.
I CAN use equivalent ratios to find the unit rate.
Preparation Information
Total Duration: 61 to 90 Minutes Materials and Resources: How Far Can You Leap? Activity Guide (found in attachments), Investigative Activity Rubric (found in attachments), Rate Exit Slip (found in attachments), Chart paper, painter's tape, Math Toolbox which includes the following: pencil, paper, graph paper, markers, scissors, glue, calculator, sticky notes Technology Resources Needed: Interactive whiteboard (Optional) with required software, document camera, projector, access to search engine (individually or whole group) Background/Preparation: The teacher must make the appropriate number of copies of the How Far Can You Leap? activity guide (found in attachments). Copies should be made so that students can work collaboratively. The teacher must make the appropriate number of copies of the Rate Exit Slip (found in attachments). Each student should have one.Teacher must prepare the appropriate number of Math Toolboxes. Teacher must mark off a starting point for the leap. In a long hallway with square tiles works best, but this can easily be modified to do an outdoor lesson. Instead of floor tiles, the teacher can mark off feet.The students must have knowledge of ratios.
Procedures/Activities:
The teacher will instruct the students to search for "fastest car in the world" (if available, students can use individual devices or this can be done in small or whole group). The teacher will ask "How do we know it is the fastest car in the world?" Students will give responses, an ideal response is "because it tells us the speed." The teacher will ask the students, "How is speed displayed?" Ideal student response "Miles per hour." The teacher will introduce speed as a rate and explain that in 1 hour that car can go n miles. The teacher will introduce rate as a ratio where a unit equals 1. The teacher will ask students for other rates we use every day. Ideal student response "dollars per hour". The teacher will show the students a rate/ratio of \$40 for five hours. The teacher will ask, "Is this a rate?" Ideal response, "No, because the hours is not equal to one." The teacher will pose an open-ended question "How can we get our hours down to one?" Allow students to give suggestions and strategies. Teacher will drive students to set up a table or use equivalent ratios. The teacher will introduce the activity How far can you leap? The teacher will demonstrate (or have a student demonstrate a leap), start with both feet together and jump with one foot. If needed, teacher will demonstrate the leap progression in the activity (the directions are on the activity guide). Students will begin investigation. As students are working, teacher will act as a facilitator or coach asking questions that drive understanding. Once adequate time (30-45 minutes) is given, the students will share their finding on the document camera. (If a document camera is not available students may present their work in the front of the class, this is where the students would need chart paper). As the students are sharing, the teacher is acting as the facilitator and coach asking questions that drive ratio understanding. "How do you know that ratio is equivalent to the first ratio?" "How did you know to do _______?" "Did someone do this differently?" Teacher will spawn debate on who has the best rate at jumping tiles. Toward the end of the class students will complete the Rate Exit Slip (found in attachments).
Attachments:**Some files will display in a new window. Others will prompt you to download. Howfarcanyouleap.pdf InvestigativeActivityRubric.pdf RateExitSlip.pdf
Assessment
Assessment Strategies Formal Formative Assessment: Rate Exit SlipFormal Assessment: Using the Investigative Activity Rubric (found in attachments) teacher will evaluate students' work.Informal Formative Assessment: As the students are working, the teacher will act as the facilitator and coach. Teacher will ask questions to evaluate students (i.e. How do you know ______?, What did you do to get that?) Teacher may pull small groups during investigation on a needs basis.
Acceleration: The investigation has an included extension on the How Far Can You Leap? Activity Sheet (found in attachments). Intervention: Because this is part of a unit, teacher may develop small groups based on the Rate Exit Slip or informal questioning as part of the investigative activity.
Each area below is a direct link to general teaching strategies/classroom accommodations for students with identified learning and/or behavior problems such as: reading or math performance below grade level; test or classroom assignments/quizzes at a failing level; failure to complete assignments independently; difficulty with short-term memory, abstract concepts, staying on task, or following directions; poor peer interaction or temper tantrums, and other learning or behavior problems. Presentation of Material Environment Time Demands Materials Attention Using Groups and Peers Assisting the Reluctant Starter Dealing with Inappropriate Behavior Be sure to check the student's IEP for specific accommodations.
| 1,847
| 8,642
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.25
| 4
|
CC-MAIN-2017-26
|
latest
|
en
| 0.897237
|
https://numbas.mathcentre.ac.uk/question/88227/mat-2b-bab-8-no-10.exam
| 1,624,506,817,000,000,000
|
text/plain
|
crawl-data/CC-MAIN-2021-25/segments/1623488550571.96/warc/CC-MAIN-20210624015641-20210624045641-00184.warc.gz
| 382,786,591
| 2,094
|
// Numbas version: exam_results_page_options {"name": "Mat 2B - Bab 8 - No 10", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Mat 2B - Bab 8 - No 10", "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "
Penyebaran populasi capung yang berada di dua kolam \$$P\$$ dan \$$Q\$$ dapat dimodelkan sebagai berikut. Setiap hari, \$$20\\%\$$ capung yang berada dalam kolam \$$P\$$ akan pindah ke kolam \$$Q\$$. Sebaliknya, \$$30\\%\$$ capung yang berada dalam kolam \$$Q\$$ akan pindah ke kolam \$$P\$$. Misalkan \$$P_t\$$ dan \$$Q_t\$$ berturut-turut menyatakan banyaknya capung pada kolam \$$P\$$ dan \$$Q\$$ setiap waktu. Maka dapat diperoleh sistem persamaan beda \$$\\begin{cases} P_{t+1}&=0.8P_t+0.3Q_t \\\\ Q_{t+1}&=0.2P_t+0.7Q_t \\end{cases}\$$.
", "advice": "", "rulesets": {}, "extensions": [], "variables": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "matrix", "useCustomName": false, "customName": "", "marks": "0.4", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "
Misalkan \$$\\mathbf{x}_t=\\begin{bmatrix} P_t \\\\Q_t \\end{bmatrix}\$$. Bentuk iterasi model tersebut dapat ditulis sebagai \$$\\mathbf{x}_{t+1}=A\\mathbf{x}_t\$$. Tentukan matriks \$$A\$$.
", "correctAnswer": "matrix([0.8,0.3],[0.2,0.7])", "correctAnswerFractions": false, "numRows": "2", "numColumns": "2", "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0}, {"type": "matrix", "useCustomName": false, "customName": "", "marks": "1.6", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "
Diketahui
\n
\n
• Nilai eigen dari matriks $A$ adalah $\\lambda_1$ dan $\\lambda_2$ dengan $\\lambda_1\\geq \\lambda_2$.
• \n
• Vektor \$$\\mathbf{v}_1=\\begin{bmatrix} 3 \\\\ a \\end{bmatrix}\$$ merupakan vektor eigen yang bersesuaian dengan $\\lambda_1$, dan vektor \$$\\mathbf{v}_2=\\begin{bmatrix} 3 \\\\ b \\end{bmatrix}\$$ merupakan vektor eigen yang bersesuaian dengan $\\lambda_2$.
• \n
• Populasi awal capung di kedua kolam tersebut adalah \$$\\mathbf{x}_0=\\begin{bmatrix} 600 \\\\ 80 \\end{bmatrix}\$$.
• \n
\n
Tentukan nilai $\\lambda_1$, $\\lambda_2$, $a$, dan $b$ berturut-turut.
", "correctAnswer": "matrix([1,0.5,2,-3])", "correctAnswerFractions": true, "numRows": 1, "numColumns": "4", "allowResize": false, "tolerance": 0, "markPerCell": true, "allowFractions": true, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "
Dalam waktu yang cukup lama, misalkan banyaknya capung pada kolam $P$ sebanyak $P_\\infty$, sedangkan banyaknya capung pada kolam $Q$ sebanyak $Q_\\infty$. Tentukan nilai dari $\\dfrac{P_\\infty}{Q_\\infty}$.
", "minValue": "1.5", "maxValue": "1.5", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": true, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "contributors": [{"name": "Meong Meong Project", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4687/"}]}]}], "contributors": [{"name": "Meong Meong Project", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4687/"}]}
| 1,517
| 4,595
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.625
| 4
|
CC-MAIN-2021-25
|
latest
|
en
| 0.179201
|
https://www.csee.umbc.edu/~chang/cs203.f10/proofs9.shtml
| 1,527,087,828,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794865679.51/warc/CC-MAIN-20180523141759-20180523161759-00582.warc.gz
| 721,572,152
| 5,101
|
# Equivalence Proofs
An equivalence proof shows that the "if and only if" relationship holds among two or more propositions. We have shown in class that for a connected undirected graph G:
G has an Euler circuit ⇐⇒ every vertex in G has even degree.
To prove this equivalence we have to show the implication in both directions. We need to show that
G has an Euler circuit ⇒ every vertex in G has even degree.
(which we did in Example 2 of the Indirect Proofs section) and we need to show that
Every vertex in G has even degree ⇒ G has an Euler circuit.
(this we will show later, when we look at loop invariants). If we did not prove this second direction, it is possible that there is a graph where every vertex has even degree, but the graph does not have an Euler circuit. It is very tempting to take a short cut and prove both directions of an equivalence proof at the same time. This almost always leads to disaster.
Another common pitfall in equivalence proofs is to prove the same direction twice. Since we can use a direct or an indirect proof for either the "if" or the "only if" direction, there are 4 possible ways we can write an equivalence proof. To prove that p ⇐⇒ q, we can use any of the following 4 methods:
1. Show pq and qp .
2. Show pq and ¬p ⇒ ¬q.
3. Show ¬q ⇒ ¬p and qp .
4. Show ¬q ⇒ ¬p and ¬p ⇒ ¬q .
However if we prove that
pq and ¬q ⇒ ¬p
then we would have actually shown the same direction twice and have neglected to show that qp.
Sometimes a theorem establishes the equivalence of several propositions p1, p2, p3, ..., pn. In this case, we do not need to prove the equivalence between every pair of statements. All we need to show is:
p1p2
p2p3
p3p4
...
pn-1pn
pnp1
Hypothetical syllogism then guarantees that every pair of propositions is equivalent.
Example 1: In graph theory, a tree is just a connected undirected graph that does not have any cycles. This type of tree might be different from ones you have encountered in other computer science classes because there is not a distinguished root vertex. Graph-theoretic trees also do not have parent-child relationships between the vertices. Trees have a nice characterization shown below:
Claim: Let G be a connected undirected graph with n vertices and e edges. Then, the following statements are equivalent:
1. G does not contain any cycles.
2. e = n − 1.
3. there is a unique path between every pair of vertices in G.
Proof:
(I ⇒ II) Suppose that G does not contain any cycles. We first argue that G must have a vertex with degree 1 as long as G has at least 2 vertices. (We do this because when we remove a degree 1 vertex, we do not disconnect the graph and we reduce the number of vertices and edges by 1.) To see this, we explore the graph along a path. We start at any vertex and move along an edge to another vertex. We keep moving along edges but we do not reuse the edge that we used to arrive at the current vertex. If we revisit a vertex at any point, then we have discovered a cycle. Since G does not have any cycles, this is impossible. Thus, each vertex we visit must be a new vertex. Since there are only n vertices in G, after at most n steps we cannot discover any more new vertices. That means, the last vertex, the one we stopped at, must have degree 1. If there is an edge other than the one that we just used to arrive at the last vertex, then that edge would either lead to a previously visited vertex (creating a cycle) or to a new vertex (which implies that G has more than n vertices) — either of which leads to a contradiction.
Now, remove the degree 1 vertex we found from G. The resulting graph is still a connected graph without any cycles, which by the argument above, must have a degree 1 vertex. We keep removing the degree 1 vertices until all we have is a single vertex and no edges. Each time we removed a degree 1 vertex, we reduce the number vertices by 1 and the number of edges by 1. Let m be the number of degree 1 vertices we removed in the process. Then, e must equal m and n must equal m + 1, since the removal process took away all of the edges and all but one of the vertices. Thus, e = n − 1.
(II ⇒ III) We prove this indirectly. That is, we show that ¬III ⇒ ¬II. In other words, we will show that if there are two paths in G between some pair of vertices u and v, then en - 1. So, suppose that there are two paths π1 and π2 from u to v:
As we proved in Example 2 of the Uniqueness Proofs section, there must be a cycle from x back to itself going through y, w and z. Here x, y, z and w are vertices on π1 and π2 shown in the figure above. We note that every vertex in this cycle has degree at least 2.
As before, we keep removing degree 1 vertices from G until there are no more degree 1 vertices. Now, we are not allowed to assume that G does not have any cycles, since in proving II ⇒ III we are not allowed to assume I. So, it might be the case that G does not have degree 1 vertices, but that's OK. This argument takes a different route. Suppose that m vertices are removed when we remove degree 1 vertices. (It is possible that m = 0.) Let G' be the resulting graph.
Consider the cycle from x to itself that we found above — the one going through y, w and z. None of the vertices in this cycle was removed. Why is this? Suppose some vertex a is the first vertex from the cycle to be removed. But, if a was first, then a was still connected to two vertices in the cycle. That means, a still had degree 2 and should not have been removed — a contradiction. Thus, the process of removing degree 1 vertices will leave at least 4 vertices in G'.
Let n' and e' be respectively the number of vertices and edges in G'. Every vertex in G' has degree at least 2, so we have:
2 e' = ∑ degree(v) v ∈ G'
and
∑ degree(v) ≥ 2 n'. v ∈ G'
We know summing the degrees always gives us twice the number of edges. That accounts for the first equation. The second comes from every vertex in G' having degree at least 2. Thus, 2 e' ≥ 2 n' which implies that e'n'. Finally, in terms of the number of vertices n and the number of edges e in the original graph G, we have e = e' + m and n = n' + m because m degree 1 vertices were removed from G to form G'. Thus,
e = e' + mn' + m = n.
Therefore, e is at least n (thus cannot be equal to n − 1) and we have shown that ¬III ⇒ ¬II.
(III ⇒ I) We prove this indirectly and and show that if G has a cycle then the path between some pair of vertices u and v is not unique. So, suppose that G has a cycle x1, x2, x3, ..., xk, x1. Then, let u = x1 and v = x2. The edge between x1 and x2 constitutes a path from u to v. Furthermore, the reversal of the portion of the cycle from x2 to xk back to x1 is also a path from u to v. Thus, there are two paths in G from u to v.
QED
Example 2: When we show that two sets A and B are equal, we are inherently arguing two implications:
xAxB
and
xBxA.
This is an equivalence proof and there is no way around having to make these two arguments separately. Trying to combine the two arguments will confuse the reader (and often the writer). Instead, a well-written proof of set equality will clearly indicate when it is proving that the left-hand side (LHS) is contained in the right-hand side (RHS) and when it is proving the other direction.
So, let's prove the following claim.
Claim: For all sets A, B and C,
( AB ) ∩ ( AC) = A ∪ ( (BC) − A ).
Proof:
(LHS ⊆ RHS) Suppose that x ∈ ( AB ) ∩ ( AC). There are two cases to consider.
Case 1: xA.
If xA, then xA ∪ ( (BC) − A ).
Case 2: xA.
Since x ∈ ( AB ) ∩ ( AC), we know that xAB and xAC. But xA and xAB, implies that xB. Similarly, xA and xAC, implies that xC. Thus, xBC.
Furthermore, xA, so x ∈ (BC) − A. Therefore, xA ∪ ( (BC) − A ).
In both cases, we've shown that xA ∪ ( (BC) − A ). Therefore, ( AB ) ∩ ( AC) ⊆ A ∪ ( (BC) − A ).
(LHS ⊆ RHS) Suppose that xA ∪ ( (BC) − A ). Then, either xA or x ∈ ( (BC) − A ). Let's consider these case separately.
Case 1: xA.
If xA, then xAB. Similarly, xAC. Thus, x ∈ ( AB ) ∩ ( AC).
Case 2: x ∈ (BC) − A.
If x ∈ (BC) − A, then If xBC and xA. Since xBC, we know that xB and xC. Thus, xAB and xAC. Therefore, x ∈ ( AB ) ∩ ( AC).
In both cases, we have shown that x ∈ ( AB ) ∩ ( AC). Therefore, A ∪ ( (BC) − A ) ⊆ ( AB ) ∩ ( AC).
QED
| 2,200
| 8,260
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.3125
| 4
|
CC-MAIN-2018-22
|
longest
|
en
| 0.945331
|
http://www.enotes.com/homework-help/what-final-volume-gas-following-problem-250246
| 1,476,992,791,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-44/segments/1476988717783.68/warc/CC-MAIN-20161020183837-00055-ip-10-171-6-4.ec2.internal.warc.gz
| 428,238,901
| 10,132
|
# What is the final volume of the gas in the following problem?If I have 21 liters of gas held at a pressure of 78 atm and a temperature of 900 K, what will be the volume of the gas if I decrease the...
What is the final volume of the gas in the following problem?
If I have 21 liters of gas held at a pressure of 78 atm and a temperature of 900 K, what will be the volume of the gas if I decrease the pressure to 45 atm and decrease the temperature to 750K?
ndnordic | High School Teacher | (Level 2) Associate Educator
Posted on
This problem can be solved using the combined gas law:
P1V1/T1 = P2V2/T2
Initial conditions are:
P1 = 78 atm
V1 = 21 l
T1 = 900 K
final conditions:
P2 = 45 atm
V2 = ?
T2 = 750 K
Substituting your values, and solving for V2 you get:
V2 = 78 * 21 * 750/ (45 * 900) = 30.33 L
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The problem states that there are initially 21 liters of a gas at a pressure of 78 atm. and the temperature of the gas is 900 K. The pressure is decreased to 45 atm. and the temperature is decreased to 750 K. The new volume as a result of the changes in the pressure and temperature has to be found.
This can be done using the ideal gas law. We have P*V = n*R*T, where P is the pressure, V is the volume, n is the amount of substance, R is a constant and T is the temperature. As the amount of the substance making up the gas n is a constant, the equation can be written as n*R = P*V/ T
Now the initial values and the final values that are known are substituted. Let the final volume be V.
21*78/900 = V*45/750
=> V = 21*78*750/900*45
=> V = 91/3 L
The required volume of the gas after the changes in pressure and temperature are made is 30.33 L.
| 492
| 1,749
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875
| 4
|
CC-MAIN-2016-44
|
latest
|
en
| 0.927947
|
https://byjus.com/rs-aggarwal-solutions/rs-aggarwal-class-6-solutions/
| 1,566,133,346,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-35/segments/1566027313889.29/warc/CC-MAIN-20190818124516-20190818150516-00259.warc.gz
| 390,523,698
| 112,697
|
RS Aggarwal Class 6 Solutions
RS Aggarwal Class 6 All Chapter Solutions in PDF Format - Download Now
RS Aggarwal Class 6 Solutions is an important tool to prepare effectively for Class 6 mathematics examination. RS Aggarwal is considered as one of the best study material to master mathematics. All the important and interesting concepts in mathematics like algebra, geometry, equations, etc are introduced in the 6th standard. CBSE Class 6 is a crucial stage in a student’s academic career and he or she must create a strong foundation in mathematics to excel in the subject. The detailed chapter-wise solution for RS Aggarwal Class 6 is given here so that students can prepare for their exams in a better way.
We at BYJU’S provide complete mathematics solutions for class 6 RS Aggarwal with free pdf download to all the students so that they can efficiently prepare the subject and clear all the basic and fundamental concepts of maths. The RS Aggarwal solution class 6 are prepared by expert teachers at BYJU’S according to the latest syllabus and in a comprehensive and interactive manner so that students do not face any difficulties while learning the concepts.
RS Aggarwal Class 6 Solutions RS Aggarwal Class 6 Solutions Chapter 1 Number System – Exercise 1-1 , 1-2 , 1-3 RS Aggarwal Class 6 SolutionsChapter 2 Factors and Multiples RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers – Exercise 3-1 , 3-2 RS Aggarwal Class 6 Solutions Chapter 4 Integers – Exercise 4-1 , 4-2 , 4-3 RS Aggarwal Class 6 Solutions Chapter 5 Fractions – Exercise 5-1 , 5-2 , 5-3 , 5-4 RS Aggarwal Class 6 Solutions Chapter 6 Simplification RS Aggarwal Class 6 Solutions Chapter 7 Decimals – Exercise 7-1 , 7-2 RS Aggarwal Class 6 Solutions Chapter 8 Algebraic Expressions – Exercise 8-1 , 8-2 RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable – Exercise 9-1 , 9-2 RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method – Exercise 10-1 , 10-2 RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line RS Aggarwal Class 6 Solutions Chapter 12 Parallel Lines RS Aggarwal Class 6 Solutions Chapter 13 Angles and Their Measurement RS Aggarwal Class 6 Solutions Chapter 14 Constructions – Exercise 14-1 RS Aggarwal Class 6 Solutions Chapter 15 Polygons RS Aggarwal Class 6 Solutions Chapter 16 Triangles RS Aggarwal Class 6 Solutions Chapter 17 Quadrilaterals RS Aggarwal Class 6 Solutions Chapter 18 Circles RS Aggarwal Class 6 Solutions Chapter 19 Three-Dimensional Shapes RS Aggarwal Class 6 Solutions Chapter 20 Two-Dimensional Reflection Symmetry RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area – Exercise 21-1 , 21-2 , 21-3 RS Aggarwal Class 6 Solutions Chapter 22 Data Handling RS Aggarwal Class 6 Solutions Chapter 23 Pictograph RS Aggarwal Class 6 Solutions Chapter 24 Bar Graphcrued liabilities, and other similar debts.
Features of RS Aggarwal Class 6 Solutions
The solutions are provided in a well-structured format with several methods of answering problems to ensure a proper understanding of concepts. Some key features of Class 6 RS Aggarwal solutions are given in the points mentioned below.
• The solutions are provided in an interactive and comprehensive manner so that students can understand each and every concept in an easy manner.
• Students will be able to access all the solutions for free and instantly clear any doubts they have.
• Students can download the chapter-wise solutions and prepare for the subject in offline mode.
• Along with solutions, a detailed explanation is provided for every concept which will help students to develop problem-solving skills.
• The best and most effective approach to every question is given which will help students to attempt questions more efficiently in the exam
• Detailed diagrams are also provided to help students visualize the questions and solutions in a better way.
In conclusion, RS Aggarwal is a crucial resource to study mathematics. With the increasing competition to get into the best engineering colleges, students must be thorough with the basics to excel in the competitive entrance exams. Stay tuned with BYJU’S to get detailed RS Aggarwal Solutions for Class 6 to 12.
| 990
| 4,230
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.625
| 4
|
CC-MAIN-2019-35
|
latest
|
en
| 0.869363
|
https://www.khanacademy.org/math/engageny-alg2/alg2-3/alg2-3e-geo-series/v/geometric-series-sum-to-figure-out-mortgage-payments
| 1,675,912,623,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764501066.53/warc/CC-MAIN-20230209014102-20230209044102-00811.warc.gz
| 850,111,793
| 107,865
|
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
# Finite geometric series word problem: mortgage
Figuring out the formula for fixed mortgage payments using the sum of a geometric series. Created by Sal Khan.
## Want to join the conversation?
• At Why do you add 1 to the interest making it 1.005. What is that 1?
• The 1 is the initial amount (200,000). 0.005 is the interest. So to find the new amount owing after one month it is the initial amount plus interest. If you just multiplied 200,000 by 0.005 you would only be left with the interest amount. When you multiply by 1.005 it adds the interest to the starting amount.
• At why is the interest per month 0.005?
• 6% per year = 6% per 12 months
therefore, 6% divided by 12 months = .5% per month
0.5% = .5 / 100 = .005
• \$1200*360 months = \$432,000 is what you end up spending for your \$200,000 house. Just wanted to make sure.
• Yes, it's true. Even with the very low interest rates we have at the moment, over a long period such as the 30 years you might have a mortgage for, the interest adds up to a lot of money and you can end up paying a total of over twice the nominal sum that you bought the house for. Think what the situation was like in the 1970s when interest rates were up to 15% per year!
• If I add additional payments(lets say a second payment each pay period) down on the principal, but everything else (interest, payment value, etc) stays the same, how could I figure out the change in time it would take to pay off the loan?
• L = 200000, P = 2400 -> S = L/P = 83.33; r = 0.995 (stays the same).
S-rS = r-r^n+1 -> r^n+1=0.5804
makes (logarithm rule) -> 0.995^n+1 = 0.995^109.1 -> n=108.1 about 9 years payments.
• Can anyone help me understand how we would extract the formula for an infinite geometric series sum (for a number, that is betwen 0 and 1, that is)? My guess is we would take the same approach, with n -> inifinity. Then for the remaining term we would take it's limit and prove it equals 0...Is this a correct approach and isn't there an easier one?
• yes, (1-r^n)/(1-r) as n-> infinity becomes 1/1-r if -1<r<1 .. easiest approach known to me too :)
• I am confused with your dividing 6% by 12 and getting 0.5%
To go from yearly to monthly do you not use( (1 + 0.06) ** (1/12) - 1) or something like that and then you end up with a monthly rate that is below 0.5%
... as per this link ...
http://www.experiglot.com/2006/06/07/how-to-convert-from-an-annual-rate-to-an-effective-periodic-rate-javascript-calculator/
• I think dividing 6% by 12 is correct. We always have annual percentage rate. It is not effective rate. If I consider rate as effective rate and do 1/12th root then there is no difference between monthly compounding and annual compounding. Both will give same answer.
• Very good video however I have one more fundamental question I have not seen any answers for when I did my research on this equation:
I understand how to find the monthly payment, P, and in this example it was about \$1200, but how do you find out how much of that payment will go towards your principle and how much will go towards paying the bank interest, because as Sal notes in previous videos, your monthly payment P on a fixed rate is always constant for 30 years, but the amount of interest and principle you pay varies significantly as time goes on. Is there a percentage of the payment that you pay towards interest in the beginning years like 80% of your monthly payment that gradually declines to say 0% after 30 years.
• What is the difference between the formula for a geometric series and the formula for a geometric series with a monthly addition?
For instance,
What is the formula for:
- Initial \$10, with monthly 10% interest added, for 10 months.
And the formula for:
- Initial \$10, with monthly 10% interest added, plus a monthly deposit of another 10\$ (after interest) for 10 months.
• Without the monthly addition, we have
after 1 month, 10 ∙ 1.1 dollars
after 2 months, (10 ∙ 1.1) ∙ 1.1 = 10 ∙ 1.1² dollars
after 3 months, 10 ∙ 1.1³ dollars
after 10 months, 10 ∙ 1.1¹⁰ dollars
With the monthly addition, each consecutive deposit has had 1 less month to grow, so after 10 months we will have
10 ∙ 1.1¹⁰ + 10 ∙ 1.1⁹ + 10 ∙ 1.1⁸ + ... + 10 ∙ 1.1² + 10 ∙ 1.1 + 10 dollars
• I have an assignment that relates to mortgage payment. My question is "Phil is trying to figure out what method his boss uses to pay him. The first week he worked, Phil was paid only \$4. The next week he was paid \$15, then \$40, then \$85, and then \$156. Determine how much Phil will be paid for each of the next two weeks?" Please try 2 answer this question!!!!
• Hi. This isn't specifically a mortgage payment, but it does have to do with geometric sequences. (Since you're not adding them, it's not a series, but a sequence).
So....
n1 is 15, n2 is 40, etc. You need to find n6 & n7. Now, you need to find the common ration, which is 40/15 which equals approximately 2.67. Then you multiply 156 by this get your answer (n6) and then you multiply n6 by 2.67 to get n7.
• What is the fraction for 0.5%?
• "Percent" means per hundred. So, .5% means .5 per hundred or .5/100. To clear the decimal out of the fraction, multiply top and bottom by 10. This gives 5/1000. Then reduce (divide top and bottom by 5) to get 1/200
| 1,509
| 5,480
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625
| 4
|
CC-MAIN-2023-06
|
longest
|
en
| 0.94545
|
https://allclearmister.com/how-many-moles-are-in-44-grams/
| 1,721,550,360,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763517648.65/warc/CC-MAIN-20240721060614-20240721090614-00541.warc.gz
| 72,648,686
| 11,197
|
# How many moles are in 44 grams?
## How many moles are in 44 grams?
ONE MOLE
Well, a mass 44⋅g mass of carbon dioxide clearly specifies ONE MOLE of carbon dioxide.
## How many grams are in a mol?
The mass of one mole of a substance is equal to that substance’s molecular weight. For example, the mean molecular weight of water is 18.015 atomic mass units (amu), so one mole of water weight 18.015 grams.
How many moles are present in 44 gram of CO2?
1 mole
Using the formula number of moles = Mass/Mr 44/44=1 mole of CO2 present. (Mr of carbon dioxide is (2*16)+12=44 Now times by Abogadros constant: 1* 6.022*10^23=6.022*10^23 molecules of CO2 are present.
### How do I find moles from grams?
Divide the mass of the substance in grams by its molecular weight. This will give you the number of moles of that substance that are in the specified mass. For 12 g of water, (25 g)/(18.015 g/mol) = 0.666 moles.
### How many moles is 50 grams?
3 moles
For example, 50 grams of oxygen is equal to 3 moles.
What does 1 gram mole mean?
A mass of a substance in grams numerically equal to its molecular weight. Example: A gram-mole of salt (NaCl) is 58.44 grams.
#### How many moles are present in 12g of carbon?
This is the theoretical atomic mass of the Carbon-12 isotope (6 protons and 6 neutrons). This means that the atomic mass or atomic weight (12 grams) of carbon is equal to exactly 1 mole of carbon.
#### How do you calculate moles to grams in chemistry?
Moles to Grams Example Problem. Calculate the molar mass by multiplying the number of atoms of each element in the compound (its subscript) times the atomic mass of the element from the periodic table. Molar mass = (2 x 1.008) + (2 x 15.999) – note the use of more significant figures for oxygen Molar mass = 34.016 grams/mol.
How do you calculate grams from moles of hydrogen peroxide?
To do this, first calculate the molar mass of a sample. Then, multiply it by the number of moles to get an answer in grams: grams of sample = (molar mass) x (moles) For example, find the number of grams in 0.700 moles of hydrogen peroxide, H 2 O 2.
## How do you convert grams to moles of CaSO4?
The weight of the CaSO4alone is 136.139 g. The weight of one-half mole of water is 18.015 x 0.5 = 9.0075 g. The molar mass of CaSO4⋅1⁄2H2O is 145.1465 g/mol. 2) Convert grams to moles: 0.900 g / 145.1465 g/mol = 0.00620 mol (to three sig figs)
## How many moles are in 5988 grams?
It’s easier to work with grams, so convert the mass: 5.988 kg = 5988 g. As you already know how the grams to moles conversion work, find the number of moles: n = 5988 g / 18.015 g/mol = 332.4 mol. You can always use our grams to moles calculator to check the result!
| 788
| 2,708
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.15625
| 4
|
CC-MAIN-2024-30
|
latest
|
en
| 0.883925
|
http://www.instructables.com/id/Conservation-of-Energy-Carnival-Games/
| 1,498,515,364,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128320865.14/warc/CC-MAIN-20170626203042-20170626223042-00345.warc.gz
| 563,399,515
| 15,819
|
The Law of Conservation of Energy says:
In physics, the law of conservation of energy states that the total energy of an isolated system remains constant—it is said to be conserved over time. Energy can neither be created nor destroyed; rather, it transforms from one form to another.
As a teacher this idea has so much that can be done with it, it starts with a class physics demonstration.
So there lots of ways you can go with this one. The same principle in the video applies to some games found at carnivals, fairs, and amusement parks known as Roller Bowler and Swinger. These games like many, are almost impossible to beat. Its not lack of skills, its trying to beat scientific laws.
## Step 1: Swinger
A wooden pin sits on a table. Above it hangs a ball suspended from a rope. The object of the game is to swing the ball past the pin and knock it over as it returns towards you.
## Step 2: Roller Bowler
The challenging game Roller Bowler is played by trying to roll the ball with just enough force so that it goes over the hump and stays in the pocket area on the other side. Just when it looks it is going to stay in the pocket area the ball comes rolling back.
## Step 3: Building the Swinger Game
Materials:
String
Ball
Pin
Table top
Find a way to attach to string to the ball and then mount it above. Depending how you mount the string and type of string used you do offer the potential to have the ball spin and add another variable making the game easier to win. But for the most part the its pretty negligible. You will want to adjust the string to appropriate height so the ball rest just over your table.
I had some light weight bowling balls and Pin that worked well courtesy of the Physical Education teachers.
## Step 4: Roller Bowler Game
For this game I had to scale it down. For starters the bowling ball used in most game is way to expensive, let alone to work to bend and weld the metal your own Roller Bowler game.
So I decide it to scale it down.
You could use a billiards ball, but I decide to go smaller. I wanted my students to be able to create their own little version at a minimal cost.
Materials:
Large Marble
3 feet of metal shelf strips
Screws
Wood board
The key to the game is to make the initial distance long enough that one can exert to much force on the marble causing the marble to eventually come back.
I think about 18 inches is good because it leaves you will enough slack to make a hill and a catch hill at the end.
Depending on how much metal you have to work with yours could vary. 18 Inches seems to work well in terms of proportion of the original game. The hill basically needs to have a slope of somewhere between 1/3 and 1/6.
Next you will want to bend the metal to create a small hill. This is tricky as you do not want to bend the metal to drastically that the marble will catch. It needs a smooth transition. The initial hill size can vary but the second hill definitely will need to larger than the first hill. I was much more successful with a smaller first hill and flatter pitch on to return the ball. It seems you pick up more friction with this design than the standard bowling ball.
Definitely a trial and error process to make this as easy or challenging as you like. I recommend two people here on the hold the metal rail and one play the game consistently. I found bending the ball on a basketball allow me to use the seams as landmarks and gave me some not subtle bends.
A few screws to hold the shelf in place and you a small scale Roller Bowler game!
<p>The pendulum demonstration is misleading. The theatrical professor gave the impression that the tremendous kinetic energy that the bowling ball achieves at the bottom of its arc would be just millimeters from impinging on his chin. But at the top of its arc, the velocity of the bowling ball is ZERO. That's ZERO kinetic energy. Even if he accidentally gave the bowling ball a small initial velocity, the kinetic energy his chin would have to absorb would also be small.</p>
<p>Correct. But it was fun to watch. And probably unforgettable. Unlike most of the physics education in school we got (admittedly I had a very good teacher for a while, but that was an exception).</p>
| 915
| 4,233
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.546875
| 4
|
CC-MAIN-2017-26
|
longest
|
en
| 0.95278
|
https://math.stackexchange.com/questions/1125366/modular-arithmetic-pairs-of-additive-inverse-pairs-and-multiplicative-inverse
| 1,566,298,248,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-35/segments/1566027315321.52/warc/CC-MAIN-20190820092326-20190820114326-00119.warc.gz
| 547,995,707
| 32,561
|
Modular Arithmetic - pairs of additive inverse pairs and multiplicative inverse pairs
I am taking a Cryptography class and we are working on modular arithmetic. I am still unsure on how to find pairs of additive inverse pairs and multiplicative inverse pairs. I've seen some videos and attempted to read about doing so but I find myself confused on what exactly I'm looking for. I've learned about the GCD, euclidean algorithm, but i just can't seem to piece it all together. Any help would be much appreciated. Thanks!
Example: 13 mod 17
How I got this: For the additive inverse, I take the number given (13) and then find the number that would add up to n (n=17), in this case it is 4. For the multiplicative inverse, I take the number given (13) and then add n to it(n=17), and then I find a number that multiples with 13 to be congruent to 1.
I picture a clock with n numbers around it. Im worried when it comes to a much bigger number such as 321^-1 mod 56709.
additive inverse:(13,4) multiplicative inverse: a x b = 1(mod 17) 13 x 4 = 1(mod 17)
I'm working on another example:
list all additive inverse pairs and multiplicative inverse pairs of the sets Z28 and Z28*. So far i have this:
Integers in the set:
Z28 = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27}
Z28* = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27}
Z28 = (0,0),(1,27),(2,26),(3,25),(4,24),(5,23),(6,22),(7,21),(8,20),(9,19),(10,18),(11,17),(12,16),(13,15),(14,14)
Z28* = (1,27),(2,26),(3,25),(4,24),(5,23),(6,22),(7,21),(8,20),(9,19),(10,18),(11,17),(12,16),(13,15),(14,14)
as for the multiplicative inverse pairs, its taking me a while to check each one. Another question is, is there a faster way to find all inverse pairs of a set?
• Show us how far you can get through an example... say, the additive and multiplicative inverses for 13 under mod 17. – Joffan Jan 29 '15 at 18:43
• 13 mod 17 additive inverse:(13,4) multiplicative inverse: a x b = 1(mod 17) 13 x 4 = 1(mod 17) – user655321 Jan 29 '15 at 18:57
• If you include how you got the answer I think it would be really useful. Apologies for choosing an example with the same result on both :-) – Joffan Jan 29 '15 at 19:00
Multiplicative inverses are pairs $(a,b)$ where $a \cdot b = 1 \text{ mod m}$. $2 \cdot 26 \text{ mod m }$ isn't $1$ but $24$.
Multiplicative inverses are thus:
Prelude> [(x,y)|x<-[1..28],y<-[1..28],x*y mod 28 == 1]
[(1,1),(3,19),(5,17),(9,25),(11,23),(13,13),(15,15),(17,5),(19,3),(23,11),(25,9),(27,27)]
If $$s \cdot r = 1$$ then $$(m-r)\cdot(m-s) = 1$$ which means that if you have a pair of multiplicative inverses you get another one for free.
$$(m-r)\cdot(m-s) = m\cdot m -rm -sm +rs = rs = 1$$
It's obvious that $1$ is always a self-inverse because $1 \cdot 1 = 1$. This also means that $m-1$ is necessarily a self-inverse. If $(3,19)$ are inverses then $(28-3,28-19) = (25,9)$ thus if $(3,19)$ is a pair of inverses then $(25,9)$ must be a pair of multiplicative inverses as well.
The extended Euclidean algorithm is probably your best tool for anything but obvious multiplicative inverses. For simple inverses it can be useful to get comfortable with using negative congruences too, so $13\equiv -4 \bmod 17$ and since $4^2\equiv -1$ then $4\cdot -4 \equiv 1 \bmod 17$.
To illustrate the extended Euclidean algorithm I'll use your example of finding $321^{-1} \bmod 56789$ (although not $56709$, since then they have a common factor of $3$). This proceeds effectively by finding a series of equations satisfying $n = 56789s +321t$, starting with $n = 56789$ and $n=321$ and proceeding to smaller values of $n$, until - if possible - $n=1$.
$\begin{array}{c|c} n & s & t & q \\ \hline 56789 & 1 & 0 & \\ 321 & 0 & 1 & 176 \\ 293 & 1 & -176 & 1 \\ 28 & -1 & 177 & 10 \\ 13 & 11 & -1946 & 2 \\ 2 & -23 & 4069 & 6 \\ 1 & 149 & -26360 & \\ \end{array}$
$q$ represents the multiplier to get a close match for the $n$ values of the previous two equations, giving the best reduction in $n$ for the next line.
The final equation here says that $1 = 56789\cdot 149 + 321\cdot(-26360)$. This of course means that $321\cdot(-26360) \equiv 1 \bmod 56789$ or, converting to a positive congruence, $321^{-1} \equiv 30429 \bmod 56789$.
The same process for finding $13^{-1} \bmod 17$ works also of course:
$\begin{array}{c|c} n & s & t & q \\ \hline 17 & 1 & 0 & \\ 13 & 0 & 1 & 1 \\ 4 & 1 & -1 & 3 \\ 1 & -3 & \color{red}{4} & \\ \end{array}$
For finding a full set of multiplicative inverses, we do have the advantage of generally only having to find less than half of them. Your listing of $\Bbb Z_{28}^\times$ includes some numbers that should not be there, since they do not have inverses - all those numbers which share a factor with $28$. The correct listing then is $\Bbb Z_{28}^\times = \{1,3,5,9,11,13,15,17,19,23,25,27\}$. $1$ is the identity already and $-1\equiv 27\bmod 28$ is self-inverse (always true for $1$ less than the modulus). Then we can look for multiples of the small numbers either side of the low multiples of $28$ (all equivalences $\bmod 28$):
$3\cdot 9 =27\equiv -1 \implies 3^{-1}\equiv -9 \equiv 19$
$5\cdot 11 =55\equiv -1 \implies 5^{-1}\equiv -11 \equiv 17$
from $3$'s result $9^{-1}\equiv -3\equiv 25$
from $5$'s result $11^{-1}\equiv -5\equiv 23$
$13\cdot 15 = (14-1)(14+1) \equiv -1 \implies 13^{-1}\equiv -15\equiv 13$
and similarly $15^{-1}\equiv -13\equiv 15$
with the other results being the other half of established inverse pairs, giving the full result:
$\{(1), (3,19), (5,17), (9,25), (11,23), (13), (15), (27)\}$
single items being self-inverse.
| 1,915
| 5,631
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.9375
| 4
|
CC-MAIN-2019-35
|
latest
|
en
| 0.855614
|
https://www.edumple.com/courses/class-2/syllabus
| 1,713,403,143,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00599.warc.gz
| 679,098,761
| 12,554
|
## Maths
• (Chapter 1 - What is Long, What is Round?); Describing shapes of different objects (edges,corners, faces)
• 2. Identifying objects that rolls, slides or both , stacking
• (Chapter 2 - Counting in Groups); Counting in groups
• 2. Ordinal Numbers
• (Chapter 3 - How Much Can You Carry?); Compare weight of two objects(Heavier-lighter) by hefting and using pan balance
• 2. Estimate what object can be lifted by whom
• (Chapter 4 - Counting in Tens); Counting in groups of 10
• (Chapter 5 - Patterns); Making and continuing repeated patterns using shapes
• 2. Continuing number patterns
• (Chapter 6 - Footprints); Tracing 3D objects to get 2D drawing (Also naming the 2D shapes)
• (Chapter 7 - Jugs and Mugs); Measures capacities of everyday objects using non standard units
• 2. Compares capacities of everyday objects using nonstandard units
• 3. Use uniform informal units to measure the capacities of containers by counting the number of times a smaller container can be emptied into bigger container
• (Chapter 8 - Tens and Ones); Decomposing a two digit number into tens and ones
• (Chapter 9 - My Funday); Days of the week (with activities related to weekdays)
• 2. Months of the year (with activities related to the months)
• (Chapter 10 - Add Our Points); Add two digit numbers without regrouping
• 2. Add two digit numbers with regrouping
• 3. Find the missing numbers when the sum is given
• 4. Solve addition problems with 3 one-digit numbers
• 5. Solve addition word problems
• (Chapter 11 - Lines and Lines); Draw different type of Straight lines
• 2. Trace and Draw curved lines
• 3. Patterns using lines
• (Chapter 12 - Give and Take); Subtract two digit numbers without regrouping
• 2. Subtract two digit numbers with regrouping
• 3. Solve subtraction word problems
• (Chapter 13 - The Longest Step); Informal ways of measuring distance
• 2. Find out the best alternative to measure the length of the given object
• Chapter 14 - Birds Come, Birds Go
• 2. Finding the addends when sum is given Finding the minuend and subtrahend when difference is given
• (Chapter 15 - How Many Ponytails?); Interprets data and answers questions on simple tables and pictographs
## English Raindrops
• Chapter 1 - Action Song (Poem)
• Chapter 2 - Our Day
• Chapter 3 - My Family
• Chapter 4 - What’s Going On?
• Chapter 5 - Mohan, The Potter
• Chapter 6 - Rain in Summer (Poem)
• Chapter 7 - My Village
• Chapter 8 - The Work People Do
• Chapter 9 - Work
• Chapter 10 - Our National Symbols
• Chapter 11 - The Festivals of India
• Chapter 12 - The Monkey and The Elephant
• Chapter 13 - Going to the fair
• Chapter 14 - Colours (Poem)
• Chapter 15 - Sikkim
## English Merigold
• 1- First Day at School
• 2- I Am Lucky
• I Want
• 3 -A Smile
• The Wine And The Sun
• 4- Rain
• Storm in the Garden
• 5 - Funny Bunny
• Zoo Manners
• 6 - Curly locks and the Three Bears
• Mr Nobody
• 7 - Make it Shorter
• On My Blackboard I can Draw
• 8- I am the Music Man
• The Mumbai Musicians
• 9- Granny Granny Please Comb my Hair
• The Magic Porridge Pot
• 10- Stranger Talk
• The Grasshopper and the Ant
## EVS
• (Chapter 1 - The invention of the Wheel); DefinItion of Wheel
• 2. Use of wheel by human being.
• (Chapter 2 - Good Habits); Introduction to Good Habit.
• 2. Use of good habits in daily life.
• 3. Importance of Safety habit
• (Chapter 3 - Means of Communication); DefinItion/ Introduction to communication
• 2. Different types or means of communications and their uses.
• (Chapter 4 - Clothes); Introduction and types of clothes
• 2. Different material used for making clothes.
• 3. Different clothes for different weather.
• (Chapter 5 - World around me); Introduction to the term World.
• 2. Different continent in the world.
• 3. Name the oceans in the world.
• (Chapter 6: Understanding Directions); Definition of directions
• 2. How do we find the direction?
• 3. Importance of direction in daily life.
• (Chapter 7: Houses); Definition and types of a house
• 2. Requirement of houses
• 3. Different types of houses for different climate
• (Chapter 8: Services In Neighbourhood); Define and types of Neighbourhood
• 2. Benefits of good neighbour
• 3. Benefits of the services.
• (Chapter 9: Time); Introduction and importance to time
• 2. Time telling/clock reading activity
• 3. 12 hrs and 24 hrs time clock.
• (Chapter 10: Air, Water, Food and Shelter); What do living things need to survive.
• 2. Lifestyle yesterday and today.
• 3. Different type shelters
• 4. Types of food required by living being and source of food.
• 5. Requirement of water and source of water
• (Chapter 11: Get To Know Me); Who am I?
• 2. What am I doing?
• 3. My daily routine
• 4. My family and friends
• 5. My home and city
• (Chapter 12: Means of Transport); Definition and uses to means of transports
• 2. Mode of transport in olden days
• 3. Need of modern day transport
• (Chapter 13: My Family); Definition and members of family
• 2. Relation between the family members
• 3. Types of families
## Hindi
• अध्याय 1 - ऊँट चला
• अध्याय 2 - भालू ने खेली फुटबॉल
• अध्याय 3 - म्याऊँ, म्याऊँ !!
• अध्याय 4 - अधिक बलवान कौन?
• अध्याय 5 - दोस्त की मदद
• अध्याय 6 - बहुत हुआ
• अध्याय 7 - मेरी किताब
• अध्याय 8 - तितली और कली
• अध्याय 9 - बुलबुल
• अध्याय 10 - मीठी सारंगी
• अध्याय 11 - टेसू राजा बीच बाजार
• अध्याय 12 - बस के नीचे बाघ
• अध्याय 13 - सूरज जल्दी आना जी
• अध्याय 14 - नटखट चूहा
• अध्याय 15 - एक्की-दोक्की
## Drawing for kids
• Butterfly
• Scenery for kids
• Jug
• Joker
• Honey bee
• Pineapple
• Big house
• Lotus
• Water melon
• Snail
• Duck
• Frog
• Penguin
• Ship
• Rose
• Bird
• Fish
## Art & Craft for Kids
• Use of colorful origami papers for Paper boat
• Use of colorful origami papers for kite
• Use of colorful origami papers for snail
• Use of colorful origami papers for tree
• Use of colorful origami papers for butterfly
• Use of ice-cream sticks for Fish
• Use of ice-cream sticks for ice-cream
• Use of ice-cream sticks for star
• Use of ice-cream sticks for bird
• Use of ice-cream sticks for basket
## Computers
• (Chapter 1: Discover Computer) ; A smart and useful Machine,Human and Computer
• 2. Uses of Computer, Types of Computer
• (Chapter 2: Parts of Computer); Main Parts of a Computer-Monitor,Mouse, Keyboard,CPU
| 1,840
| 6,252
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.25
| 4
|
CC-MAIN-2024-18
|
latest
|
en
| 0.7931
|
https://metanumbers.com/150403
| 1,624,151,685,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623487653461.74/warc/CC-MAIN-20210619233720-20210620023720-00178.warc.gz
| 354,761,621
| 10,885
|
## 150403
150,403 (one hundred fifty thousand four hundred three) is an odd six-digits composite number following 150402 and preceding 150404. In scientific notation, it is written as 1.50403 × 105. The sum of its digits is 13. It has a total of 4 prime factors and 8 positive divisors. There are 135,520 positive integers (up to 150403) that are relatively prime to 150403.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 13
• Digital Root 4
## Name
Short name 150 thousand 403 one hundred fifty thousand four hundred three
## Notation
Scientific notation 1.50403 × 105 150.403 × 103
## Prime Factorization of 150403
Prime Factorization 113 × 113
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 1243 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 150,403 is 113 × 113. Since it has a total of 4 prime factors, 150,403 is a composite number.
## Divisors of 150403
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 166896 Sum of all the positive divisors of n s(n) 16493 Sum of the proper positive divisors of n A(n) 20862 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 387.818 Returns the nth root of the product of n divisors H(n) 7.20942 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 150,403 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 150,403) is 166,896, the average is 20,862.
## Other Arithmetic Functions (n = 150403)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 135520 Total number of positive integers not greater than n that are coprime to n λ(n) 67760 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 13843 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 135,520 positive integers (less than 150,403) that are coprime with 150,403. And there are approximately 13,843 prime numbers less than or equal to 150,403.
## Divisibility of 150403
m n mod m 2 3 4 5 6 7 8 9 1 1 3 3 1 1 3 4
150,403 is not divisible by any number less than or equal to 9.
• Arithmetic
• Deficient
• Polite
• Frugal
## Base conversion (150403)
Base System Value
2 Binary 100100101110000011
3 Ternary 21122022111
4 Quaternary 210232003
5 Quinary 14303103
6 Senary 3120151
8 Octal 445603
10 Decimal 150403
12 Duodecimal 73057
20 Vigesimal ig03
36 Base36 381v
## Basic calculations (n = 150403)
### Multiplication
n×i
n×2 300806 451209 601612 752015
### Division
ni
n⁄2 75201.5 50134.3 37600.8 30080.6
### Exponentiation
ni
n2 22621062409 3402275649500827 511712464511872883281 76963089799979217264112243
### Nth Root
i√n
2√n 387.818 53.1805 19.6931 10.8505
## 150403 as geometric shapes
### Circle
Diameter 300806 945010 7.10662e+10
### Sphere
Volume 1.42514e+16 2.84265e+11 945010
### Square
Length = n
Perimeter 601612 2.26211e+10 212702
### Cube
Length = n
Surface area 1.35726e+11 3.40228e+15 260506
### Equilateral Triangle
Length = n
Perimeter 451209 9.79521e+09 130253
### Triangular Pyramid
Length = n
Surface area 3.91808e+10 4.00962e+14 122804
## Cryptographic Hash Functions
md5 e94a47ca37c29b6dfd4ae9b8a7c2f730 d327a022f3648fe4da63dd5bd69b9f4e52eb6fe9 84c7a72f53c2223bcfff1ad05fe4b8d568ebd0357c5b24e0c839da92136d86f4 4d3843a9753ebd44ce778b4a184c393600bfa9ca1516f21538706a4bfcf4538c27e99449cf7b96493f1e127792fbd84756b4af3e7df2bc1cac00a89434fa475f af08c0dbb8257cf4868ae3733208e9b891f506d9
| 1,420
| 4,109
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.71875
| 4
|
CC-MAIN-2021-25
|
latest
|
en
| 0.825397
|
https://calculator.tutorpace.com/comparing-fractions-online-tutoring
| 1,623,663,466,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623487611641.26/warc/CC-MAIN-20210614074543-20210614104543-00272.warc.gz
| 169,883,222
| 8,667
|
#### A PHP Error was encountered
Severity: Warning
Message: count(): Parameter must be an array or an object that implements Countable
Filename: controllers/calculator.php
Line Number: 37
Comparing Fractions - Online Calculator - Tutorpace
# Comparing Fractions
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
A fraction is a form where the numbers are written in the numerator and the denominator. When two fractions are compared with each other, we check which fraction is greater and which fraction is smaller among the two. Comparing fractions calculator is an easy and a fun tool to compare fractions.
Example 1: The fraction 1/7 is greater than the fraction 1/9. True or False?
Given two fractions: 1/7 and 1/9
In order to compare two fractions, first their denominators have to be the same.
Multiply the first fraction 1/7 with the denominator of the second fraction
1/7 = (1 * 9)/ (7 * 9) = 9/63
Similarly ? Multiply the second fraction 1/9 with the denominator of the first fraction
1/9 = (1 * 7)/ (9 * 7) = 7/ 63
Now clearly 9 > 7 hence 9/63 > 7/63
1/7 > 1/9 ?True
Example 2: Find the greater fraction by the comparing the two fractions, 2/5 and 3/4.
Given two fractions: 2/5 and 3/4
In order to compare two fractions, first their denominators have to be the same.
Multiply the first fraction 2/5 with the denominator of the second fraction
2/5 = (2 * 4)/ (5 * 4) = 8/20
Similarly ? Multiply the second fraction 3/4 with the denominator of the first fraction
3/4 = (3 * 5)/ (4 * 5) = 15/ 20
Now clearly 15 > 8 hence 15/20 > 8/20
Hence? 3/4 > 2/5
| 471
| 1,654
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.953125
| 4
|
CC-MAIN-2021-25
|
longest
|
en
| 0.859318
|
https://www.enotes.com/homework-help/teacher-has-27-students-her-class-she-asks-540378
| 1,498,639,305,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128323588.51/warc/CC-MAIN-20170628083538-20170628103538-00295.warc.gz
| 884,681,689
| 12,562
|
# a teacher has 27 students in her class she asks the students to form as many groups of 4 as possible. how many students will not be in a group
embizze | High School Teacher | (Level 2) Educator Emeritus
Posted on
There are 27 students in the class. The maximum number of groups of 4 is 6, since 6x4=24 and 7x4=28. (There cannot be 7 groups as there aren't enough students, but there are 6 groups.)
Once 6 groups have been formed, 24 students will belong to some group. (6x4=24). This leaves 3 students without a group (since 27-24=3.)
There will be 3 students who are not in a group of 4.
preethi315 | eNotes Newbie
Posted on
Number of students in the class =24
Number of students required to form a group= 4
Total number of groups that can be formed→ divide 27 by 4 →answer 6 and remainder 3.
Thus six groups will be formed and three students will be left out.
| 245
| 875
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.78125
| 4
|
CC-MAIN-2017-26
|
longest
|
en
| 0.959803
|
https://socratic.org/questions/how-do-you-simplify-3n-2-and-write-it-using-only-positive-exponents
| 1,695,661,885,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233509023.57/warc/CC-MAIN-20230925151539-20230925181539-00551.warc.gz
| 588,948,478
| 5,649
|
How do you simplify 3n^-2 and write it using only positive exponents?
$= \frac{3}{n} ^ 2$
$3 {n}^{-} 2$
$= \frac{3}{n} ^ 2$
| 55
| 124
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.671875
| 4
|
CC-MAIN-2023-40
|
latest
|
en
| 0.461303
|
robertdkirkby.github.io
| 1,721,925,595,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00079.warc.gz
| 429,932,455
| 10,163
|
An Introduction to Chebyshev polynomials and Smolyak grids.
This is an 'interactive' introduction to learn about Chebyshev polynomials and Smolyak Grids. It aims to both teach the concepts, and give an idea how to code them in practice.
To use you will need to also download: https://au.mathworks.com/matlabcentral/fileexchange/50963-smolyak-anisotropic-grid and add it to the Matlab path (or just put them in your active folder).
Proceeds as follows. First: look at some graphs of Chebyshev polynomials (just to have seen what they look like) Second: introduce the idea of curve-fitting. Do this first with standard polynomials as it is likely more intuitive, then with Chebyshev polynomials. Third: how can we do this in higher dimensions? look at inner-product, then discuss limitations of this and how Smolyak grids help us.
1. Chebyshev polynomials of the first kind are defined as Tn(x) = cos(n*arccos(x)).
Can also define them by a recursive formula which makes computing them much faster.
chebyshevT(n,x) represents the nth degree Chebyshev polynomial of the first kind at the point x
Matlab has built in commands for Chebyshev polynomials. We will start with these just to see what Chebyshev polynomials but later switch to the codes of Judd, Maliar, Maliar & Valero (2014).
```chebyshevT(3,0)
% the values of the 0th to 4th order polynomials evaluated at 0
chebyshevT([0,1,2,3,4],0)
% the values of the 3rd order polynomial evaluated at 0,1,2,3
chebyshevT(3,[0,1,2,3])
```
```ans =
0
ans =
1 0 -1 0 1
ans =
0 1 26 99
```
Plot first five Chebyshev polynomials of the first kind
```polyorders=5;
xvalues=-1.5:0.1:1.5;
figure(1)
subplot(2,1,1); plot(xvalues, chebyshevT(0,xvalues))
hold on
for ii=1:polyorders
subplot(2,1,1); plot(xvalues,chebyshevT(ii,xvalues))
end
hold off
axis([-1.5 1.5 -2 2])
grid on
ylabel('T_n(x)')
legend('T_0(x)', 'T_1(x)', 'T_2(x)', 'T_3(x)', 'T_4(x)', 'T_5(x)', 'Location', 'Best')
title('Chebyshev polynomials of the first kind')
```
Chebyshev polynomials of the second kind are defined as follows: U(n,x)=sin((n+1)*a*cos(x))/sin(a*cos(x))
chebyshevU(n,x) represents the nth degree Chebyshev polynomial of the second kind at the point x
```chebyshevU(3,0)
% the values of the 0th to 4th order polynomials evaluated at 0
chebyshevU([0,1,2,3,4],0)
% the values of the 3rd order polynomial evaluated at 0,1,2,3
chebyshevU(3,[0,1,2,3])
```
```ans =
0
ans =
1 0 -1 0 1
ans =
0 4 56 204
```
Plot first five Chebyshev polynomials of the second kind
```polyorders=5;
xvalues=-1.5:0.1:1.5;
subplot(2,1,2); plot(xvalues, chebyshevU(0,xvalues))
hold on
for ii=1:polyorders
subplot(2,1,2); plot(xvalues,chebyshevU(ii,xvalues))
end
hold off
axis([-1.5 1.5 -2 2])
grid on
ylabel('T_n(x)')
legend('T_0(x)', 'T_1(x)', 'T_2(x)', 'T_3(x)', 'T_4(x)', 'T_5(x)', 'Location', 'Best')
title('Chebyshev polynomials of the second kind')
```
2. Say we have some points and we want to fit a curve to them.
```figure (2)
% Specifically, let's take the example of exp(x) on the domain [-1,1]
xgrid=-1:0.1:1; xgrid=xgrid';
ygrid=exp(xgrid);
% We could fit a polynomial of order m.
m=5;
fittedpolynomialcoeffs = polyfit(xgrid,ygrid,m);
% [Normally not such a good choice in practice (for computational reasons), it is just a more
% intuitive example of the concept that we will see with Chebyshev polynomials.]
% Take a look at the fit
% Evaluate the fitted polynomial on our xgrid
ygrid_fittedpolynomial=polyval(fittedpolynomialcoeffs,-1:0.1:1);
% Now graph
subplot(2,1,1); plot(xgrid,ygrid,'*',-1:0.1:1,ygrid_fittedpolynomial,'-')
title('Fitted polynomial')
legend('Original function', 'Fitted polynomial approximation')
% Another alternative, we know that any polynomial of order m can be
% represented as a weighted sum of chebyshev polynomials (of first kind) of
% order m.
% [https://en.wikipedia.org/wiki/Chebyshev_polynomials#Polynomial_in_Chebyshev_form]
% So lets fit a chebyshev polynomial of order m instead (unfortunately
% matlab does not yet have inbuilt function for fitting chebyshev polynomials)
% Implementation follows: https://stackoverflow.com/questions/11993722/need-to-fit-polynomial-using-chebyshev-polynomial-basis
% Fit the chebyshev polynomial of order m
numdatapoints=length(xgrid);
A(:,1) = ones(numdatapoints,1);
if m > 1
A(:,2) = xgrid;
end
if m > 2
for k = 3:m+1
A(:,k) = 2*xgrid.*A(:,k-1) - A(:,k-2); %% recurrence relation
end
end
fittedchebyshevcoeffs = A \ ygrid;
% Take a look at the fit
% Evaluate the fitted chebyshev polynomial on our xgrid
% Follows: [ https://people.sc.fsu.edu/~jburkardt/m_src/chebyshev_polynomial/t_project_value.m ]
b1 = zeros(numdatapoints,1);
b0 = zeros(numdatapoints,1);
for jj=m:-1:0
b2=b1;
b1=b0;
b0=fittedchebyshevcoeffs(jj+1)+2*xgrid.*b1-b2;
end
ygrid_fittedchebyshev= 0.5*(fittedchebyshevcoeffs(1)+b0-b2);
% Now graph
subplot(2,1,2); plot(xgrid,ygrid,'*',-1:0.1:1,ygrid_fittedchebyshev,'-')
title('Fitted Chebyshev polynomial')
legend('Original function', 'Fitted chebyshev approximation')
%
% Side note: in cases such as this example where the actual function exp(x)
% to be evaluated is known we shouldn't just be using our arbitrary xgrid
% to fit the chebyshev polynomial, we can use a better xgrid, or at least a
% faster implementation.
% See implementation in: https://people.sc.fsu.edu/~jburkardt/m_src/chebyshev/chebyshev_coefficients.m
%
% Why would we want to approximate the function using chebyshev polynomials
% rather than just normal polynomials? Both polynomials and chebyshev polynomials
% form a complete basis for the set of infinitely continuously differentiable
% functions on the real line (or more precisely in the case of the chebyshev
% polynomials on the interval [0,1], as this is where they are typically defined,
% but we can always transform the real line to [0,1] and vice-versa anyway.)
% From the perspective of numerical computation Chebyshev polynomials have
% the advantage of being an orthogonal basis (while normal polynomials are
% not) which means that the algebra tends to be simpler and so the
% computation can be done faster. Computation with orthogonal bases also
% tends to be more stable.
% More on nice properties of Chebyshev polynomials in terms of
% Approximation Theory: https://en.wikipedia.org/wiki/Approximation_theory
%
% If for some reason you ever wanted to swich from Chebyshev polynomials
% into just a standard polynomials you could easily calculate the
% coefficients. See pg 197 of
% https://www2.units.it/ipl/students_area/imm2/files/Numerical_Recipes.pdf
% But in practice you are unlikely to ever want to do so.
%
% Remark: As well as standard polynomials, other choices include
% polynomials in logs, and Hermite polynomials (the later are orthogonal
% with respect to normally distributed shocks). Since none of these are
% orthogonal you cannot just fit them as a matrix operation like we did
% here, instead you have to use some kind of distance metric, like
% OLS for standard polynomials, or non-linear least squares for
% polynomials. These other types of polynomials are quite common in
% Economics when performing Parametrized Expectations Approach.
```
Fitting Chebyshev polynomials when the domain is [a,b] rather than [-1,1]
What about if your data are not originally on the interval, [-1,1]? Then just add an initial step to get your data from their existing interval [a,b] onto the interval [-1,1]. zgrid = ((xgrid-min(xgrid))-(max(xgrid)-xgrid))/(max(xgrid)-min(xgrid)); Then just do all the same as above using zgrid and ygrid.
```figure(3)
clear A % just need to clean this out as otherwise the A created above would cause an error
xgrid=linspace(0,3,21)'; % Turn up the interval to [0,10] or [0,100] and 5th order Chebyshev will start to struggle to approximate.
ygrid=exp(xgrid);
% xgrid=-1:0.1:1; xgrid=xgrid';
% ygrid=exp(xgrid);
numdatapoints=length(xgrid);
% Create zgrid by moving xgrid from [a,b] onto [-1,1]
a=min(xgrid); b=max(xgrid);
zgrid = (2*xgrid-a-b)/(b-a);
% Fit chebyshev polynomial of order m
A(:,1) = ones(numdatapoints,1);
if m > 1
A(:,2) = zgrid;
end
if m > 2
for k = 3:m+1
A(:,k) = 2*zgrid.*A(:,k-1) - A(:,k-2); %% recurrence relation
end
end
fittedchebyshevcoeffs = A \ ygrid;
% Note: other than adding zgrid line this is unchanged (A depends on zgrid
% instead of xgrid, but this is largely irrelevant).
% Evaluate the fitted chebyshev polynomial on our zgrid
b1 = zeros(numdatapoints,1);
b0 = zeros(numdatapoints,1);
for jj=m:-1:0
b2=b1;
b1=b0;
b0=fittedchebyshevcoeffs(jj+1)+2*zgrid.*b1-b2; % only change is now uses zgrid
end
ygrid_fittedchebyshev= 0.5*(fittedchebyshevcoeffs(1)+b0-b2);
% Now graph to take a look at the fit
plot(xgrid,ygrid,'*',xgrid,ygrid_fittedchebyshev,'-')
title('Fitted Chebyshev on interval [a,b]')
legend('Original function', 'Fitted chebyshev approximation')
%
% switching problems from [a,b] to [-1,1] and solving there, then switching
% back. It has the advantage that all your codes written to solve models
% can just work with the [-1,1] hypercube (or more accurately, the obvious
% extension of this to [-1,1]^d in higher dimensions, where d is number of dimensions).
```
Higher dimensions
Until now the function to be approximated as just one-dimensional. What about higher dimensional functions?
One basic approach would just be to use the tensor-product of Chebyshev polynomials. This can work fine, but only for a few dimensions. After that the curse of dimensionality kicks in too hard. [eg. say we want a 5th order chebyshev polynomial in each dimension. Then for four dimensions we will have 5^4=625 coefficients, and for 10 dimensions we would have 9765625=5^10 coefficients.]
We can use 'other products' of chebyshev polynomials to reduce the number of coefficients required. Smolyak grids are one approach to doing this. Smolyak grids have a 'level of approximation' parameter, mu, that determines how many grid points (and hence coefficients) will be required. [eg. for the two dimensional case, Smolyak grid has 1,5, and 13 points for mu=0,1,2 respectively. The tensor product of 5 points in each dimension would involve 25=5^2 points.]
It turns out that we can combine the Smolyak grid concept of how to create higher dimensional grids with the Chebyshev polynomial approximation for a single dimension (loosely, a 5th order Chebyshev polynomial can be thought of as 5 grid points used to approximate a single dimensional function)
While the concept is simple enough the algebra is hard and the final formulae are convoluted. Judd, Maliar, Maliar & Valero (2014) provide the formulae.
Lets create a two dimensional example based on just basic Tensor product.
http://ice.uchicago.edu/2011_presentations/Judd/Approximation_ICE11.pdf (pg 22 of 25) NOT IMPLEMENTED SAVING figure(5) FOR THIS
Formula for coeffs is also on bottom of 2nd page of http://icaci.org/files/documents/ICC_proceedings/ICC2009/html/nonref/10_3.pdf
Now a two dimensional example using the smolyak grid with chebyshev polynomials.
Rather than just using tensor product of the 1D chebyshev polynomials. Use JMMV (2014) codes: https://au.mathworks.com/matlabcentral/fileexchange/50963-smolyak-anisotropic-grid An anisotropic grid is used, meaning that the 'level of approximation' for the Smolyak grid can be set differently for different dimensions.
```% 1. Smolyak anisotropic method for 2 dimensions;
% -----------------------------------------------
vector_mus_dimensions = [5,5]; % Introduce the level of approximation in every dimension from 1 to 10; see Section 4 of JMMV (2014)
d=length(vector_mus_dimensions); % Number of dimensions
mu_max = max(vector_mus_dimensions); % Compute the maximum level of approximation across all dimensions
Smolyak_elem_iso = Smolyak_Elem_Isotrop(d,mu_max);
% Construct the matrix of indices of multidimesional Smolyak elements (grid points and polynomial basis functions) that satisfy the usual
% isotropic Smolyak rule for the approximation level equal to "mu_max"
Smol_elem_ani = Smolyak_Elem_Anisotrop(Smolyak_elem_iso,vector_mus_dimensions);
% Select from the matrix of isotropic indices "Smol elem_iso" a subset of indices that correspond to the given anisotropic "vector_mus_dimensions"
Smol_grid_ani = Smolyak_Grid(d,mu_max,Smol_elem_ani);
% Construct the Smolyak grid for the given subindices of anisotropic Smolyak elements "Smol_elem_ani"
Smol_polynom = Smolyak_Polynomial(Smol_grid_ani,d,mu_max,Smol_elem_ani);
% Matrix of the polynomial basis functions evaluated in the grid points
if d==2
figure(6)
scatter(Smol_grid_ani(:,1),Smol_grid_ani(:,2),'filled'),title('Smolyak grid') % Enable it to draw the Smolyak grid
end
% 2. Interpolation of a function y=2*x1 .*exp(-4*x1.^2-16*x2.^2);
% ---------------------------------------------------------------
y_Smolyak = 2*Smol_grid_ani(:,1) .* exp(-4*Smol_grid_ani(:,1).^2 - 16*Smol_grid_ani(:,2).^2);
% Evaluate the function on the Smolyak grid
b = Smol_polynom\y_Smolyak;
% Compute the coefficients of Smolyak interpolating polynomial
size(b)
% Take a look to see how many coefficients this involves
% 3. Compare the true and interpolated functions on a dense grid
% ---------------------------------------------------------------
[x1,x2] = meshgrid(-1:0.05:1, -1:0.1:1);
% Create a uniformly spaced grid of 41 points on [-1,1]x[-1,1]
y_true = 2*x1.* exp(-4*x1.^2 - 16*x2.^2);
% Evaluate the true function on the grid
figure(7)
subplot(1,2,1), surf(x1,x2,y_true),title('True function')
% Plot the true function
for j=1:size(x1(1,:),2)
y_fitted(:,j) = Smolyak_Polynomial([x1(:,j),x2(:,j)],d,mu_max,Smol_elem_ani)*b;
% Evaluate Smolyak interpolating polynomial on the grid
end
subplot(1,2,2), surf(x1,x2,y_fitted),title('Smolyak interpolation')
% Plot Smolyak interpolation
%
% Note: the commands being called here to do the Smolyak anisotropic grids
% based on chebyshev polynomials are not limited to the two dimensions that
% we use here. They work for any number 'd' of dimensions.
%
```
```ans =
145 1
```
A bit of related Theory explaining Smolyak grids and sparse grids in general.
http://web.stanford.edu/~paulcon/slides/Oxford_2012.pdf
How should we do integrals when our functions are in Smolyak-Chebyshev form?
Welcome to the bleeding edge of numerical computation. Quite simply this is a hard problem and one to which we do not yet know the answer.
One obvious answer would just be to evaluate the entire function on a non-sparse grid and then just take the integral there using standard quadrature (or monte-carlo) integration methods. While easy to code it does of course defeat the entire point of using a sparse grid in the first place.
Current approaches include: Gauss-Hermite quadrature: JMM2011 (and JMMV2014) use Guass-Hermite quadrature to take integrals with respect to shocks. Problem is that it only really works if the shocks are independent. Monomials: JMM2011. This is essentially about applying the sparse grids idea to the quadrature approach. Just do quadrature at some points on a sparse grid rather than all the points on a 'product grid'.
Monte-Carlo integration methods do not appear to perform as well as the above mentioned (according to JMM2011).
```% JMM2011: Numerically stable and accurate stochastic simulation approaches for solving dynamic economic models
% JMMV2014: Smolyak method for solving dynamic economic models: Lagrange interpolation, anisotropic grid and adaptive domain
```
Anisotropic grids.
The codes of JMMV2014 allow for the grids to be anisotropic, meaning that the degree of the Chebyshev polynomials can be different in each dimension. Conceptually this is an easy idea to get your head around, but the actual implementation/algebra is tricky.
| 4,486
| 15,735
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.921875
| 4
|
CC-MAIN-2024-30
|
latest
|
en
| 0.875863
|
https://vdocuments.mx/simple-linear-regression1.html
| 1,675,465,727,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500076.87/warc/CC-MAIN-20230203221113-20230204011113-00158.warc.gz
| 601,353,144
| 28,747
|
# simple linear regression[1]
Post on 14-Apr-2018
234 views
Category:
## Documents
Embed Size (px)
TRANSCRIPT
• 7/30/2019 Simple Linear Regression[1]
1/50
1
Linear Regressionand
Correlation Analysis
• 7/30/2019 Simple Linear Regression[1]
2/50
2
Chapter Goals
To understand the methods for displaying and describingrelationship among two variables
Models
Linear regression
Correlations
Frequency tables
Methods for Studying Relationships
Graphical
Scatter plots
Line plots 3-D plots
• 7/30/2019 Simple Linear Regression[1]
3/50
3
Two Quantitative Variables
The response var iable, also called thedependent var iable, is the variable we wantto predict, and is usually denoted by y.
Theexplanatory variab le, also called theindependent var iable, is the variable thatattempts to explain the response, and isdenoted by x.
• 7/30/2019 Simple Linear Regression[1]
4/50
4
Scatter Plots and Correlation
A scatter plot (or scatter diagram) is usedto show the relationship between twovariables
Correlation analysis is used to measurestrength of the association (linearrelationship) between two variables
Only concerned with strength of therelationship
No causal effect is implied
• 7/30/2019 Simple Linear Regression[1]
5/50
5
Example
The following graphshows the scatter plotof Exam 1 score (x)and Exam 2 score (y)for 354 students in aclass.
Is there a relationshipbetween x and y?
• 7/30/2019 Simple Linear Regression[1]
6/50
6
Scatter Plot Examples
y
x
y
x
y
y
x
x
Linear relationships Curvilinear relationships
• 7/30/2019 Simple Linear Regression[1]
7/50
7
Scatter Plot Examples
y
x
y
x
No relationship
(continued)
• 7/30/2019 Simple Linear Regression[1]
8/50
8
Correlation Coefficient
The population correlation coefficient (rho) measures the strength of theassociation between the variables
The sample correlation coefficient ris an estimate of and is used tomeasure the strength of the linear
relationship in the sample observations
(continued)
• 7/30/2019 Simple Linear Regression[1]
9/50
9
Features of and r Unit free
Range between -1 and 1
The closer to -1, the stronger thenegative linear relationship
The closer to 1, the stronger the positivelinear relationship
The closer to 0, the weaker the linearrelationship
• 7/30/2019 Simple Linear Regression[1]
10/50
10
Examples of Approximate r Values
yy
x
y
x
x
y
x
y
x
Tag with appropriate value: -1, -.6, 0, +.3, 1
• 7/30/2019 Simple Linear Regression[1]
11/50
11
Earlier Example
Correlations
1 .400**
.000
366 351
.400** 1
.000
351 356
Pearson Correlation
Sig. (2-tailed)
N
Pearson Correlation
Sig. (2-tailed)
N
Exam1
Exam2
Exam 1 Exam 2
Correlation is significant at the 0.01 level**.
• 7/30/2019 Simple Linear Regression[1]
12/50
12
Questions?
What kind of relationship would you expectin the following situations:
Age (in years) of a car, and its price.
Number of calories consumed per day andweight.
Height and IQ of a person.
• 7/30/2019 Simple Linear Regression[1]
13/50
13
Exercise
Identify the two variables that vary and decidewhich should be the independent variable andwhich should be the dependent variable.Sketch a graph that you think best representsthe relationship between the two variables.
1. The size of a persons vocabulary over his orher lifetime.
2. The distance from the ceiling to the tip of theminute hand of a clock hung on the wall.
• 7/30/2019 Simple Linear Regression[1]
14/50
14
Introduction to Regression Analysis
Regression analysis is used to:
Predict the value of a dependent variable basedon the value of at least one independent variable.
Explain the impact of changes in an independentvariable on the dependent variable.
Dependent variable: the variable we wish to
explain.Independent variable: the variable used to
explain the dependent variable.
• 7/30/2019 Simple Linear Regression[1]
15/50
15
Simple Linear RegressionModel
Only one independent variable, x.
Relationship between x and y isdescribed by a linear function.
Changes in y are assumed to becaused by changes in x.
• 7/30/2019 Simple Linear Regression[1]
16/50
16
Types of Regression ModelsPositive Linear Relationship
Negative Linear Relationship
Relationship NOT Linear
No Relationship
• 7/30/2019 Simple Linear Regression[1]
17/50
17
ni ,...,2,1; ii10i XY
Linear component
Population Linear Regression
The population regression model:
Populationy intercept
PopulationSlopeCoefficient
RandomError term,
or residual
Dependent
Variable
IndependentVariable
Random Errorcomponent
• 7/30/2019 Simple Linear Regression[1]
18/50
18
Linear Regression Assumptions
The Error terms i, i=1, 2. , n areindependent and i ~ Normal (0,
2).
The Error terms i, i=1, 2. , n have constantvariance 2.
The underlying relationship between the Xvariable and the Y variable is linear.
• 7/30/2019 Simple Linear Regression[1]
19/50
19
Population Linear Regression(continued)
Random Errorfor this xi value
Y
X
Observed Valueof yi for xi
Predicted Valueof yi for xi
ii10i XY
xi
Slope = 1
Y-Intercept = 0
i
• 7/30/2019 Simple Linear Regression[1]
20/50
20
i10i XbbY
The sample regression line provides an estimateof the population regression function.
Estimated Regression Model
Estimate ofthe regressiony-intercept
Estimate of theregression slopeEstimated
(or predicted)y value Independent
variable
The individual random error terms ei have a mean of zero
The Regression Function: iiii XEXYE 1010 )()(
• 7/30/2019 Simple Linear Regression[1]
21/50
21
Earlier Example
• 7/30/2019 Simple Linear Regression[1]
22/50
22
ResidualA residualis the difference between the observed
response yiand the predicted response i. Thus, for
each pair of observations (xi, yi), the ith residual isei= yi i= yi (b0+ b1xi)
Least Squares Criterion b0 and b1 are obtained by finding the
values of b0 and b1 that minimize the sumof the squared residuals.
2
10
22
x))b(b(y
)y(ye
• 7/30/2019 Simple Linear Regression[1]
23/50
23
b0 is the estimated average value of
y when the value of x is zero.
b1 is the estimated change in the
average value of y as a result of a
one-unit change in x.
Interpretation of theSlope and the Intercept
• 7/30/2019 Simple Linear Regression[1]
24/50
24
The Least Squares Equation
The formulas for b1
and b0
are:
algebraic equivalent:
nxx
n
yxxy
b2
2
1
)(
21 )(
))((
xx
yyxxb
xbyb 10 and
• 7/30/2019 Simple Linear Regression[1]
25/50
25
Finding the Least Squares Equation
The coefficients b0 and b1 willusually be found using computersoftware, such as Excel, Minitab, or
SPSS.
Other regression measures will alsobe computed as part of computer-based regression analysis.
Si l Li R i
• 7/30/2019 Simple Linear Regression[1]
26/50
26
Simple Linear RegressionExample
A real estate agent wishes to examine therelationship between the selling price of a
home and its size (measured in square feet)
A random sample of 10 houses is selected
Dependent variable (y) = house price in
\$1000s
Independent variable (x) = square feet
• 7/30/2019 Simple Linear Regression[1]
27/50
27
Sample Data for House PriceModel
House Price in \$1000s(y)
Square Feet(x)
245 1400
312 1600
279 1700
308 1875
199 1100
219 1550405 2350
324 2450
319 1425
255 1700
• 7/30/2019 Simple Linear Regression[1]
28/50
28
Model Summary
.762a .581 .528 41.33032Model1
R R Square
R Square
Std. Error of
the Estimate
Predictors: (Constant), Square Feeta.
Coefficientsa
98.248 58.033 1.693 .129
.110 .033 .762 3.329 .010
(Constant)
Square Feet
Model
1
B Std. Error
UnstandardizedCoefficients
Beta
StandardizedCoefficients
t Sig.
Dependent Variable: House Pricea.
SPSS Output
The regression equation is:feet)(square0.11098.248pricehouse
• 7/30/2019 Simple Linear Regression[1]
29/50
29
0
50
100
150
200
250300
350
400
450
0 500 1000 1500 2000 2500 3000
Square Feet
HousePrice(\$1
000s)
Graphical Presentation
House price model: scatter plot andregression line
feet)(square0.11098.248pricehouse
Slope
= 0.110
Intercept
= 98.248
• 7/30/2019 Simple Linear Regression[1]
30/50
30
Interpretation of the Intercept, b0
b0 is the estimated average value of Y when
the value of X is zero (if x = 0 is in the range
of observed x values)
Here, no houses had 0 square feet, so
b0 = 98.24833 just indicates that, for houses
within the range of sizes observed,
\$98,248.33 is the portion of the house price
not explained by square feet
feet)(square0.11098.248pricehouse
Interpretation of the
• 7/30/2019 Simple Linear Regression[1]
31/50
31
Interpretation of theSlope Coefficient, b1
b1 measures the estimated change in
the average value of Y as a result ofa one-unit change in X
Here, b1 = .10977 tells us that the averagevalue of a house increases by .10977(\$1000)
= \$109.77, on average, for each additional
one square foot of size
feet)(square0.1097798.24833pricehouse
• 7/30/2019 Simple Linear Regression[1]
32/50
32
Least Squares Regression Properties
The sum of the residuals from the least
squares regression line is 0 ( ).
The sum of the squared residuals is a
minimum (minimized ).
The simple regression line always passes
through the mean of the y variable and the
mean of the x variable.
The least squares coefficients b0 and b1 are
unbiased estimates of 0 and 1.
0)( yy
2)( yy
• 7/30/2019 Simple Linear Regression[1]
33/50
33
Exercise
The growth of children from early childhood through
adolescence generally follows a linear pattern. Data onthe heights of female Americans during childhood, fromfour to nine years old, were compiled and the leastsquares regression line was obtained as = 32 + 2.4xwhere is the predicted height in inches, andxis age inyears.
Interpret the value of the estimated slope b1= 2. 4. Would interpretation of the value of the estimated
y-intercept, b0= 32, make sense here? What would you predict the height to be for a female
American at 8 years old? What would you predict the height to be for a femaleAmerican at 25 years old? How does the quality of thisanswer compare to the previous question?
• 7/30/2019 Simple Linear Regression[1]
34/50
34
The coefficient of determination is theportion of the total variation in thedependent variable that is explained byvariation in the independent variable.
The coefficient of determination is alsocalled R-squared and is denoted as R2
Coefficient of Determination, R2
1R0 2
• 7/30/2019 Simple Linear Regression[1]
35/50
35
Coefficient of Determination, R2
(continued)
Note: In the single independent variable case, the coefficientof determination is
where:R2 = Coefficient of determinationr = Simple correlation coefficient
22
rR
E l f A i t
• 7/30/2019 Simple Linear Regression[1]
36/50
36
Examples of ApproximateR2 Values
y
x
y
x
y
x
y
x
E l f A i t
• 7/30/2019 Simple Linear Regression[1]
37/50
37
Examples of ApproximateR2 Values
R2 = 0
No linear relationshipbetween x and y:
The value of Y does not
depend on x. (None of thevariation in y is explainedby variation in x)
y
xR2 = 0
SPSS O t t
• 7/30/2019 Simple Linear Regression[1]
38/50
38
SPSS OutputModel Summary
.762a .581 .528 41.33032
Model
1
R R Square
R Square
Std. Error of
the Estimate
Predictors: (Constant), Square Feeta.
ANOVAb
18934.935 1 18934.935 11.085 .010a
13665.565 8 1708.196
32600.500 9
Regression
Residual
Total
Model
1
Sum of
Squares df Mean Square F Sig.
Predictors: (Constant), Square Feeta.
Dependent Variable: House Priceb.
Coefficientsa
98.248 58.033 1.693 .129
.110 .033 .762 3.329 .010
(Constant)
Square Feet
Model1
B Std. Error
UnstandardizedCoefficients
Beta
StandardizedCoefficients
t Sig.
Dependent Variable: House Pricea.
• 7/30/2019 Simple Linear Regression[1]
39/50
39
Standard Error of Estimate
The standard deviation of the variation ofobservations around the regression line iscalled the standard error of estimate
The standard error of the regressionslope coefficient (b1) is given by sb1
s
• 7/30/2019 Simple Linear Regression[1]
40/50
40
Model Summary
.762a .581 .528 41.33032
Model
1
R R Square
R Square
Std. Error of
the Estimate
Predictors: (Constant), Square Feeta.
Coefficientsa
98.248 58.033 1.693 .129
.110 .033 .762 3.329 .010
(Constant)
Square Feet
Model
1
B Std. Error
Unstandardized
Coefficients
Beta
Standardized
Coefficients
t Sig.
Dependent Variable: House Pricea.
SPSS Output
41.33032s
0.03297s1b
• 7/30/2019 Simple Linear Regression[1]
41/50
41
Comparing Standard Errors
y
y y
x
x
x
y
x
1bssmall
1
bslarge
ssmall
slarge
Variation of observed y valuesfrom the regression line Variation in the slope of regressionlines from different possible samples
I f b t th Sl t T t
• 7/30/2019 Simple Linear Regression[1]
42/50
42
Inference about the Slope: t-Test t-test for a population slope
Is there a linear relationship between X and Y? Null and alternate hypotheses
H0: 1 = 0 (no linear relationship)
H1: 1 0 (linear relationship does exist)
Test statistic:
Degree of Freedom:
1b
11
s
bt
2nd.f.
where:
b1 = Sample regression slopecoefficient
1
= Hypothesized slope
sb1 = Estimator of the standarderror of the slope
• 7/30/2019 Simple Linear Regression[1]
43/50
43
House Pricein \$1000s
(y)
Square Feet(x)
245 1400
312 1600
279 1700
308 1875
199 1100
219 1550
405 2350
324 2450
319 1425
255 1700
(sq.ft.)0.109898.25pricehouse
Estimated Regression Equation:
The slope of this model is 0.1098
Does square footage of the houseaffect its sales price?
Example: Inference about the Slope:t Test
(continued)
I f b t th Sl
• 7/30/2019 Simple Linear Regression[1]
44/50
44
Inferences about the Slope:tTest Example - Continue
H0: 1 = 0
HA: 1 0
Test Statistic: t = 3.329
There is sufficient evidencethat square footage affects
house price
From Excel output:
Coeff icients Standard Error t Stat P-value
Intercept 98.24833 58.03348 1.69296 0.12892Square Feet 0.10977 0.03297 3.32938 0.01039
1bs t*
b1
Decision: Reject H0
Conclusion:
Reject H0Reject H0
a/2=.025
-t/2Do not reject H0
0t(1-/2)
a/2=.025
-2.3060 2.3060 3.329
d.f. = 10-2 = 8
• 7/30/2019 Simple Linear Regression[1]
45/50
45
Confidence Interval for the Slope
Confidence Interval Estimate of the Slope:
Excel Printout for House Prices:
At 95% level of confidence, the confidence interval forthe slope is (0.0337, 0.1858)
1b/211sb a t
Coeff icients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 98.24833 58.03348 1.69296 0.12892 -35.57720 232.07386
Square Feet 0.10977 0.03297 3.32938 0.01039 0.03374 0.18580
d.f. = n - 2
C f f S
• 7/30/2019 Simple Linear Regression[1]
46/50
46
Confidence Interval for the Slope
Since the units of the house price variable is\$1000s, we are 95% confident that the averageimpact on sales price is between \$33.70 and\$185.80 per square foot of house size
Coeff ic ient
sStandard
Error t Stat P-value Low er 95% Upper
95%
Intercept 98.24833 58.03348 1.69296 0.12892 -35.57720 232.07386
Square Feet 0.10977 0.03297 3.32938 0.01039 0.03374 0.18580
This 95% confidence interval does not include 0.
Conclusion: There is a significant relationship betweenhouse price and square feet at the .05 level of significance
R id l A l i
• 7/30/2019 Simple Linear Regression[1]
47/50
47
Residual Analysis
Purposes
Examine for linearity assumption
Examine for constant variance for alllevels of x
Evaluate normal distribution assumption
Graphical Analysis of Residuals
Can plot residuals vs. x Can create histogram of residuals tocheck for normality
• 7/30/2019 Simple Linear Regression[1]
48/50
48
Residual Analysis for Linearity
Not Linear Linear
x
residua
ls
x
y
x
y
x
residua
ls
• 7/30/2019 Simple Linear Regression[1]
49/50
49
Residual Analysis for Constant Variance
Non-constant variance Constant variance
x x
y
x x
y
residuals
residuals
• 7/30/2019 Simple Linear Regression[1]
50/50
50
House Price Model Residual Plot
-60
-40
-20
0
20
40
60
80
0 1000 2000 3000
Square Feet
Residuals
Example: Residual Output
RESIDUAL OUTPUTPredicted
House
Price Residuals
1 251.92316 -6.923162
2 273.87671 38.123293 284.85348 -5.853484
4 304.06284 3.937162
5 218.99284 -19.99284
6 268.38832 -49.38832
7 356.20251 48.79749
8 367.17929 -43.17929
9 254.6674 64.33264
10 284.85348 -29.85348
| 4,863
| 16,487
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.09375
| 4
|
CC-MAIN-2023-06
|
latest
|
en
| 0.815802
|
https://stillproud.org/consider-the-internal-reflection-of-light-at-the-interface-between-water-and-ice/
| 1,656,465,321,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656103619185.32/warc/CC-MAIN-20220628233925-20220629023925-00639.warc.gz
| 595,442,616
| 5,329
|
theta two. And also we're gonna be addressing for theta one, and also we're gonna speak to it theta c, crucial angle. Let's division both sides by n one here, and also we have sine the the crucial angle climate is the table of contents of refraction that the second medium, which is ice, times sine that the edge of refraction, which is 90 degrees, and divide the by the index of refraction of the early medium, i beg your pardon is water. And sine that 90 is one, and so we'll just say, n 2 over n one here. And also so, the an important angle is the station sine that the index of refraction of ice separated by the table of contents of refraction the water. For this reason that's station sine of 1.309 separated by 1.333, which gives an edge of 79.11 degrees.">
You are watching: Consider the internal reflection of light at the interface between water and ice.
This is college Physics Answers through Shaun Dychko. We want to understand at what minimum angle will we get full internal reflection at the interface in between water and ice. So it"s water here, and ice down listed below here. And, complete internal reflection, that just starts to take place at this vital angle. Strict speaking, crucial angle method the angle of refraction is 90 levels parallel come the interface, and it"s just past that the you get complete internal reflection. So this represents basically the border and also we"ll speak to that the minimum angle such that you obtain this total internal reflection. Snell"s legislation says the the initial index of refraction times sine of the edge of incidence — i beg your pardon is the crucial angle in this case — amounts to the 2nd index that refraction multiply by sine of the angle of refraction. And the angle of refraction is 90 degrees, for this reason that"s theta two. And we"re gonna be solving for theta one, and we"re gonna call it theta c, critical angle. Let"s divide both sides by n one here, and we have sine of the crucial angle then is the table of contents of refraction that the second medium, i m sorry is ice, times sine the the edge of refraction, which is 90 degrees, and also divide that by the index of refraction the the early medium, i beg your pardon is water. And also sine the 90 is one, and so we"ll just say, n two over n one here. And so, the an important angle is the train station sine of the index of refraction the ice divided by the table of contents of refraction that water. For this reason that"s station sine the 1.309 divided by 1.333, which offers an angle of 79.11 degrees.
1PE2PE3PE4PE5PE6PE7PE8PE9PE10PE11PE12PE13PE14PE15PE16PE17PE19PE20PE21PE22PE23PE24PE25PE26PE27PE28PE29PE30PE31PE32PE33PE34PE35PE36PE37PE38PE39PE40PE41PE42PE43PE44PE45PE46PE47PE48PE49PE50PE51PE52PE53PE54PE55PE56PE57PE58PE59PE60PE61PE62PE63PE
| 709
| 2,792
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.125
| 4
|
CC-MAIN-2022-27
|
latest
|
en
| 0.933002
|
https://www.4options.in/2022/07/real-number-class-10.html
| 1,718,697,233,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00400.warc.gz
| 539,039,201
| 36,646
|
Type Here to Get Search Results !
# REAL NUMBER-CLASS 10
CONTENT LIST
## REAL NUMBER
### Introduction
A set of all rational number & irretional number is called real number.
All the arithmetic operations can be performed on real numbers.
It can be represented in the number line.
It can be both positive or negative and are denoted by the symbol “R”.
Set of real numbers consists of:
1. Natural Numbers: All numbers such as 1, 2, 3, 4, 5, 6,…..…
2. Whole Numbers: All natural numbers including 0.
3. Integers : All whole numbers and negative of all natural numbers.
4. Rational Numbers :Numbers which can be written in the form of p/q, where q≠0 such as 1,3/5 etc
5. Irrational Numbers : Numbers that are not rational and cannot be written in the form of p/q like √2,√3
### Euclid’s Division Lemma
#### What is Lemma?
A lemma is a proven statement used for proving another statement.
#### What is Algorithm?
An algorithm is a series of well defined steps which gives a procedure for solving a type of problem.
In Mathematics, lemma represent Dividend.
We know that , Dividend = (Divisor × Quotient) + Remainder.
For example, for two positive numbers 58 and 8, Euclid's division lemma holds true in the form of 58 = (7 × 8) + 2.
Euclid's Division Lemma says, if we have two positive integers a and b, then there would be whole numbers q and r that satisfy the equation: a = bq + r, (where 0 ≤ r < b)
Here,
a is called as dividend.
b is called as divisor.
q is called as quotient and r is the remainder
#### How to find HCF using Euclid’s Algorithm?
Follow the simple and easy procedures to find the HCF using Euclid’s Algorithm of positive integers a, b where a>b.
Step 1: On applying Euclid’s division lemma to integers a and b, we get other two whole numbers q and r such that, a = bq+r ; 0 r < b.
Step 2: If r =0, then b is the HCF of a, b. If r ≠ 0 then apply Euclid’s Division Lemma to b and r.
Step 3: Continue the steps until the remainder is zero (r =0).
Step 4: When r =0, the divisor at this stage is called the HCF of given numbers a and b.
### The Fundamental Theorem of Arithmetic
The fundamental theorem of arithmetic states that, every composite number can be expressed (factorised) as the product of primes and this factorisation is unique, apart from the order in which the prime factors occur.
For example, the number 45 can be written in the form of its prime factors as:
45 = 3×3×5 =3 ²X5
Here, 3 and 5 are the prime factors of 45
### Revisiting Rational Numbers and Their Decimal Expansions
Any number x is said to be an irrational number if it cannot be expressed in the form of pq where q≠0.
√5, √7, √2 etc. are irrational numbers.
Irrational numbers have non-terminating and non-repeating decimal representation.
### Properties of irrestional numbers
The sum or difference of any two irrational numbers is also rational or an irrational number.
The sum or difference of any rational and any irrational number is also an irrational number.
Product of any rational and an irrational number is also an irrational number.
Product of any two irrational numbers is also rational or an irrational number.
### Theorem:
If p be a prime number. If p divides a², then p also divides a, where a is a positive integer.
### Decimal Representation of Rational Numbers
Any rational number can have two types of decimal representations :
1. Terminating
2. Non-terminating but repeating
#### How to find a fraction is terminating or repeating without long devision method.
Let we have a the rational number p/q, where q ≠ 0.
Take the prime factorization of denominator (q).
If we can write the denominator (q) in terms of 2^m x 5^n or 2^m or 5^n only , then the rational number p/q will be in the form as shown below and have terminating else repeating .
| 934
| 3,812
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.46875
| 4
|
CC-MAIN-2024-26
|
latest
|
en
| 0.899217
|
https://www.math-dictionary.com/zero-property-of-addition.html
| 1,669,917,352,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710829.5/warc/CC-MAIN-20221201153700-20221201183700-00194.warc.gz
| 938,245,704
| 7,826
|
# Zero Property of Addition - Definition and Examples
The zero property of addition, also called identity property for addition, states that the sum of any number and zero is equal to the number.
In other words, for any number a, a + 0 = a
## Examples showing the zero property of addition
5 + 0 = 5
-20 + 0 = -20
2/3 + 0 = 2/3
89584 + 0 = 89584
The zero property of addition can also apply to numerical and algebraic expressions.
The sum of any numerical or algebraic expression and 0 is equal to the numerical expression or the algebraic expression.
## More examples showing the zero property of addition
(25 + 8) + 0 = (25 + 8)
(6x4 - 12x3 + x + 14) + 0 = (6x4 - 12x3 + x + 14)
(10 ÷ 2 + 5) + 0 = (10 ÷ 2 + 5)
(e4xy - y) + 0 = (e4xy - y)
The zero property of addition is also called addition property of zero.
| 264
| 827
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.96875
| 4
|
CC-MAIN-2022-49
|
longest
|
en
| 0.838624
|
https://www.physicsforums.com/threads/adding-moments-together.312199/
| 1,652,881,488,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662522270.37/warc/CC-MAIN-20220518115411-20220518145411-00340.warc.gz
| 1,095,394,414
| 15,252
|
Hi people,
I'm a bit confused as to when we can and cannot add 2 moments together. From what I understand, we CAN add them together if they are both about the same origin but CANNOT if they are about different origins.
But when drawing and calculating bending moment diagrams, you have to find the bending moment at each point/section of the structure but doing so seems to require you to add external moments and internal moments together even though they do not rotate about the same point.
Could someone please clarify this for me?
Thanks
tiny-tim
Homework Helper
bump?
A torque is the vector cross product of a lever arm and a force. If the force had a component parallel to the lever arm, that part would not be a torque. So the vector cross product is the right equation. Thus said, having two moments about the same origin is insufficient. They must be on the same axis (defined by the vector cross product).
If you are calculating the torque about a roof overhang, a weight at the end of the overhang puts twice as much torque at the roof support as it does half way out. Thus a 10 Newton weight at the end of a 10 meter overhang produces 100 Newton-meters of torque at the roof support, but only 5 N-m half way out. Another 10 Newton weight half way out produces an additional 5 N-m at the roof support, but none half way out. So here we have 15 N-m at the roof support, and only 5 N-m half way out.
Mapes
Homework Helper
Gold Member
Hi deufo, welcome to PF. If an external moment is specified to be applied to the structure (e.g., 5 N-m at a point), it is added to the bending moment and applies to the whole structure. Note the difference between this scenario and that of an internal force that acts via a lever arm, producing a different moment in different parts of the structure. Does this answer your question?
Hi..
1.Understand while finding Bending moment on beams at different locations,you are actually finding out bending moment at that point because of remote load.
You should add moments on beam if it is already experiencing a moment component throughout.
Ex:
Bending moment because of remote load =P*L
Uniform bending moment on beam =M
Total moment =M+/-PL
Where M is uniform bending moment on beam
2.You can add moments if they are in the same plane only.
Last edited:
| 515
| 2,304
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.640625
| 4
|
CC-MAIN-2022-21
|
latest
|
en
| 0.936142
|
https://math.stackexchange.com/questions/1620949/compound-distribution-normal-distribution-with-log-normally-distributed-varian
| 1,717,085,804,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00409.warc.gz
| 316,385,815
| 36,385
|
# Compound Distribution — Normal Distribution with Log Normally Distributed Variance
Could someone please point me to a source or suggest ways in which we can obtain the Distribution, Density Functions, Expected Value, etc. of a Normal Distribution whose variance is distributed Log Normally.
# Given,
$$X \sim N[\mu_{X},e^{Y}]$$
$$Y \sim N[\mu_{Y},\sigma^2_{Y}]$$
# To Determine,
$$f_{X}(x), F_{X}(x), E(X), E(X^{2})$$
Related Question when Mean is Normal
Compound Distribution --- Normal Distribution with Normally Distributed Mean
Related General Question
Starting with the above special case, it quickly becomes apparent there are many combinations possible. Hence was wondering if there were general techniques to derive the density, distribution function, expected value, higher moments, conditional expectations etc. of compound distributions and some source where certain combinations and results therein were given with detailed steps and complete proofs: https://math.stackexchange.com/questions/1614212/compound-distributions-basic-techniques-and-key-general-results-from-first-p
• See comment there.
– Did
Jan 22, 2016 at 7:11
Hint:
$X \mid Y \sim \mathcal{N}\left(\mu_{X}, Y\right)$, no?
So $$f_{X}(x) = \int_{-\infty}^{\infty}f_{X\mid Y}(x \mid y)f_{Y}(y)\text{ d}y$$
$F_{X}$ can be easily found from this.
In general, $$\mathbb{E}[g(X)] = \mathbb{E}\left[\mathbb{E}[g(X) \mid Y]\right] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(x)f_{X \mid Y}(x \mid y)f_{Y}(y)\text{ d}x\text{ d}y$$
• Just how does one propose to deal with the fact that when $Y \le 0$, $X \mid Y$ is not well-defined? Jan 21, 2016 at 12:50
• @heropup You know, that's an interesting question. I'm not sure. And now that I think about it, I wonder if the OP's question is even valid. I'm looking back at some conjugate prior notes, and I remember now that the additional distribution assumption is done on the mean, rather than the variance. Jan 21, 2016 at 12:53
• @Clarinetist Thanks for the pointer. Perhaps we can set the variance to be log normal. Jan 21, 2016 at 12:59
• @user249613 No matter what you end up doing, you're likely going to have a very disgusting expression that you'll likely have to numerically integrate. This isn't going to be pretty. Jan 21, 2016 at 13:00
• @user249613 Google normal-normal conjugate prior. Jan 21, 2016 at 13:02
| 665
| 2,362
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.828125
| 4
|
CC-MAIN-2024-22
|
latest
|
en
| 0.873874
|
https://en.academic.ru/dic.nsf/enwiki/1261022/Fermat_point
| 1,582,457,649,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00439.warc.gz
| 364,225,477
| 17,058
|
# Fermat point
Fermat point
In geometry, the first Fermat point, or simply the Fermat point, also called Torricelli point, is the solution to the problem of finding a point F inside a triangle ABC such that the total distance from the three vertices to point F is the minimum possible. It is so named because this problem is first raised by Fermat in a private letter.
Construction
To locate the Fermat point of a triangle with largest angle at most 120°:
#Construct three regular triangles out of the three sides of the given triangle.
#For each new vertex of the regular triangle, draw a line from it to the opposite triangle's vertex.
#These three lines intersect at the Fermat point.
The reader should be careful not to confuse the Fermat point with the first isogonic center also known as "X"(13).Entry X(13) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers] ] Whilst the two points are one and the same provided no angle of the triangle exceeds 120° the coincidence ends there. When a triangle has an angle greater than 120° the first isogonic center always lies outside the triangle whereas the Fermat point is sited at the obtuse angled vertex. This means that in such a triangle the two points are never the same.
Geometry
Given any Euclidean triangle ABC and an arbitrary point P let d(P) = PA+PB+PC. The aim of this section is to identify a point P0 such that d(P0) < d(P) for all P ≠ P0. If such a point exists then it will be the Fermat point. In what follows Δ will denote the points inside the triangle and will be taken to include its boundary Ω.
A key result that will be used is the dogleg rule which asserts that if a triangle and a polygon have one side in common and the rest of the triangle lies inside the polygon then the triangle has a shorter perimeter than the polygon.
[ If AB is the common side extend AC to cut the polygon at X. Then by the triangle inequality the polygon perimeter > AB+AX+XB = AB+AC+CX+XB ≥ AB+AC+BC. ]
Let P be any point outside Δ. Associate each vertex with its remote zone; that is, the half-plane beyond the (extended) opposite side. These 3 zones cover the entire plane except for Δ itself and P clearly lies in either one or two of them. If P is in two (say the B and C zones) then setting P' = A implies d(P') = d(A) < d(P) by the dogleg rule. Alternatively if P is in only one zone, say the A-zone, then d(P') < d(P) where P' is the intersection of AP and BC. So for every point P outside Δ there exists a point P' in Ω such that d(P') < d(P).
Case 1. The triangle has an angle ≥ 120o.
Without loss of generality suppose that the angle at A is ≥ 120o. Construct the equilateral triangle AFB and for any point P in Δ (except A itself) construct Q so that the triangle AQP is equilateral and has the orientation shown. Then the triangle ABP is a 60o rotation of the triangle AFQ about A so these two triangles are congruent and it follows that d(P) = CP+PQ+QF which is simply the length of the path CPQF. As P is constrained to lie within ABC, by the dogleg rule the length of this path exceeds AC+AF = d(A). Therefore d(A) < d(P) for all P є Δ, P ≠ A. Now allow P to range outside Δ. From above a point P' є Ω exists such that d(P') < d(P) and as d(A) ≤ d (P') it follows that d(A) < d(P) for all P outside Δ. Thus d(A) < d(P) for all P ≠ A which means that A is the Fermat point of Δ. In other words the Fermat point lies at the obtuse angled vertex.
Case 2. The triangle has no angle ≥ 120o.
Let P be any point inside Δ and construct the equilateral triangle CPQ. Then CQD is a 60o rotation of CPB about C so d(P) = PA+PB+PC = AP+PQ+QD which is simply the length of the path APQD. Let P0 be the point where AD and CF intersect. This point is commonly called the first isogonic center. By the angular restriction P0 lies inside Δ moreover BCF is a 60o rotation of BDA about B so Q0 must lie somewhere on AD. Since CDB = 60o it follows that Q0 lies between P0 and D which means AP0Q0D is a straight line so d(P0) = AD. Moreover if P ≠ P0 then either P0 or Q0 won't lie on AD which means d(P0) = AD < d(P). Now allow P to range outside Δ. From above a point P' є Ω exists such that d(P') < d(P) and as d(P0) ≤ d(P') it follows that d(P0) < d(P) for all P outside Δ. That means P0 is the Fermat point of Δ. In other words the Fermat point is coincident with the first isogonic center.
Derivation
Since the time the problem first appeared, many methods to arrive at the solution have been developed. One method is to simply rotate BEC, where E is an arbitrary point, 60º counter-clockwise. Now the distance to minimize is the same as the path AEE'C'. Obviously the solution is when it is a straight line, from which the construction method can be derived.
Proof
This proof will show that the three lines are concurrent. One proof, using properties of concyclic points, is as follows:
Suppose RC and BQ intersect at F, and two lines, AF and AP, are drawn. We aim to prove that AFP is a straight line.
Because AR = AB and AC = AQ by construction,
: $angle RAC = angle RAB + angle BAC$
: $angle BAQ = angle BAC + angle CAQ$
Since $angle RAB$ and $angle CAQ$ equal 60º, which are interior angles of an equilateral triangle, $angle RAC = angle BAQ$. This implies that triangles RAC and BAQ are congruent. Hence $angle ARF = angle ABF$ and $angle AQF = angle ACF$. By converse of angle in the same segment, ARBF and AFCQ are both concyclic.
Thus $angle AFB = angle AFC = angle BFC = 120$º. Because $angle BFC$ and $angle BPC$ add up to 180º, BPCF is also concyclic. Hence $angle BFP = angle BCP = 60$º. Because $angle BFP + angle BFA = 180$º, AFP is a straight line.
Q.E.D.
Another proof
Another approach to find a point within the triangle, from where sum of the distances to the vertices of triangle is minimum, is to use one of the optimization (mathematics) methods. In particular, method of the lagrange multipliers and the law of cosines.
We draw lines from the point within the triangle to its vertices and call them X, Y and Z. Also, let the lengths of these lines be x, y, and z, respectively. Let the acute angle between X and Y be α, Y and Z be β. Then the angle between X and Z is (2π − α − β). Using the method of lagrange multipiers we have to find the minimum of the lagrangian, which is expressed as:
: "x" + "y" + "z" + : "λ"1 ("x"2 + "y"2 − 2"xy" cos("α") − "a"2) + : "λ"2 ("y"2 + "z"2 − 2"yz" cos(β) − "b"2) + : "λ"3 ("z"2 + "x"2 − 2"zx" cos("α" + "β") − "c"2) .
where "a", "b" and "c" are the lengths of the sides of the triangle.
Calculating the partial derivatives δ/δx, δ/δy, δ/δz, δ/δα, δ/δβ gives a system of 5 equations:
: δ/δx: 1 + λ1(2x − 2y cos(α)) + λ3(2x − 2z cos(α + β)) = 0
: δ/δy: 1 + λ1(2y − 2x cos(α)) + λ2(2y − 2z cos(β)) = 0
: δ/δz: 1 + λ2(2z − 2y cos(β)) + λ3(2z − 2x cos(α + β)) = 0
: δ/δα: λ1y sin(α) + λ3z sin(α + β) = 0
: δ/δβ: λ2y sin(β) + λ3x sin(α + β) = 0
After some algebraic manipulations equations for α and β separate from the rest of the parameters, giving:
: sin(α) = sin(β)
: sin(α + β) = −sin(β)
that gives: α = β = 120o
Q.E.D.
Note: if one of the vertices of triangle has angle not less than 120o, than Fermat point is at that vertex.
X(13) properties
* In case the largest angle of the triangle is not larger than 120º, the point minimizes the total distance from the three vertices to this point.
* The internal angles brought about by this point, that is, $angle AFB$, $angle BFC$, and $angle CFA$, are all equal to 120º.
* The circumcircles of the three regular triangles in the construction intersect at this point.
* Trilinear coordinates for the 1st isogonic center, "X"(13)::csc("A" + π/3) : csc("B" + π/3) : csc("C" + π/3), or, equivalently,:sec("A" − π/6) : sec("B" − π/6) : sec("C" − π/6).
* Trilinear coordinates for the 2nd isogonic center, "X"(14)::csc("A" − π/3) : csc("B" − π/3) : csc("C" − π/3), or, equivalently,:sec("A" + π/6) : sec("B" + π/6) : sec("C" + π/6). [Entry X(14) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers] ]
* The isogonal conjugate of the 1st isogonic center is the 1st isodynamic point, "X"(15)::sin("A" + π/3) : sin("B" + π/3) : sin("C" + π/3). [Entry X(15) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers] ]
* The isogonal conjugate of the 2nd isogonic center is the 2nd isodynamic point, "X"(16)::sin("A" − π/3) : sin("B" − π/3) : sin("C" − π/3). [Entry X(16) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers] ]
* The following triangles are equilateral::antipedal triangle of "X"(13):antipedal triangle of "X"(14):pedal triangle of the "X"(15):pedal triangle of the "X"(16):circumcevian triangle of "X"(15):circumcevian triangle of "X"(16)
* The lines "X"(13)"X"(15) and "X"(14)"X"(16) are parallel to the Euler line. The three lines meet at the Euler infinity point, "X"(30).
* The 1st isogonic center, 2nd isogonic center, circumcenter, nine-point center lie on a Lester circle.
History
This question was proposed by Fermat, as a challenge to Evangelista Torricelli. He solved the problem in a similar way to Fermat's, albeit using intersection of the circumcircles of the three regular triangles instead. His pupil, Viviani, published the solution in 1659. [MathWorld|urlname=FermatPoints |title=Fermat Points]
ee also
*Geometric median or Fermat–Weber point, the point minimizing the sum of distances to more than three given points.
*Lester's theorem
References
*" [http://demonstrations.wolfram.com/FermatPoint/ Fermat Point] " by Chris Boucher, The Wolfram Demonstrations Project.
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Point de Fermat — Le point de Fermat, du nom du mathématicien français Pierre de Fermat est un point remarquable d un triangle en géométrie euclidienne. Il est également appelé point de Torricelli, son existence correspond au théorème de Schruttka (aussi connu… … Wikipédia en Français
• FERMAT (P. de) — Le mathématicien français Pierre de Fermat est à l’origine des branches les plus fécondes des mathématiques: géométrie analytique, dont il découvre le principe indépendamment de Descartes, calcul infinitésimal, calcul des probabilités, théorie… … Encyclopédie Universelle
• Fermat's theorem (stationary points) — Fermat s theorem is a theorem in real analysis, named after Pierre de Fermat. It gives a method to find local maxima and minima of differentiable functions by showing that every local extremum of the function is a stationary point (the function… … Wikipedia
• Fermat's Last Theorem — is the name of the statement in number theory that:: It is impossible to separate any power higher than the second into two like powers,or, more precisely:: If an integer n is greater than 2, then the equation a^n + b^n = c^n has no solutions in… … Wikipedia
• Fermat's factorization method — is based on the representation of an odd integer as the difference of two squares: :N = a^2 b^2. That difference is algebraically factorable as (a+b)(a b); if neither factor equals one, it is a proper factorization of N .Each odd number has such… … Wikipedia
• Fermat's principle — In optics, Fermat s principle or the principle of least time is the idea that the path taken between two points by a ray of light is the path that can be traversed in the least time. This principle is sometimes taken as the definition of a ray of … Wikipedia
• Fermat's principle — Optics. the law that the path taken by a ray of light in going from one point to another point will be the path that requires the least time. [1885 90; named after P. de FERMAT] * * * ▪ optics in optics, statement that light traveling… … Universalium
• Fermat (computer algebra system) — Infobox Software name = Fermat caption = developer = Robert H. Lewis latest release version = 3.9.7 latest release date = May 6 2008 programming language = C operating system = Mac OS X, Mac OS, Linux, Unix, Windows genre = Computer algebra… … Wikipedia
• Fermat (crater) — lunar crater data caption=Location of Fermat crater. latitude=22.6 N or S=S longitude=19.8 E or W=E diameter=39 km depth=2.0 km colong=20 eponym=Pierre de FermatFermat is a lunar impact crater located to the west of the Rupes Altai escarpment. To … Wikipedia
• fermat's principle — (ˈ)fer|mäz noun Usage: usually capitalized F Etymology: after Pierre de Fermat died 1665 French mathematician, its formulator : a statement in optics: the path actually followed by a ray of light undergoing reflection or refraction is one of… … Useful english dictionary
### Share the article and excerpts
Do a right-click on the link above
| 3,634
| 12,725
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.53125
| 5
|
CC-MAIN-2020-10
|
latest
|
en
| 0.898326
|
https://www.onlinemath4all.com/sat-math-questions-on-absolute-value.html
| 1,721,815,015,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763518198.93/warc/CC-MAIN-20240724075911-20240724105911-00097.warc.gz
| 800,181,404
| 7,974
|
# SAT MATH QUESTIONS ON ABSOLUTE VALUE
Question 1 :
If |1 - x| > 4 and x is positive, what is one possible value of x?
|1 - x| > 4
1 - x > 4 or 1 - x < -4
1 - x > 4(1 - x) - 1 > 4 - 11 - x - 1 > 4 - 1-x > 3x < -3 1 - x < -4(1 - x) - 1 < -4 - 11 - x - 1 < -5-x < -5x > 5
Since x is positive, one possible value of x satisfying x > 5 is 6. That is
x = 6
Question 2 :
0 < |5/x| < 1
Which of the following values of x satisfies the inequality above?
A) 2
B) 3
C) -3
D) -8
0 < |5/x| < 1
A) 2 :
Substitute x = 2 in |5/x|.
|5/2| = 5/2 > 1
When x = 2, the value of |5/x| is not between 0 and 1.
2 does not work.
B) 3 :
Substitute x = 3 in |5/x|.
|5/3| = 5/3 > 1
When x = 3, the value of |5/x| is not between 0 and 1.
3 does not work.
C) -3 :
Substitute x = -3 in |5/x|.
|5/(-3)| = |-5/3|
= 5/3 > 1
When x = -3, the value of |5/x| is not between 0 and 1.
-3 does not work.
D) -8 :
Substitute x = 8 in |5/x|.
|5/8| = |5/8|
= 5/8
The value 5/8 is between 0 and 1, that is
0 < 5/8 < 1
When x = -8, the value of |5/x| is between 0 and 1.
-8 works.
The correct answer choice is (D).
Question 3 :
A bakery standardizes muffins to weigh between 1¾ and 2¼ounces. If m is the weight of muffin from this bakery, which of the following inequalities expresses the following possible values of m?
A) |m - 1¾| < ¼
B) | m - 2| < ¼
C) |m - 2| < ½
D) |m - 1¾| < ½
It's kind of solving absolute value inequality in reverse order. Here, the boundary values of m are given and we have to find the absolute value inequality which gives the given boundary values ¾ and 2¼.
Find the middle value of the two boundary values.
Middle value of 1¾ and 2¼ is their average :
= (1¾ + 2¼÷ 2
= 4 ÷ 2
= 2
Since 1¾ and 2¼ are the boundary values of m,
1¾ < m < 2¼
Subtract the middle value.
(1¾ - 2) < (m - 2) < (2¼ - 2)
-¼ < (m - 2) < ¼
We know that
-A < x < A ----> |x| < A
So, we have
< (m - 2) < ¼ ----> |m - 2| < ¼
The correct answer choice is (B).
Question 4 :
If |x + 3| < 2, which of the following could be the value of |x|?
A) 1
B) 4
C) 6
D) 10
|x + 3| < 2
x + 3 < 2 or x + 3 > -2
x < -1 or x > -5
Combined inequality of x < -1 and x > -5 :
-5 < x < -1
The possible integer values of x are -4, -3, -2.
Substitute x = -4, -3, -2, -1 in |x|.
|-4| = 4
|-3| = 3
|-2| = 2
The correct answer choice is (B).
Question 5 :
Which of the following could be the equation of the function graphed in the xy-plane above?
A) y = -|x - 1| - 2
B) y = |x - 1| - 2
C) y = |x - 1| + 2
D) y = |x + 1| - 2
The graph of y = |x| is a V-shape centered at the origin.
The graph above is also V-shaped. But, it is shifted 1 unit to the right and 2 units down.
After shifting 1 unit to the right, y = |x| becomes
y = |x - 1|
Further shifting of 2 units down,
y = |x - 1| - 2
The correct answer choice is (B).
Question 6 :
If |x - 10| = y, x < 10, then which of the following is equivalent to (y - x)?
A) -10
B) 10
C) 2y - 10
D) 10 - 2y
Assume some value for x such that x < 10, say x = 3.
In |x - 10| = y, substitute x = 3 and find the value of y.
y = |3 - 10|
y = |-7|
y = 7
Find the value of (y - x).
y - x = 7 - 3
y - x = 4
Among the given answer choices, pick the one which gives the result 4 when y = 7.
The correct answer choice is (C).
Note :
Instead of assuming x = 3, you can assume any value for x such that x < 10 and solve this problem. You will get the same answer.
Kindly mail your feedback to v4formath@gmail.com
## Recent Articles
1. ### Order of Operations Worksheet
Jul 24, 24 03:02 AM
Order of Operations Worksheet
2. ### Order of Operations
Jul 24, 24 03:01 AM
Order of Operations - Concept - Solved Examples
| 1,443
| 3,690
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.3125
| 4
|
CC-MAIN-2024-30
|
latest
|
en
| 0.814344
|
https://minuteshours.com/339-69-hours-in-hours-and-minutes
| 1,716,788,895,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00231.warc.gz
| 343,287,064
| 6,066
|
339.69 hours in hours and minutes
Result
339.69 hours equals 339 hours and 41.4 minutes
You can also convert 339.69 hours to minutes.
How to convert 339.69 hours to hours and minutes?
In order to convert 339.69 hours to hours and minutes we can take the decimal part of 339.69 hours and convert it into minutes. In this case we need to convert 0.69 hours to minutes.
We know that 1 hours equals 60 minutes, therefore to convert 0.69 hours to minutes we simply multiply 0.69 hours by 60 minutes:
0.69 hours × 60 minutes = 41.4 minutes
Finally, we can say that 339.69 hours in hours and minutes is equivalent to 339 hours and 41.4 minutes:
339.69 hours = 339 hours and 41.4 minutes
Three hundred thirty-nine point six nine hours is equal to three hundred thirty-nine hours and forty-one point four minutes.
Conversion table
For quick reference purposes, below is the hours and hours to minutes conversion table:
hours(hr) hours(hr) minutes(min)
340.69 hours 340 hours 41.4 minutes
341.69 hours 341 hours 41.4 minutes
342.69 hours 342 hours 41.4 minutes
343.69 hours 343 hours 41.4 minutes
344.69 hours 344 hours 41.4 minutes
345.69 hours 345 hours 41.4 minutes
346.69 hours 346 hours 41.4 minutes
347.69 hours 347 hours 41.4 minutes
348.69 hours 348 hours 41.4 minutes
349.69 hours 349 hours 41.4 minutes
Units definitions
The units involved in this conversion are hours and minutes. This is how they are defined:
Hours
An hour (symbol: h, also abbreviated hr) is a unit of time conventionally reckoned as 1⁄24 of a day and scientifically reckoned between 3,599 and 3,601 seconds. In the modern metric system, hours are an accepted unit of time defined as 3,600 atomic seconds. There are 60 minutes in an hour, and 24 hours in a day.
Minutes
The minute is a unit of time usually equal to 1⁄60 (the first sexagesimal fraction) of an hour, or 60 seconds. In the UTC time standard, a minute on rare occasions has 61 seconds, a consequence of leap seconds (there is a provision to insert a negative leap second, which would result in a 59-second minute, but this has never happened in more than 40 years under this system). Although not an SI unit, the minute is accepted for use with SI units. The SI symbol for minute or minutes is min (without a dot).
| 600
| 2,267
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.828125
| 4
|
CC-MAIN-2024-22
|
latest
|
en
| 0.86701
|
https://en.academic.ru/dic.nsf/enwiki/408491/Annualized_failure_rate
| 1,597,300,665,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439738960.69/warc/CC-MAIN-20200813043927-20200813073927-00276.warc.gz
| 280,873,723
| 13,096
|
Annualized failure rate
Annualized failure rate
Annualized failure rate (AFR) is the relation between the mean time between failure (MTBF) and the assumed hours that a device is run per year, expressed in percent.
Example
A disk drive's MTBF number may be 1,200,000 hours and the disk drive may be running 24 hours a day, seven days a week [Annualized Failure Rate (AFR) and Mean Time Between Failures (MTBF) in: [http://www.seagate.com/staticfiles/support/disc/manuals/enterprise/Barracuda%20ES/SATA/100424667b.pdf Seagate Barracuda ES SATA Product Manual] , Page 29, Chapter 2.12: Reliability] . One year has 8,760 hours.
:$frac\left\{1,200,000,hours\right\}\left\{8760 , hours / year\right\}=136.9863 , years$
then take the reciprocal of 136.9863 years
:$frac\left\{1 , failure\right\}\left\{136.9863 , years\right\} imes 100%=0.73%$
You can expect about 0.73 percent of the population of these disk drives to fail in the average year.
Another example: A disk drive's MTBF number may be 700,000 hours and the disk drive may be running 2400 hours a year [Annualized Failure Rate (AFR) and Mean Time Between Failures (MTBF) in: [http://www.seagate.com/staticfiles/support/disc/manuals/desktop/Barracuda%207200.10/100402371f.pdf Seagate Barracuda 7200.10 SATA Product Manual] , Page 42, Chapter 2.12: Reliability] .
:$frac\left\{700,000 , hours\right\}\left\{2400 , hours / year\right\}=291.6667 , years$
then take the reciprocal of 291.6667 years
:$frac\left\{1 , failure\right\}\left\{291.6667 , years\right\} imes 100%=0.34%$
You can expect about 0.34 percent of the population of these disk drives to fail in the average year.
Now assuming you let the same disk run 24 hours a day, 7 days a week:
:$frac\left\{700,000 , hours\right\}\left\{8760 hours / year\right\}=79.9087 , years$
then take the reciprocal of 79.9087 years
:$frac\left\{1 , failure\right\}\left\{79.9087 , years\right\} imes 100%=1.25%$
You can expect about 1.25 percent of the population of these disk drives to fail in the average year.
ee also
* Failure rate
References
Wikimedia Foundation. 2010.
Look at other dictionaries:
• Failure rate — is the frequency with which an engineered system or component fails, expressed for example in failures per hour. It is often denoted by the Greek letter λ (lambda) and is important in reliability engineering. The failure rate of a system usually… … Wikipedia
• Hard disk drive — Hard drive redirects here. For other uses, see Hard drive (disambiguation). Hard disk drive Mechanical interior of a modern hard disk drive Date invented 24 December 1954 [1] … Wikipedia
• AFR — may stand for:* AFR, a fictional documentary depicting the death of the Danish Prime Minister Anders Fogh Rasmussen * Africa, UNESCO region * The ICAO code for Air France * Air Fuel Ratio * Alternate Frame Rendering * American Family Radio *… … Wikipedia
• AFR — steht als Abkürzung für: Alternate Frame Rendering (Technik), Multi GPU Technik zur Lastverteilung der Rechenarbeit auf mehrere Grafikchips Alternate Frame Rendering (Modus), Render Modus, der bei Multi GPU Verfahren eingesetzt wird Automatische… … Deutsch Wikipedia
• AFR — aeronaut. abbr. Air Force Regulation engin. abbr. Amplitude Frequency Responses milit. abbr. Air Force Regulation telecom. abbr. Amplitude Frequency Responses abbr. Annualized Failure Rate (Maxtor, HDD) … United dictionary of abbreviations and acronyms
• Economic Affairs — ▪ 2006 Introduction In 2005 rising U.S. deficits, tight monetary policies, and higher oil prices triggered by hurricane damage in the Gulf of Mexico were moderating influences on the world economy and on U.S. stock markets, but some other… … Universalium
• Force of mortality — In actuarial science, force of mortality represents the instantaneous rate of mortality at a certain age measured on an annualized basis. It is identical in concept to failure rate, also called hazard function, in reliability theory.In a life… … Wikipedia
• Toughbook — is the trademarked brand name owned by Panasonic and marketed by their international brand name Panasonic. Toughbook refers to its line of semi rugged and rugged laptop computers. In 2005, Panasonic added the Toughbook Arbitrator mobile digital… … Wikipedia
• japan — japanner, n. /jeuh pan /, n., adj., v., japanned, japanning. n. 1. any of various hard, durable, black varnishes, originally from Japan, for coating wood, metal, or other surfaces. 2. work varnished and figured in the Japanese manner. 3. Japans,… … Universalium
• Japan — /jeuh pan /, n. 1. a constitutional monarchy on a chain of islands off the E coast of Asia: main islands, Hokkaido, Honshu, Kyushu, and Shikoku. 125,716,637; 141,529 sq. mi. (366,560 sq. km). Cap.: Tokyo. Japanese, Nihon, Nippon. 2. Sea of, the… … Universalium
| 1,293
| 4,807
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.796875
| 4
|
CC-MAIN-2020-34
|
latest
|
en
| 0.757963
|
https://www.shaalaa.com/question-bank-solutions/factorising-polynomial-completely-after-obtaining-one-factor-factor-theorem-if-x-1-x-2-are-factors-x3-a-1-x2-b-2-x-6-find-values-b-then-factorise-given-expression-completely_29323
| 1,563,534,354,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-30/segments/1563195526210.32/warc/CC-MAIN-20190719095313-20190719121313-00466.warc.gz
| 844,214,816
| 10,813
|
Share
Books Shortlist
If (X + 1) and (X – 2) Are Factors of X3 + (A + 1)X2 – (B – 2)X – 6, Find the Values of a and B. and Then, Factorise the Given Expression Completely. - ICSE Class 10 - Mathematics
ConceptFactorising a Polynomial Completely After Obtaining One Factor by Factor Theorem
Question
If (x + 1) and (x – 2) are factors of x3 + (a + 1)x2 – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.
Solution
Let f(x) = x3 + (a + 1)x2 – (b – 2)x – 6
Since, (x + 1) is a factor of f(x).
Remainder = f(-1) = 0
(-1)3 + (a + 1)(-1)2 – (b – 2) (-1) – 6 = 0
-1 + (a + 1) + (b – 2) – 6 = 0
a + b – 8 = 0 …(i)
Since, (x – 2) is a factor of f(x).
Remainder = f(2) = 0
(2)3 + (a + 1) (2)2 – (b – 2) (2) – 6 = 0
8 + 4a + 4 – 2b + 4 – 6 = 0
4a – 2b + 10 = 0
2a – b + 5 = 0 …(ii)
Adding (i) and (ii), we get,
3a – 3 = 0
a = 1
Substituting the value of a in (i), we get,
1 + b – 8 = 0
b = 7
∴ f(x) = x3 + 2x2 – 5x – 6
Now, (x + 1) and (x – 2) are factors of f(x). Hence, (x + 1) (x – 2) = x2 – x – 2 is a factor of f(x).
f(x) = x3 + 2x2 – 5x – 6 = (x + 1) (x – 2) (x + 3)
Is there an error in this question or solution?
APPEARS IN
Solution If (X + 1) and (X – 2) Are Factors of X3 + (A + 1)X2 – (B – 2)X – 6, Find the Values of a and B. and Then, Factorise the Given Expression Completely. Concept: Factorising a Polynomial Completely After Obtaining One Factor by Factor Theorem.
S
| 646
| 1,429
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.5625
| 5
|
CC-MAIN-2019-30
|
longest
|
en
| 0.78964
|
https://nrich.maths.org/public/leg.php?code=71&cl=4&cldcmpid=6382
| 1,508,272,776,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187822488.34/warc/CC-MAIN-20171017200905-20171017220905-00863.warc.gz
| 813,840,176
| 10,137
|
# Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to Mind Your Ps and Qs:
Filter by: Content type:
Stage:
Challenge level:
### There are 184 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Mind Your Ps and Qs
##### Stage: 5 Short Challenge Level:
Sort these mathematical propositions into a series of 8 correct statements.
### Notty Logic
##### Stage: 5 Challenge Level:
Have a go at being mathematically negative, by negating these statements.
### Picturing Pythagorean Triples
##### Stage: 4 and 5
This article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.
### Proof of Pick's Theorem
##### Stage: 5 Challenge Level:
Follow the hints and prove Pick's Theorem.
### Whole Number Dynamics II
##### Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### Whole Number Dynamics III
##### Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
### Whole Number Dynamics V
##### Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
##### Stage: 5 Challenge Level:
Find all positive integers a and b for which the two equations: x^2-ax+b = 0 and x^2-bx+a = 0 both have positive integer solutions.
### Where Do We Get Our Feet Wet?
##### Stage: 5
Professor Korner has generously supported school mathematics for more than 30 years and has been a good friend to NRICH since it started.
### Long Short
##### Stage: 4 Challenge Level:
What can you say about the lengths of the sides of a quadrilateral whose vertices are on a unit circle?
### To Prove or Not to Prove
##### Stage: 4 and 5
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### Sperner's Lemma
##### Stage: 5
An article about the strategy for playing The Triangle Game which appears on the NRICH site. It contains a simple lemma about labelling a grid of equilateral triangles within a triangular frame.
### Proof Sorter - Quadratic Equation
##### Stage: 4 and 5 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
### Binomial
##### Stage: 5 Challenge Level:
By considering powers of (1+x), show that the sum of the squares of the binomial coefficients from 0 to n is 2nCn
### Magic Squares II
##### Stage: 4 and 5
An article which gives an account of some properties of magic squares.
### Square Mean
##### Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
### Target Six
##### Stage: 5 Challenge Level:
Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.
### Euclid's Algorithm II
##### Stage: 5
We continue the discussion given in Euclid's Algorithm I, and here we shall discover when an equation of the form ax+by=c has no solutions, and when it has infinitely many solutions.
### Direct Logic
##### Stage: 5 Challenge Level:
Can you work through these direct proofs, using our interactive proof sorters?
### Iffy Logic
##### Stage: 4 and 5 Challenge Level:
Can you rearrange the cards to make a series of correct mathematical statements?
### The Great Weights Puzzle
##### Stage: 4 Challenge Level:
You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?
### Contrary Logic
##### Stage: 5 Challenge Level:
Can you invert the logic to prove these statements?
### The Clue Is in the Question
##### Stage: 5 Challenge Level:
This problem is a sequence of linked mini-challenges leading up to the proof of a difficult final challenge, encouraging you to think mathematically. Starting with one of the mini-challenges, how. . . .
### Dodgy Proofs
##### Stage: 5 Challenge Level:
These proofs are wrong. Can you see why?
### Advent Calendar 2011 - Secondary
##### Stage: 3, 4 and 5 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
### Multiplication Square
##### Stage: 4 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Some Circuits in Graph or Network Theory
##### Stage: 4 and 5
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
### Impossible Sandwiches
##### Stage: 3, 4 and 5
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
### Integral Inequality
##### Stage: 5 Challenge Level:
An inequality involving integrals of squares of functions.
### Mouhefanggai
##### Stage: 4
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
### Unit Interval
##### Stage: 4 and 5 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
### Sprouts Explained
##### Stage: 2, 3, 4 and 5
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
### Water Pistols
##### Stage: 5 Challenge Level:
With n people anywhere in a field each shoots a water pistol at the nearest person. In general who gets wet? What difference does it make if n is odd or even?
### More Sums of Squares
##### Stage: 5
Tom writes about expressing numbers as the sums of three squares.
##### Stage: 5 Challenge Level:
Find all real solutions of the equation (x^2-7x+11)^(x^2-11x+30) = 1.
### An Introduction to Number Theory
##### Stage: 5
An introduction to some beautiful results of Number Theory
### Find the Fake
##### Stage: 4 Challenge Level:
There are 12 identical looking coins, one of which is a fake. The counterfeit coin is of a different weight to the rest. What is the minimum number of weighings needed to locate the fake coin?
### Tetra Inequalities
##### Stage: 5 Challenge Level:
Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.
### Diverging
##### Stage: 5 Challenge Level:
Show that for natural numbers x and y if x/y > 1 then x/y>(x+1)/(y+1}>1. Hence prove that the product for i=1 to n of [(2i)/(2i-1)] tends to infinity as n tends to infinity.
### Zig Zag
##### Stage: 4 Challenge Level:
Four identical right angled triangles are drawn on the sides of a square. Two face out, two face in. Why do the four vertices marked with dots lie on one line?
### Tree Graphs
##### Stage: 5 Challenge Level:
A connected graph is a graph in which we can get from any vertex to any other by travelling along the edges. A tree is a connected graph with no closed circuits (or loops. Prove that every tree has. . . .
##### Stage: 4 and 5 Challenge Level:
Which of these roads will satisfy a Munchkin builder?
### Continued Fractions II
##### Stage: 5
In this article we show that every whole number can be written as a continued fraction of the form k/(1+k/(1+k/...)).
### Modulus Arithmetic and a Solution to Dirisibly Yours
##### Stage: 5
Peter Zimmerman from Mill Hill County High School in Barnet, London gives a neat proof that: 5^(2n+1) + 11^(2n+1) + 17^(2n+1) is divisible by 33 for every non negative integer n.
### Proofs with Pictures
##### Stage: 5
Some diagrammatic 'proofs' of algebraic identities and inequalities.
### Composite Notions
##### Stage: 4 Challenge Level:
A composite number is one that is neither prime nor 1. Show that 10201 is composite in any base.
### Mediant
##### Stage: 4 Challenge Level:
If you take two tests and get a marks out of a maximum b in the first and c marks out of d in the second, does the mediant (a+c)/(b+d)lie between the results for the two tests separately.
### A Computer Program to Find Magic Squares
##### Stage: 5
This follows up the 'magic Squares for Special Occasions' article which tells you you to create a 4by4 magicsquare with a special date on the top line using no negative numbers and no repeats.
### Transitivity
##### Stage: 5
Suppose A always beats B and B always beats C, then would you expect A to beat C? Not always! What seems obvious is not always true. Results always need to be proved in mathematics.
| 2,229
| 9,307
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.3125
| 4
|
CC-MAIN-2017-43
|
longest
|
en
| 0.868055
|
https://socratic.org/questions/how-do-you-solve-16-0-2-10-x-35
| 1,579,949,586,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579251672440.80/warc/CC-MAIN-20200125101544-20200125130544-00419.warc.gz
| 652,240,762
| 5,865
|
# How do you solve -16+0.2(10^x)=35?
$x = \log 255 \approx 2.407$
#### Explanation:
$- 16 + 0.2 \left({10}^{x}\right) = 35$
Let's move all the non-x terms to the right side:
$0.2 \left({10}^{x}\right) = 51$
${10}^{x} = \frac{51}{0.2} = 255$
Now we can take the log of both sides:
$\log {10}^{x} = \log 255$
$x \log 10 = \log 255$
Remember that $L o {g}_{10} 10 = 1$, so:
$x = \log 255 \approx 2.407$
| 175
| 410
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.375
| 4
|
CC-MAIN-2020-05
|
longest
|
en
| 0.52551
|
http://mathhelpforum.com/geometry/124472-golden-rectangle-print.html
| 1,529,475,723,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267863463.3/warc/CC-MAIN-20180620050428-20180620070428-00581.warc.gz
| 211,734,399
| 5,344
|
# Golden rectangle.
• Jan 19th 2010, 02:44 PM
artofstoo
Golden rectangle.
Hello all,
I'm wondering if someone here could kindly give me a hand to solve a small problem.
I'm trying to draft a golden spiral but I keep finding myself snagged mid-way through the drawing.
I'll explain my procedure first, leading into the issue i'm having.
I start by creating the largest square of the sequence using inches, I pick any size to fit within the page i'm working on... so anywhere from 4"x4" - 9x9.
Once i've done that I divide the length in half and draw my diagonal line from the mid-point of the length to one of the corners on the adjacent side. Then while using a compass pivoting on the mid-point, I create my arc starting at the corner traveling down to level with the length of the square.
Now, to my understanding i've created the golden rectangle.
So, I then proceed on to creating the second, smaller square by using the measurement of the length I had just created (the golden mean of the first square)
Now i'm left with two squares, one big and one small and one small golden rectangle. Using the same method as in creating the second square, I create the third square within the smaller rectangle, which then leaves me with an even smaller rectangle.
Repeating this pattern once or twice more I hit a point where the rectangle i'm left with is too narrow to continue on ( it's proportions are outside of being a golden rectangle ) ...For example. The next square in the sequence I have to create may be 3"x3".. the rectangle would be 3"x8". leaving me with a 5 unit remainder and if I was to treat the 3"x3" square as the 1,1 of the fibonacci sequence then I would create another 3"x3" square however, that creates a 6"x3" rectangle, still leaving me with a 3"x2" square...
*the measurements I used are just for a rough example.
-So, what the heck am i'm doing wrong? lol. What i'm hoping to get is a 9 square sequence for this spiral.
I would also like to mention that i've gone over my measurements many times now and all of my boxes are completely square.... my only guess is that my arc may be too wide but I don't understand how that would be if I was accurate with the measurement of the first square....
Any help will be greatly appreciated :) !!!
• Jan 19th 2010, 07:19 PM
Soroban
Hello, artofstoo!
Quote:
I'm trying to draft a golden spiral but I keep finding myself snagged
mid-way through the drawing.
I'll explain my procedure first, leading into the issue i'm having.
I start by creating the largest square of the sequence using inches.
I pick any size to fit within the page i'm working on, anywhere from 4"x4" - 9x9.
Once i've done that I divide the length in half and draw my diagonal line from the midpoint
of the side to one of the corners on the adjacent side.
Then while using a compass pivoting on the midpoint, I create my arc
starting at the corner traveling down to level with the length of the square.
Now, to my understanding i've created the golden rectangle.
So, I then proceed on to creating the second, smaller square by using
the measurement of the length I had just created (the golden mean of the first square)
Now i'm left with two squares, one big and one small and one small golden rectangle.
Using the same method as in creating the second square, I create the third square
within the smaller rectangle, which then leaves me with an even smaller rectangle.
Repeating this pattern once or twice more I hit a point where the rectangle i'm left with
is too narrow to continue on. (It's proportions are outside of being a Golden Rectangle).
Code:
D 1 C * - - - - - * | /| * | / | * 1 | / | * | / | | / | * * - - * - - * - - * - - A M B P
We have square $\displaystyle ABCD\!:\;\;AB = BC = CD = DA = 1$
$\displaystyle M$ is the midpoint of $\displaystyle AB\!:\;\;MB = \tfrac{1}{2}$
In right triangle $\displaystyle CBM\!:\!:\;MC^{\:\!2} \:=\:1^2 + \left(\tfrac{1}{2}\right)^2 \:=\:\frac{5}{4} \quad\Rightarrow\quad MC = \tfrac{\sqrt{5}}{2}$
With center $\displaystyle M$ and radius $\displaystyle MC$, draw arc $\displaystyle CP$, cutting $\displaystyle AB$ extended at $\displaystyle P.$
Therefore: .$\displaystyle MP \:=\:\tfrac{1}{2} + \tfrac{\sqrt{5}}{2} \;=\;\frac{1+\sqrt{5}}{2}$ . . . the Golden Mean.
There should be no integers after the initial square.
$\displaystyle \text{We have a }\text{1-by-}\phi\text{ rectangle }APQD\;\text{ . . . a Golden Rectangle.}$
Code:
: - - - - - φ - - - - - : D C Q - *-------------*---------* : | | | : | | | 1 | | | : | R *---------* S : | | | - *-------------*---------* A B P
Cut off square $\displaystyle ABCD$
and the remaining rectangle $\displaystyle CQPB$ is a Golden Rectangle.
Cut off square $\displaystyle CQSR$
and the remaining rectangle $\displaystyle RSPB$ is a Golden Rectangle.
In theory, we can repeat this process forever.
. . The ratio of the sides will always be $\displaystyle 1\!:\!\phi.$
• Jan 19th 2010, 11:20 PM
artofstoo
Thanks for the response Soroban, Fancy ASCII illustration too btw haha.
The problem i am having though is when I do repeat the process.
Lets say I cut off R,S,B,P to get:
Code:
: - - - - - φ - - - - - : D C Q - *-------------*---------* : | | | : | | | 1 | | t | : | R *---------* S : | | | | - *-------------*---------* A B u P
Now I have R,T,B,U .. The problem i'm getting usually happens while squaring off this section on either this step or on the following step ( R,T,"V,W"? ) ...somehow I keep finding myself left with more space then what I should have.
I assume, obviously, that my measurements have to be off in order for this to happen. What I don't get is how they could be off in the first place unless the "negative space" ,i'll call it, is the summation of some earlier lines being short a millimeter or two.
I'll try to illustrate what i'm being left with -
Code:
^ ^ ^ ^ C* *Q | | R T | *---------*-------------------*S | | | X*----------*Y | | | | | | | V*----------*W | | | | | | | <<*------------ *---------*-------------------* A B U P
Alright, So this illustration shows the rectangle zoomed in to the R,T,B,U sector.
now, lets say I had just created the cut from T-U and I shifted my attention to R,T,B,U...the next line I need to create is V,W, so I take the length of B,U and bring it over to B,R to get B,V and I create my square V,W,B,U.... Now, notice R,T,V,W... it's an usual shape.
-The cut for V,W should actually be where I placed X,Y... giving me the horizontal rectangle R,T,X,Y which actually looks a lot closer to the dimensions of a golden rectangle even in this illustration.
However, I keep getting that V,W line...which I should call "checkmate" lol because I can't follow it up with another cut. When I do I end up creating an adjacent square to V,W,B,U and of the same dimension, leaving me with the "negative space" I was explaining earlier, similar to the shape of R,T,X,Y...
...it's definitely not an infinite spiral i'm drawing lol. If anything B,U,V,W and X,Y,V,W should both = 1 leaving me with nothing correct?
So whats potentially causing this problem?
my lines are accurate to 1/16 of an inch.
• Jan 19th 2010, 11:40 PM
artofstoo
yes code reformatting!.
-i'll have to edit that later.
| 2,110
| 8,537
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.515625
| 4
|
CC-MAIN-2018-26
|
latest
|
en
| 0.940119
|
https://www.jamiletheteacher.com/geometry/faq-how-to-name-a-segment-in-geometry.html
| 1,653,114,491,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662538646.33/warc/CC-MAIN-20220521045616-20220521075616-00782.warc.gz
| 947,187,933
| 9,494
|
## FAQ: How To Name A Segment In Geometry?
A segment is named by its two endpoints, for example, ¯AB. A ray is a part of a line that has one endpoint and goes on infinitely in only one direction.
## What is a segment in geometry terms?
Geometry. a part cut off from a figure, especially a circular or spherical one, by a line or plane, as a part of a circular area contained by an arc and its chord or by two parallel lines or planes. Also called line segment.
## What is used to name a line in geometry?
Points are used to name lines in geometry. Because there are an infinite number of points in a line, the points you draw can be at any place on the line. A dot indicates where a point is on the line. Name each point with a capital letter.
## How do you name a segment?
A segment is named by its two endpoints, for example, ¯AB. A ray is a part of a line that has one endpoint and goes on infinitely in only one direction. You cannot measure the length of a ray. A ray is named using its endpoint first, and then any other point on the ray (for example, →BA ).
## How do you write a segment?
In geometry, you will write a line segment using the letters for each of the end points and a line over the top of the letters. For example, if your end points were A and B, then you would write your line segment AB with a line over the top.
You might be interested: Question: What Is Easier Algebra Or Geometry?
## What is a segment in a triangle?
Univ. of Wisconsin. A midsegment is the line segment connecting the midpoints of two sides of a triangle. Since a triangle has three sides, each triangle has three midsegments. A triangle midsegment is parallel to the third side of the triangle and is half of the length of the third side.
## What is a segment of a circle in geometry?
From Wikipedia, the free encyclopedia. In geometry, a circular segment (symbol: ⌓) is a region of a circle which is “cut off” from the rest of the circle by a secant or a chord.
## How would you write the name of a segment differently than the name of a line Brainly?
How would you write the name of a segment differently than a line what symbols would you use? You can write both the name of a segment and the name of a line using two points; for the line, it is two points on the line; for the segment, it is the two endpoints of the segment.
## How did you determine a line segment?
If you mark two points A and B on it and pick this segment separately, it becomes a line segment. This line segment has two endpoints A and B whose length is fixed. The length of this line segment is the distance between its endpoints A and B. So, a line segment is a piece or part of a line having two endpoints.
| 617
| 2,699
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375
| 4
|
CC-MAIN-2022-21
|
latest
|
en
| 0.945731
|
http://jeff.cs.mcgill.ca/~godfried/teaching/cg-projects/97/Ian/twoears.html
| 1,508,777,523,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187826210.56/warc/CC-MAIN-20171023164236-20171023184236-00488.warc.gz
| 178,805,767
| 3,079
|
The Two-Ears Theorem
The Two-Ears Theorem was developed and proven by Gary H. Meisters at the University of Nebraska in 1975 [4].
### Proof #1: By Gary H. Meisters
The proof by Meisters is by induction on the number of vertices, n, in the simple polygon P. It is quite elegant.
Base Case:
n = 4. The simple polygon P is a quadrilateral, which has two ears.
Induction:
Let P be a simple polygon with at least 4 vertices. Vertex pi is a vertex of P where the interior angle formed by pi-1, pi, and pi+1 is less than 180o.
Case 1:
Polygon P has an ear at pi. If this ear is removed, the remaining polygon P' is a simple polygon with more than three vertices, but with one less vertex than P. By the induction hypothesis, there are two non-overlapping ears E1 and E2 for P'. Since they are non-overlapping, at least one of these two ears, say E1, is not at either of the vertices pi-1 or pi+1. Since all ears of P' are also ears of P, polygon P has two ears: E1 and the ear at pi.
Case 2:
Polygon P does not have an ear at pi. So, the triangle formed by pi-1, pi, and pi+1 contains at least one vertex of P in its interior. Let q be the vertex in the interior of this triangle such that the line L through q and parallel to the line through pi-1 and pi+1 is as close to pi as possible.
Let a and b be the points of intersection of line L with the polygon edges (pi, pi+1) and (pi-1, pi), respectively. The triangle formed by pi, a, and b does not contain in its interior any vertex of P (or else our choice of vertex q would be incorrect).
The line segment Q from vertex q to pi can be constructed without intersecting any edges of P. Line segment Q divides P into two simple polygons: P1 (that contains vertices pi, pi+1,..., q) and P2 (that contains vertices pi, pi-1,..., q).
There are now two sub-cases to consider:
Case 2a:
Polygon P1 is a triangle. So, P1 is an ear for polygon P. By the induction hypothesis, polygon P2 must have at least two non-overlapping ears, E1 and E2 (or else it too would be a triangle and polygon P would have only four vertices). Since they are non-overlapping, at least one, say E1, is at neither vertices pi nor q. E1 does not overlap with the ear formed by polygon P1, so it is the second ear in polygon P.
Case 2b:
Polygon P1 is not a triangle. So, by the induction hypothesis, polygon P1 has two ears, E11 and E12 and polygon P2 has two ears, E21 and E22. Since they are non-overlapping, at least one of the ears in P1, say E11, is not at vertex pi or q. Similarly, at least one of the ears in P2, say E21, is not at vertex pi or q. These two ears, E11 and E21 will be non-overlapping ears for polygon P
Q.E.D.
### Proof #2: By Joseph O'Rourke[6]
It is known that a simple polygon can always be triangulated. Leaves in the dual-tree of the triangulated polygon correspond to ears and every tree of two or more nodes must have at least two leaves.
Q.E.D.
### Some Examples:
This page was last updated on Wednesday, December 10th, 1997.
© 1997 Ian Inc.
| 801
| 3,012
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.46875
| 4
|
CC-MAIN-2017-43
|
latest
|
en
| 0.961237
|
https://www.onlinecalculatorsfree.com/math/permutation-and-combination-calculator.html
| 1,695,698,328,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510130.53/warc/CC-MAIN-20230926011608-20230926041608-00726.warc.gz
| 1,006,395,513
| 14,357
|
# Permutation and Combination Calculator
Permutation and Combination Calculator - Easily calculate permutations and combinations for any set of elements. Find the number of possible arrangements and selections. Fast, accurate, and efficient tool.
Total Amount in a Set (n) Amount in each Sub-Set (r)
Related
Permutations and combinations are part of a branch of mathematics called combinatorics, which involves studying finite, discrete structures. Permutations are specific selections of elements within a set where the order in which the elements are arranged is important, while combinations involve the selection of elements without regard for order. A typical combination lock for example, should technically be called a permutation lock by mathematical standards, since the order of the numbers entered is important; 1-2-9 is not the same as 2-9-1, whereas for a combination, any order of those three numbers would suffice. There are different types of permutations and combinations, but the calculator above only considers the case without replacement, also referred to as without repetition. This means that for the example of the combination lock above, this calculator does not compute the case where the combination lock can have repeated values, for example, 3-3-3.
### Permutations
The calculator provided computes one of the most typical concepts of permutations where arrangements of a fixed number of elements r, are taken from a given set n. Essentially this can be referred to as r-permutations of n or partial permutations, denoted as nPr, nPr, P(n,r), or P(n,r) among others. In the case of permutations without replacement, all possible ways that elements in a set can be listed in a particular order are considered, but the number of choices reduces each time an element is chosen, rather than a case such as the "combination" lock, where a value can occur multiple times, such as 3-3-3. For example, in trying to determine the number of ways that a team captain and goalkeeper of a soccer team can be picked from a team consisting of 11 members, the team captain and the goalkeeper cannot be the same person, and once chosen, must be removed from the set. The letters A through K will represent the 11 different members of the team:
A B C D E F G H I J K 11 members; A is chosen as captain
B C D E F G H I J K 10 members; B is chosen as keeper
As can be seen, the first choice was for A to be captain out of the 11 initial members, but since A cannot be the team captain as well as the goalkeeper, A was removed from the set before the second choice of the goalkeeper B could be made. The total possibilities if every single member of the team's position were specified would be 11 × 10 × 9 × 8 × 7 × ... × 2 × 1, or 11 factorial, written as 11!. However, since only the team captain and goalkeeper being chosen was important in this case, only the first two choices, 11 × 10 = 110 are relevant. As such, the equation for calculating permutations removes the rest of the elements, 9 × 8 × 7 × ... × 2 × 1, or 9!. Thus, the generalized equation for a permutation can be written as:
nPr
n! (n - r)!
Or in this case specifically:
11P2
11! (11 - 2)!
=
11! 9!
= 11 × 10 = 110
Again, the calculator provided does not calculate permutations with replacement, but for the curious, the equation is provided below:
nPr = nr
### Combinations
Combinations are related to permutations in that they are essentially permutations where all the redundancies are removed (as will be described below), since order in a combination is not important. Combinations, like permutations, are denoted in various ways, including nCr, nCr, C(n,r), or C(n,r), or most commonly as simply
( n ) r
. As with permutations, the calculator provided only considers the case of combinations without replacement, and the case of combinations with replacement will not be discussed. Using the example of a soccer team again, find the number of ways to choose 2 strikers from a team of 11. Unlike the case given in the permutation example, where the captain was chosen first, then the goalkeeper, the order in which the strikers are chosen does not matter, since they will both be strikers. Referring again to the soccer team as the letters A through K, it does not matter whether A and then B or B and then A are chosen to be strikers in those respective orders, only that they are chosen. The possible number of arrangements for all n people, is simply n!, as described in the permutations section. To determine the number of combinations, it is necessary to remove the redundancies from the total number of permutations (110 from the previous example in the permutations section) by dividing the redundancies, which in this case is 2!. Again, this is because order no longer matters, so the permutation equation needs to be reduced by the number of ways the players can be chosen, A then B or B then A, 2, or 2!. This yields the generalized equation for a combination as that for a permutation divided by the number of redundancies, and is typically known as the binomial coefficient:
nCr
n! r! × (n - r)!
Or in this case specifically:
11C2
11! 2! × (11 - 2)!
=
11! 2! × 9!
= 55
It makes sense that there are fewer choices for a combination than a permutation, since the redundancies are being removed. Again for the curious, the equation for combinations with replacement is provided below:
nCr
(r + n -1)! r! × (n - 1)!
### Permutation and Combination Calculator Example
Imagine you have a set of letters: A, B, C.
To calculate permutations, which refers to the number of possible arrangements where order matters, we can use the permutation formula. The number of permutations can be calculated by multiplying the total number of elements by one less than each subsequent element, and so on. In this case, since we have 3 elements, the number of permutations would be 3! = 3 x 2 x 1 = 6.
The six possible permutations for the given set of letters are:
1. ABC
2. ACB
3. BAC
4. BCA
5. CAB
6. CBA
To calculate combinations, which refers to the number of possible selections regardless of order, we can use the combination formula. The number of combinations can be calculated by dividing the number of permutations by the factorial of the selected elements. In this case, if we want to select 2 elements from the set of 3, the number of combinations would be 3C2 = 3! / (2! * (3-2)!) = 3.
The three possible combinations for selecting 2 elements from the given set of letters are:
1. AB
2. AC
3. BC
Therefore, using a permutation and combination calculator with the provided table, you can easily determine the number of permutations and combinations for any given set of elements.
| 1,527
| 6,711
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625
| 4
|
CC-MAIN-2023-40
|
latest
|
en
| 0.927244
|
https://brainmass.com/statistics/probability/probability-small-shop-large-shop-no-shop-95505
| 1,477,580,639,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-44/segments/1476988721347.98/warc/CC-MAIN-20161020183841-00559-ip-10-171-6-4.ec2.internal.warc.gz
| 824,283,717
| 19,309
|
Share
Explore BrainMass
# Probability - small shop, large shop, or no shop
Jerry can open up a small shop, large shop, or no shop. There will be a 5 year lease on a building he wants to make the correct decision. Jerry is also thinking about hiring a consultant to conduct a market research study. If the study is conducted, the results could be either favorable or unfavorable.
Develop a decsion tree for Jerry.
Part II
Options Favorable unfavorable
Large shop \$60000 -\$40000
Small shop \$30000 -\$10000
Jerry believes there is a 50-50 chance that the market will be favorable.
The consultant will charge him \$5000 for the marketing research. It is estimated that there is a 0.6 probability that the survey will be favorable. There is a 0.9 probability that they market will be favorable given a favorable outcome from the study. The consultant warned Jerry that there is only a probability of 0.12 of a favorable market if the marketing research results are not favorable. Jerry is confused.
A. Should Jerry use the marketing research? Why
B. Jerry is unsure the 0.6 probability of a favorable marketing research study is correct. How sensitive is Jerry's decision to this probabilty value?
C. How far can this probability value deviate from 0.6 without causing Jerry to change his decision?
#### Solution Preview
Hello
Please see the attached files
1. Word file contains ...
#### Solution Summary
1. Word file contains decision tree with all details, computations, and explanation to all answers
2. Excel file contains expected value calculations, formulas sheet, and sensitivity analysis graph.
\$2.19
| 352
| 1,621
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.515625
| 4
|
CC-MAIN-2016-44
|
longest
|
en
| 0.922397
|
edthecoder.dev
| 1,722,676,488,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00535.warc.gz
| 170,406,871
| 4,588
|
How hexadecimal helped me understand binary, part 1
This week the CS50x course I am working through is teaching (more accurately getting me to learn about) pointers in C, and using images to help understand the concept. I've been looking through bitmap files and getting to grips with how computers make images, something I've always been curious about. Three weeks ago I also looked at binary and how computers use only 1's and 0's to get stuff done. I thought I'd got to grips with it, but looking at hexadecimal, which is used for images, really cemented this understanding.
Instead of going up to 10 in each digit you go up to 16.
In decimal we count, 0, 1, 2, ... 8, 9 and then we have to add another digit to get to 10 (two digits). Whereas in hexadecimal when we get past nine we use the letters a to f (which take up one digit (or bit in a computers perspective)). See this table for a clearer picture of this:
Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Hexadecimal 0 1 2 3 4 5 6 7 8 9 A B C D E F
This means we can write bigger numbers in a shorter number of bits (digits).
`F` in hexadecimal would equal to 15. But what would `FF` equal to? Can you work it out?
In decimal the first digit of 35 is 3, but we know it represents 30 (3 * 10). 10 here is the base (decimal). We then add the next digit to the first (30 + 5 = 35).
In hexadecimal the first digit of `FF` is 15, but it actually represents 240 (15 * 16). 16 here is the base (hexadecimal). We then add the next digit to the first (240 + `F` = 255). Remember F = 15.
Let's clarify this a little with one more example using a table. The number is 345. Here it is in decimal (base 10):
Decimal number: 3 4 5 Base to the power of each digit: 10^2 10^1 10^0 Which equals: 100 10 1 Total: 300 + 40 + 5 = 345
Whereas if we had 345 in hexadecimal (base 16) it would look like this:
Hexadecimal number: 3 4 5 Base to the power of each digit: 16^2 16^1 16^0 Which equals: 256 16 1 Total: 768 + 64 + 5 = 837
See if you can work out the hexadecimal versions of these decimal numbers: 11, 78, 613. Use the tables to help.
I hope this has helped spread at least a little light on the hexadecimal! In my next post I'll let you know why this is so helpful in computers, and why this has helped me with binary.
You can find part 2 here.
| 670
| 2,310
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.21875
| 4
|
CC-MAIN-2024-33
|
latest
|
en
| 0.874024
|
http://www.euclideanspace.com/maths/algebra/clifford/algebra/functions/index.htm
| 1,686,067,847,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224652959.43/warc/CC-MAIN-20230606150510-20230606180510-00499.warc.gz
| 59,709,341
| 6,959
|
Maths - Multivector Functions
We will derive multivector equivalents to the quaternion and complex number idea of the conjugate, so lets start by reviewing that concept:
Comparison to Quaternions
On the Quaternion and Complex number pages (here) we described the conjugate. It is a quaternion with the same magnitudes but with the sign of the imaginary parts changed, so:
• conj(a + b i + c j + d k) = a - b i - c j - d k
The notation for the conjugate of a quaternion 'q' is either of the following:
• conj(q)
• q'
The conjugate is useful because it has the following properties:
• qa' * qb' = (qb*qa)' In this way we can change the order of the multiplicands.
• q * q' = a2 + b2 + c2 + d2 = real number. Multiplying a quaternion by its conjugate gives a real number. This makes the conjugate useful for finding the multiplicative inverse. For instance, if we are using a quaternion q to represent a rotation then conj(q) represents the same rotation in the reverse direction.
• Pout = q * Pin * q' We use this to calculate a rotation transform.
It would be useful to be able to do similar things with multivectors and so we investigate the following functions:
Inverse
To investigate how to calculate the multiplicative inverse of multivector 'm', that is '1/m' then we will start from the simplest case and gradually work up to more complex cases.
In standard vector algebra there is not an inverse, in geometric algebra vectors do have an inverse, but not all multivectors have an inverse, as we shall see later.
So, first we will take the inverse of a base vector 1/e1,
We multiply top and bottom by the same vector, since the square of the vector is a scalar value, the bottom part cancels out and the vector is its own inverse.
We can also do the same when the vector is the linear combination of vector bases.
So again the vector is its own inverse, with a scaling factor being the division by the scalar (a² + b²).
So lets move on to pure bivectors. We can invert a basis bivector by multiplying the top & bottom by its reverse, in other words we reverse the order of its terms, the reverse function is discussed below.
This reversal makes it very easy to cancel out the denominator as like terms are adjacent and as the middle term cancels out the outer terms come together and can be cancelled out. For the rules used when calculating the results of products see this page. (adjacent like term cancel out and reversing the order changes the sign)
We can also calculate the inverse of a linear combination of bivector bases as follows.
So to take the inverse we can separately take the reverse of each term (and apply the scaling factor).
There is a problem though, the non scalar terms only cancel out because e12 and e23 have a common term, otherwise there will be a non scalar term in the denominator. In the case of 3 dimensional vectors this is not a problem, because there will always be a common term, but four dimensional bivectors don't in general have a multiplicative inverse.
So what about higher order terms and multivectors containing a mixture of different blades?
For more information about the inverse see this page.
Reverse
The reverse function of a multivector reverses the order of its factors, including the order of the base values within a component. The reverse function is denoted by †, so the reversal of A is denoted by A†.
so for 3D vectors, if A = B† then:
a.e = b.e321 = -b.e123
a.e1 = b.e32 = -b.e23
a.e2 = b.e13 = -b.e31
a.e3 = b.e21 = -b.e12
a.e12 = b.e3
a.e31 = b.e2
a.e23 = b.e1
a.e123 = b.e
The reversal function is important for a number of reasons, one reason is that it can map a multiplication into another multiplication with the order of the multiplicands reversed:
(A * B)† = B†* A†
We can think of this as a morphism where † maps to an equivalent expression with order of multiplication reversed.
Another application for the reversal function is to specify a transformation from one vector field to another:
pout = A pin A†
In other words, if pin is a pure vector (i.e. real, bivector and tri-vector parts are all zero) then pout will also be a pure vector. I would appreciate any proof of this.If you have a proof, that I could add to this page, please let me know, In the 3D case I guess I could try the brute force approach of multiplying out the terms for a general expression A and p.
I think this may also apply to:
pout = A pin A -1
but not every multivector is invertible, one condition that should ensure that a multivector is invertible is:
A A† = 1
Dual
The dual of a grade-r multivector Ar is the product of Arwith the grade-n pseudoscalar I. This has grade n-r.
Ar* = dual of Ar which has grade n-r.
Also known as Hodge Dual
Let I = the pseudoscalar
And if this is normalised:
|I2| = 1
The textbooks (such as Hestenes) choose 'I' such that it has the orientation specified by a right-handed set of vectors in 3D, the rest of this site uses a left hand coordinate system, so what should we use?
I is known as the dextral unit psudoscalar
Identities
Ar* = I Ar = (-1)r(n-1)Ar I
Ar • (Bs I) = (Ar Bs) I
where:
• Ar = blade r of multivector A
• n = grade of multivector
• I = the pseudoscalar
Conjugate
This reverses all directions in space
A~ denotes the conjugate of A
conjugate, reverse and dual are related as follows.
A~= (A†)* = (A*)†
identities
(A * B)~ = B~* A~
Norm
under construction
Exponential
We define the exponential of multivector a as ea º åi=0¥ ai / i ! .
In particular:
ea is the traditional scalar exponential function
For any pure square multivector [ a2 = ±|a|2 ] we have ea = cos|a| + a~ sin|a| if a2 < 0 ;
cosh|a| + a~ sinh|a| if a2 > 0 ;
1+a if a2 = 0.
For unit multivector a:
eafeay = ea(f+y)
(d/df) elaf =lea(lf+p/2) =laelaf
Involution
a# = sum k=0(N (-1)k a <k> = a<+> - a<-> .
Determinant
The determinant of a multivector can be calculated from the table shown for the multipication of each dimentional vectors.
metadata block
see also:
Correspondence about this page
Book Shop - Further reading.
Where I can, I have put links to Amazon for books that are relevant to the subject, click on the appropriate country flag to get more details of the book or to buy it from them.
Clifford Algebra to Geometric Calculus: A Unified Language for Mathematics and Physics (Fundamental Theories of Physics). This book is intended for mathematicians and physicists rather than programmers, it is very theoretical. It covers the algebra and calculus of multivectors of any dimension and is not specific to 3D modelling.
Specific to this page here:
This site may have errors. Don't use for critical systems.
Copyright (c) 1998-2023 Martin John Baker - All rights reserved - privacy policy.
| 1,702
| 6,743
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.53125
| 5
|
CC-MAIN-2023-23
|
latest
|
en
| 0.888989
|
https://www.open.edu/openlearn/mod/oucontent/view.php?id=82662§ion=3
| 1,726,310,184,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00091.warc.gz
| 852,126,218
| 23,056
|
### Become an OU student
Teaching mathematics
Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.
# 3 Connecting ratio and proportional reasoning
According to the Oxford English Dictionary (OED) Online, ratio is defined as ‘the quantitative relation between two amounts showing the number of times one value contains or is contained within the other’.
For example, the ratio of male to female professors is 4 to 1.
In contrast, proportion is described by the OED as a part, share, or number considered in comparative relation to a whole.
For example, the proportion of greenhouse gases in the atmosphere is rising.
In other words, a ratio is used to describe a comparison between two (or more) objects, amounts or values, and a proportion is used to describe one object, amount or value, using a comparison to a whole.
For example if the ratio of boys to girls in the class is 2 to 3, or 2:3, the proportion of boys is 2/5.
Since the proportion of boys in the class is expressed as a fraction of a whole, it is not necessary to know the proportion of girls. If there are 30 children in the class and 2/5 of them are boys, it can be inferred that 3/5, or 18, of them are girls.
In contrast, when expressed as a ratio, the ‘2’ is meaningless without the comparison to the ‘3’.
Proportional thinking and reasoning is knowing the multiplicative relationship between the base ratio and the proportional situation to which it is applied.
Understanding ratio and proportion is more than merely being able to perform appropriate calculations, being able to apply rules and formulae or manipulating numbers and symbols in proportion equations.
| 374
| 1,758
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.65625
| 4
|
CC-MAIN-2024-38
|
latest
|
en
| 0.940442
|
https://gmatclub.com/forum/from-the-bark-of-the-paper-birch-tree-the-menomini-crafted-a-canoe-abo-85481.html
| 1,490,640,122,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-13/segments/1490218189495.77/warc/CC-MAIN-20170322212949-00281-ip-10-233-31-227.ec2.internal.warc.gz
| 800,287,934
| 68,330
|
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club
It is currently 27 Mar 2017, 11:42
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# From the bark of the paper birch tree the Menomini crafted a canoe abo
Author Message
TAGS:
### Hide Tags
Manager
Joined: 13 Jun 2009
Posts: 108
Location: Tu
Schools: Chicago b.
Followers: 11
Kudos [?]: 367 [0], given: 92
From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
19 Oct 2009, 05:22
3
This post was
BOOKMARKED
00:00
Difficulty:
55% (hard)
Question Stats:
50% (01:57) correct 50% (01:04) wrong based on 265 sessions
### HideShow timer Statistics
From the bark of the paper birch tree the Menomini crafted a canoe about twenty feet long and two feet wide, with small ribs and rails of cedar, which could carry four persons or eight hundred pounds of baggage so light that a person could easily portage it around impeding rapids.
(A) baggage so light
(B) baggage being so light
(C) baggage, yet being so light
(D) baggage, and so light
(E) baggage yet was so light
[Reveal] Spoiler: OA
Last edited by carcass on 18 Jan 2017, 17:05, edited 3 times in total.
Edited by Carcass
If you have any questions
New!
Manager
Joined: 28 Jul 2009
Posts: 126
Followers: 2
Kudos [?]: 26 [0], given: 12
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
19 Oct 2009, 22:05
Option C and E both have a conjunction "yet". The means whatever is after this conjugation should be parallel with the subject in number and in tense.
E is in past tense which is required here.
Also C uses "being" so it should be avoided if there is another similar option available.
BSchool Forum Moderator
Joined: 02 Oct 2009
Posts: 604
GMAT 1: 530 Q47 V17
GMAT 2: 710 Q50 V36
Followers: 37
Kudos [?]: 328 [0], given: 411
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
19 Oct 2009, 23:39
Think this Q tests on modifiers
its clearly between C & E
baggage, yet being so light here baggage is modified it means baggage is so light
E uses it properly
Senior Manager
Joined: 25 Jun 2009
Posts: 294
Followers: 1
Kudos [?]: 158 [0], given: 4
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
15 Nov 2009, 12:22
This was a tough one. I originally thought it was A, but upon closer inspection, E does look better.
Manager
Joined: 11 Jul 2009
Posts: 57
Followers: 1
Kudos [?]: 27 [0], given: 19
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
15 Nov 2009, 20:49
Ayrish wrote:
Please can anyone help? I chose C.
Cedar, which could carry baggage yet being so light that a person could portage it.
320. From the bark of the paper birch tree the Menomini crafted a canoe about twenty feet long and two feet wide, with small ribs and rails of cedar, which could carry four persons or eight hundred pounds of baggage so light that a person could easily portage it around impeding rapids.
(A) baggage so light
(B) baggage being so light
(C) baggage, yet being so light
(D) baggage, and so light
(E) baggage yet was so light
E is a right one.
IMO E
yet shows contrast so it is preferred over and.
Being can be removed from c without affecting the meaning. hence. E.
SVP
Joined: 07 Nov 2007
Posts: 1821
Location: New York
Followers: 36
Kudos [?]: 900 [1] , given: 5
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
15 Nov 2009, 21:00
1
KUDOS
Ayrish wrote:
Please can anyone help? I chose C.
Cedar, which could carry baggage yet being so light that a person could portage it.
320. From the bark of the paper birch tree the Menomini crafted a canoe about twenty feet long and two feet wide, with small ribs and rails of cedar, which could carry four persons or eight hundred pounds of baggage so light that a person could easily portage it around impeding rapids.
(A) baggage so light
(B) baggage being so light
(C) baggage, yet being so light
(D) baggage, and so light
(E) baggage yet was so light
E is a right one.
Please Don't post the OA right away.
Examine the Main sub and verb.
Menomini crafted a canoe <about...blah blah blah>, which could carry... yet was so light
Canoe could carry .. Yet <canoe> was so light.
So light should modify "Canoe" not "baggage"
_________________
Smiling wins more friends than frowning
Senior Manager
Joined: 25 Jun 2009
Posts: 294
Followers: 1
Kudos [?]: 158 [0], given: 4
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
15 Nov 2009, 22:14
That makes a lot more sense. thanks.
Senior Manager
Joined: 25 Jun 2009
Posts: 294
Followers: 1
Kudos [?]: 158 [0], given: 4
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
15 Nov 2009, 23:04
Where did this problem come from?
Forum Moderator
Status: mission completed!
Joined: 02 Jul 2009
Posts: 1425
GPA: 3.77
Followers: 180
Kudos [?]: 869 [1] , given: 621
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
17 Apr 2010, 11:42
1
KUDOS
1
This post was
BOOKMARKED
.... Menomini crafted a canoe about twenty feet long and two feet wide [,with small ribs and rails of cedar, {which could carry four persons or eight hundred pounds of baggage}] yet was so light that a person could easily portage it around impeding rapids.
here first bolded area tells about how big was a canoe, the second part contrasts that with such huge dimensions it was yet was so light
_________________
Audaces fortuna juvat!
GMAT Club Premium Membership - big benefits and savings
Intern
Joined: 20 Apr 2010
Posts: 38
Followers: 0
Kudos [?]: 1 [0], given: 7
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
20 Oct 2010, 17:17
Dude - Don't indicate your choice or the correct answer like that in the original post.
Manager
Joined: 19 Sep 2010
Posts: 182
Followers: 2
Kudos [?]: 91 [0], given: 18
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
08 Dec 2010, 09:05
@Pkit..
I think the contrast is between...which could carry four persons....yet was so light
Forum Moderator
Status: mission completed!
Joined: 02 Jul 2009
Posts: 1425
GPA: 3.77
Followers: 180
Kudos [?]: 869 [0], given: 621
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
08 Dec 2010, 09:23
Raths wrote:
@Pkit..
I think the contrast is between...which could carry four persons....yet was so light
I think that "which could carry four persons" statement additionally supports the idea that the boat was very large, yet so light.
_________________
Audaces fortuna juvat!
GMAT Club Premium Membership - big benefits and savings
Manager
Joined: 26 Dec 2009
Posts: 147
Location: United Kingdom
Concentration: Strategy, Technology
GMAT 1: 500 Q45 V16
WE: Consulting (Computer Software)
Followers: 2
Kudos [?]: 12 [0], given: 10
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
29 Dec 2010, 05:06
yea E for me. yet is required here. E is the right choice. being doesnt fit well in C
Manager
Joined: 08 Jun 2010
Posts: 172
Followers: 1
Kudos [?]: 30 [0], given: 15
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
07 Jan 2011, 10:55
its clearly between C & E........E from me
Manager
Joined: 19 Dec 2010
Posts: 145
Followers: 2
Kudos [?]: 30 [0], given: 12
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
25 Mar 2011, 03:44
I chose C first but upon closer inspection E is the only other right choice. If you think about it, C has the word "Being", this is not encouraged in GMAT grammar.
Senior Manager
Joined: 29 Oct 2008
Posts: 410
Location: United States
Concentration: Marketing, Technology
Followers: 9
Kudos [?]: 280 [0], given: 34
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
25 Apr 2011, 06:34
x2suresh wrote:
Ayrish wrote:
Please can anyone help? I chose C.
Cedar, which could carry baggage yet being so light that a person could portage it.
320. From the bark of the paper birch tree the Menomini crafted a canoe about twenty feet long and two feet wide, with small ribs and rails of cedar, which could carry four persons or eight hundred pounds of baggage so light that a person could easily portage it around impeding rapids.
(A) baggage so light
(B) baggage being so light
(C) baggage, yet being so light
(D) baggage, and so light
(E) baggage yet was so light
E is a right one.
Please Don't post the OA right away.
Examine the Main sub and verb.
Menomini crafted a canoe <about...blah blah blah>, which could carry... yet was so light
Canoe could carry .. Yet <canoe> was so light.
So light should modify "Canoe" not "baggage"
I liked E for obvious reasons. Wouldnt sentence have been better if we have obviated [highlight]"was"[/highlight]?
Is "was" specifically needed in Choice E?
_________________
If you know what you're worth, then go out and get what you're worth. But you gotta be willing to take the hits, and not pointing fingers saying you ain't where you wanna be because of anybody! Cowards do that and You're better than that!
The path is long, but self-surrender makes it short; the way is difficult, but perfect trust makes it easy.
Fire the final bullet only when you are constantly hitting the Bull's eye, till then KEEP PRACTICING.
Failure establishes only this, that our determination to succeed was not strong enough.
Getting defeated is just a temporary notion, giving it up is what makes it permanent.
http://gmatclub.com/forum/1000-sc-notes-at-one-place-in-one-document-with-best-of-explanations-192961.html
Press +1 Kudos, if you think my post gave u a tiny tip.
Manager
Joined: 11 Apr 2011
Posts: 108
Followers: 0
Kudos [?]: 30 [0], given: 16
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
29 Apr 2011, 11:46
Can anyone explain why a comma is not right after baggage in the choice E
(E) baggage yet was so light ==> (E) baggage, yet was so light
Senior Manager
Joined: 29 Oct 2008
Posts: 410
Location: United States
Concentration: Marketing, Technology
Followers: 9
Kudos [?]: 280 [1] , given: 34
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
29 Apr 2011, 15:33
1
KUDOS
pkmme wrote:
Can anyone explain why a comma is not right after baggage in the choice E
(E) baggage yet was so light ==> (E) baggage, yet was so light
There could or could not be a comma around usage of "yet", when acting as a coordinating conjunction. Here is one reference for this:
http://www.english-test.net/forum/ftopic9434.html#25363
_________________
If you know what you're worth, then go out and get what you're worth. But you gotta be willing to take the hits, and not pointing fingers saying you ain't where you wanna be because of anybody! Cowards do that and You're better than that!
The path is long, but self-surrender makes it short; the way is difficult, but perfect trust makes it easy.
Fire the final bullet only when you are constantly hitting the Bull's eye, till then KEEP PRACTICING.
Failure establishes only this, that our determination to succeed was not strong enough.
Getting defeated is just a temporary notion, giving it up is what makes it permanent.
http://gmatclub.com/forum/1000-sc-notes-at-one-place-in-one-document-with-best-of-explanations-192961.html
Press +1 Kudos, if you think my post gave u a tiny tip.
Intern
Joined: 29 Jul 2010
Posts: 44
Followers: 0
Kudos [?]: 1 [0], given: 1
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
11 Jul 2011, 18:56
_________________
"Not everyone who works hard succeeds. But all those who succeeded have worked hard!" ~~ Coach Kamogawa
Best,
ItsNotOver
Manager
Joined: 06 Jul 2011
Posts: 129
Followers: 0
Kudos [?]: 76 [0], given: 240
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink]
### Show Tags
29 May 2013, 18:09
We have to show contrast so C or E.E is correct because of past tense verb "crafted" and was..reject being as it is not happening right now.
Re: From the bark of the paper birch tree the Menomini crafted a canoe abo [#permalink] 29 May 2013, 18:09
Go to page 1 2 Next [ 21 posts ]
Similar topics Replies Last post
Similar
Topics:
1 From the bark of the paper birch tree the Menomini crafted a 3 30 Nov 2010, 17:13
11 From the bark of the paper birch tree the Menomini crafted a 9 22 Oct 2010, 01:09
2 From the bark of the paper birch tree the Menomini crafted a 10 01 Oct 2009, 01:40
From the bark of the paper birch tree the Menomini crafted a 9 18 Jan 2008, 03:59
From the bark of the paper birch tree the Menomini crafted a 14 20 Aug 2007, 22:46
Display posts from previous: Sort by
| 3,892
| 13,762
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.65625
| 4
|
CC-MAIN-2017-13
|
longest
|
en
| 0.885545
|
http://www.ck12.org/physics/Capacitors-in-Series-and-Parallel/?difficulty=basic&by=ck12
| 1,493,065,440,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-17/segments/1492917119782.43/warc/CC-MAIN-20170423031159-00287-ip-10-145-167-34.ec2.internal.warc.gz
| 470,518,003
| 15,407
|
# Capacitors in Series and Parallel
## A group of capacitors in series all have the same stored charge, a group of capacitors in parallel all have the same voltage.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
• Video
## RC Circuits 1: Capacitors in Series - Overview
by CK-12 //basic
Overview
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
## RC Circuits 2: Capacitors in Parallel - Overview
by CK-12 //basic
Overview
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
## RC Circuits 1: Capacitors in Series - Example 2
by CK-12 //basic
Determining the total charge and the charge of each capacitor arranged in a series circuit
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
## RC Circuits 1: Capacitors in Series - Example 3
by CK-12 //basic
Determining the total voltage and voltage drop across each capacitor in a series circuit
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
## RC Circuits 2: Capacitors in Parallel - Example 2
by CK-12 //basic
Determining the total charge and the charge of each capacitor arranged in a parallel circuit
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
## RC Circuits 2: Capacitors in Parallel - Example 3
by CK-12 //basic
Determining the total voltage and voltage drop across each capacitor in a parallel circuit
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
## RC Circuits 1: Capacitors in Series - Example 1
by CK-12 //basic
Determining the equivalent capacitance and the capacitance of each capacitor arranged in a series circuit using the equation 1/C[s] = 1/C[1] + 1/C[2]
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
## RC Circuits 2: Capacitors in Parallel - Example 1
by CK-12 //basic
Determining the equivalent capacitance and the capacitance of each capacitor arranged in aparallel circuit using the equation C[s ]= C[1 ]+ C[2]
MEMORY METER
This indicates how strong in your memory this concept is
0
| 608
| 2,597
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.671875
| 4
|
CC-MAIN-2017-17
|
latest
|
en
| 0.869539
|
shop78900.pages10.com
| 1,601,515,998,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600402130531.89/warc/CC-MAIN-20200930235415-20201001025415-00353.warc.gz
| 97,398,477
| 5,638
|
# What Will calculate percentages Be Like in 100 Years?
If you have actually ever discovered yourself gazing at a half-eaten pie, wondering how the part that's left compares to the size of the original pie, congratulations: You have actually been pondering percentages. Although technically the term "portion" refers to a part out of 100, in real-world terms it truly handles how a portion of something-- state, that half-eaten pie-- compares to the entire. For instance, half amounts to half, or 50 out of 100. You can utilize a calculator to quickly work out portions.
The three terms in a portion computation are the part, the whole, and the percentage. In the formula: 25% of 40 = 10, 10 is the part, 40 is the whole, and 25 is the percentage. In the mathematics world, working out portions usually indicates that a person of those terms is missing and you require to discover it. If the question is "What percentage of 40 is 10?" you have the part (10) and the entire (40 ), so the left out term is the percentage. If the question is "What is 25 percent of 40?" you have the portion (25) and the whole (40 ), so the missing term is the part. Utilizing the exact same reasoning, if the question is "10 is 25 percent of what?" the the term is the whole.
If the omitted term is the percentage, divide the part by the whole utilizing your calculator to identify the response. For the example formula, this is 10 ÷ 40 = 0.25. If your calculator has a percentage button, press it to identify the percentage. If your calculator does not have such a button, increase your previous answer by 100 to determine the portion: 0.25 x 100 = 25%.
VIDEO OF THE DAY
If the left out term is the part, utilize the calculator to multiply the entire by the percentage to identify the response. If your calculator has a portion button, the estimation is as follows: 40 x 25% = 10. If your calculator does not have a percentage button, you must percent calculator first divide the portion by 100: 25 ÷ 100 = 0.25. You can then multiply this response by the entire to determine the part: 0.25 x 40 = 10.
If the omitted term is the whole, divide the part by the percentage to determine the answer. If your calculator has a percentage button, the calculation is as follows: 10 ÷ 25% = 40. If your calculator does not have a percentage button, you must divide the percentage by 100 before completing the calculation: 25 ÷ 100 = 0.25. You can then divide the part by this answer to determine the entire: 10 ÷ 0.25 = 40. Computing portions can be a simple job. There are various percentage calculators online that can assist with job by simply looking for "portion calculator." However, there might be a time when (nevertheless, unlikely it sounds) you may need to be able to calculate percentages without any digital assistance.
Before you can compute a portion, you ought to first understand precisely what a portion is.
The word portion comes from the word percent. If you split the word percent into its root words, you see "per" and "cent." Cent is an old European word with French, Latin, and Italian origins suggesting "hundred". So, percent is equated straight to "per hundred." If you have 87 percent, you literally have 87 per 100. If it snowed 13 times in the last 100 days, it snowed 13 percent of the time.
The numbers that you will be converting into percentages can be provided to you in 2 different formats, decimal and fraction. Decimal format is much easier to compute into a portion. Transforming a decimal to a percentage is as basic as multiplying it by 100. To convert.87 to a percent, merely several
If you are given a portion, convert it to a portion by dividing the top number by the bottom
Then, follow the actions above for transforming a decimal to a percent.
The more challenging job comes when you need to understand a percentage when you are offered numbers that don't fit so nicely into 100.
The majority of the time, you will be given a percentage of a given number. For instance, you may understand that 40 percent of your income will go to taxes and you wish to learn just how much cash that is. To determine the portion of a specific number, you initially convert the portion number to a decimal.
Once you have the decimal variation of your percentage, merely multiply it by the offered number. In this case, the amount of your income. If your paycheck is \$750, you would multiply 750 by.40.
Let's attempt another example. You need to save 25 percent of your income for the next 6 months to pay for an upcoming trip. If your paycheck is \$1500, just how much should you save?
| 1,058
| 4,591
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375
| 4
|
CC-MAIN-2020-40
|
latest
|
en
| 0.92452
|
https://puzzling.stackexchange.com/questions/24013/a-prime-number-game
| 1,720,979,601,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00191.warc.gz
| 426,079,869
| 41,855
|
A prime number game
Alice and Bob play the following game on a $9\times11$ chessboard.
• First Alice divides the chessboard into $33$ smaller rectangles of dimensions $1\times3$ and $3\times1$.
• Then Bob labels each of the $99$ squares with one of the integers $0,1,2,3,4,5$, such that in each of Alice's smaller rectangles the sum of the labels equals $5$.
• Bob wins the game, if in each of the nine rows the sum of the labels is a prime number. Otherwise Alice wins.
Question: Which player is going to win this game? (As usual, we assume that Alice and Bob both use optimal strategies.)
Bob can always win by labeling the squares using the following scheme:
$31131131131 \quad 19$
$13113113113 \quad 19$
$11311311311 \quad 17$
$31131131131 \quad 19$
$13113113113 \quad 19$
$11311311311 \quad 17$
$31131131131 \quad 19$
$13113113113 \quad 19$
$11311311311 \quad 17$
The sum of each row is a prime number. Any horizontal or vertical combination of $3$ squares has the sum $5$.
• Can you please tell us how you came up with this answer ? What was your intuition/thought process ? Thanks Commented Jul 21, 2022 at 20:36
• @HemantAgarwal As far as I remember there wasn't really a big thought process. I wanted to generate a regular pattern which always generates a sum of 5 for three consecutive fields. This can be done in several ways (e.g. 500 or 410), but using 311 has the additional property of giving a prime sum for the rows. Commented Jul 23, 2022 at 3:56
Bob will win.
This is because it is impossible for Alice to force him to make a non-prime sum in a row.
I will go a step further and say it is always possible to fill the rows so that 6 rows have the sum 19 and 3 rows have the sum 17.
To prove this i will solve it with an easy board and show how every other combination of rectangles can be made from this board with simple transformations and still show the same solution.
Our easy board will have the first two colums filled with 3 vertical rectangles per column.
The rest of the board will be filled with horizontal rectangles for an additional 3 horizontal rectangles per row:
┌─┬─┬─────┬─────┬─────┐
│ │ │ │ │ │
│ │ ├─────┼─────┼─────┤
│ │ │ │ │ │
│ │ ├─────┼─────┼─────┤
│ │ │ │ │ │
├─┼─┼─────┼─────┼─────┤
│ │ │ │ │ │
│ │ ├─────┼─────┼─────┤
│ │ │ │ │ │
│ │ ├─────┼─────┼─────┤
│ │ │ │ │ │
├─┼─┼─────┼─────┼─────┤
│ │ │ │ │ │
│ │ ├─────┼─────┼─────┤
│ │ │ │ │ │
│ │ ├─────┼─────┼─────┤
│ │ │ │ │ │
└─┴─┴─────┴─────┴─────┘
Now each row will have a sum of 15 (for the three horizontal boxes) plus a small amount from the vertical boxes.
I will fill it in as easy as possible:
┌─┬─┬─────┬─────┬─────┐
│1│1│5 0 0│5 0 0│5 0 0│ 17
│ │ ├─────┼─────┼─────┤
│2│2│0 5 0│0 5 0│0 5 0│ 19
│ │ ├─────┼─────┼─────┤
│2│2│0 0 5│0 0 5│0 0 5│ 19
├─┼─┼─────┼─────┼─────┤
│1│1│5 0 0│5 0 0│5 0 0│ 17
│ │ ├─────┼─────┼─────┤
│2│2│0 5 0│0 5 0│0 5 0│ 19
│ │ ├─────┼─────┼─────┤
│2│2│0 0 5│0 0 5│0 0 5│ 19
├─┼─┼─────┼─────┼─────┤
│1│1│5 0 0│5 0 0│5 0 0│ 17
│ │ ├─────┼─────┼─────┤
│2│2│0 5 0│0 5 0│0 5 0│ 19
│ │ ├─────┼─────┼─────┤
│2│2│0 0 5│0 0 5│0 0 5│ 19
└─┴─┴─────┴─────┴─────┘
Now we can already guess where this is going.
3 Transformations are possible/needed:
1. substituting 3 horizontal rectangles by 3 vertical rectangles.
This is done easily as we already have our 5s filled in perfectly
┌─┬─┬─────┬─────┬─────┐
│1│1│5 0 0│5 0 0│5 0 0│
│ │ ├─────┼─────┼─────┤
│2│2│0 5 0│0 5 0│0 5 0│
│ │ ├─────┼─────┼─────┤
│2│2│0 0 5│0 0 5│0 0 5│
├─┼─┼─┬─┬─┼─────┼─────┤
│1│1│5│0│0│5 0 0│5 0 0│
│ │ │ │ │ ├─────┼─────┤
│2│2│0│5│0│0 5 0│0 5 0│
│ │ │ │ │ ├─────┼─────┤
│2│2│0│0│5│0 0 5│0 0 5│
├─┼─┼─┴─┴─┼─────┼─────┤
│1│1│5 0 0│5 0 0│5 0 0│
│ │ ├─────┼─────┼─────┤
│2│2│0 5 0│0 5 0│0 5 0│
│ │ ├─────┼─────┼─────┤
│2│2│0 0 5│0 0 5│0 0 5│
└─┴─┴─────┴─────┴─────┘
This can actually be done at any position we wish as long as we have 3 rectangles in a square formation.
1. moving vertically
When we have 3 vertical rectangles in a square formation and a horizontal rectangle directly above or below we can switch them.
┌─┬─┬─────┬─────┬─────┐
│1│1│5 0 0│5 0 0│5 0 0│
│ │ ├─────┼─────┼─────┤
│2│2│0 5 0│0 5 0│0 5 0│
│ │ ├─┬─┬─┼─────┼─────┤
│2│2│5│0│0│0 0 5│0 0 5│
├─┼─┤ │ │ ├─────┼─────┤
│1│1│0│5│0│5 0 0│5 0 0│
│ │ │ │ │ ├─────┼─────┤
│2│2│0│0│5│0 5 0│0 5 0│
│ │ ├─┴─┴─┼─────┼─────┤
│2│2│0 0 5│0 0 5│0 0 5│
├─┼─┼─────┼─────┼─────┤
│1│1│5 0 0│5 0 0│5 0 0│
│ │ ├─────┼─────┼─────┤
│2│2│0 5 0│0 5 0│0 5 0│
│ │ ├─────┼─────┼─────┤
│2│2│0 0 5│0 0 5│0 0 5│
└─┴─┴─────┴─────┴─────┘
We dont need to care for the first 2 columns for now as whenever this movement is possible there will always be at least 3 vertical 5-rectangles in the same 3 rows.
1. horizontal movement
When we have 3 horizontal rectangles in a square formation and a vertical rectangle directly left or right we can switch them.
┌─┬─┬─────┬─────┬─────┐
│1│1│5 0 0│5 0 0│5 0 0│
│ │ ├─────┼─────┼─────┤
│2│2│0 5 0│0 5 0│0 5 0│
│ │ ├─┬─┬─┼─────┼─────┤
│2│2│5│0│0│0 0 5│0 0 5│
├─┼─┤ │ │ ├─────┼─────┤
│1│1│0│5│0│5 0 0│5 0 0│
│ │ │ │ │ ├─────┼─────┤
│2│2│0│0│5│0 5 0│0 5 0│
│ │ ├─┴─┴─┼─────┼─────┤
│2│2│0 0 5│0 0 5│0 0 5│
├─┼─┴───┬─┼─────┼─────┤
│1│5 0 0│1│5 0 0│5 0 0│
│ ├─────┤ ├─────┼─────┤
│2│0 5 0│2│0 5 0│0 5 0│
│ ├─────┤ ├─────┼─────┤
│2│0 0 5│2│0 0 5│0 0 5│
└─┴─────┴─┴─────┴─────┘
This can be done as often as needed without changing the sums in the rows.
By using these transformations as often as needed we can reach all possible combinations of rectangles.
• You're probably right, but could you explain how or prove that all possible combinations can be reached with these 3 transformations?
– Ivo
Commented Nov 17, 2015 at 10:53
• @Ivo Beckers I think it can be seen like that and I actually don't know how i could prove it to you rather than solving each and every possible combination wich my 3 transformations. As a compromise i am willing to solve a single combination that you think might be impossible. For this i would suggest making a community wiki answer and then commenting again so i get a message and can do a step by step solution. Commented Nov 17, 2015 at 10:59
• @ivobeckers Every row has to have 2+3*n vertical rectangles, Every horizontal rectangle you remove forces you to remove 3, or the gap can't be filled in. Now you might slide things around, but in the end, you will always have to make a 3x3 'hole' to begin putting in vertical rectangles. And you can always fill in these vertical rectangles so they have a 5 in each column. Since we're only concerned about the rows, sliding around horizontally has no effect on the result, Therefore all possible combinations are basically one and the same. Commented Nov 17, 2015 at 13:15
| 2,911
| 6,766
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.890625
| 4
|
CC-MAIN-2024-30
|
latest
|
en
| 0.902165
|
http://experiment-ufa.ru/(5x-6)+89+57=180
| 1,527,045,544,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794865411.56/warc/CC-MAIN-20180523024534-20180523044534-00081.warc.gz
| 101,443,882
| 6,554
|
(5x-6)+89+57=180
Simple and best practice solution for (5x-6)+89+57=180 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.
If it's not what You are looking for type in the equation solver your own equation and let us solve it.
Solution for (5x-6)+89+57=180 equation:
Simplifying
(5x + -6) + 89 + 57 = 180
Reorder the terms:
(-6 + 5x) + 89 + 57 = 180
Remove parenthesis around (-6 + 5x)
-6 + 5x + 89 + 57 = 180
Reorder the terms:
-6 + 89 + 57 + 5x = 180
Combine like terms: -6 + 89 = 83
83 + 57 + 5x = 180
Combine like terms: 83 + 57 = 140
140 + 5x = 180
Solving
140 + 5x = 180
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-140' to each side of the equation.
140 + -140 + 5x = 180 + -140
Combine like terms: 140 + -140 = 0
0 + 5x = 180 + -140
5x = 180 + -140
Combine like terms: 180 + -140 = 40
5x = 40
Divide each side by '5'.
x = 8
Simplifying
x = 8`
| 379
| 1,029
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875
| 4
|
CC-MAIN-2018-22
|
latest
|
en
| 0.854312
|
https://math.stackexchange.com/questions/141764/convergence-of-this-integral
| 1,579,724,080,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579250607407.48/warc/CC-MAIN-20200122191620-20200122220620-00048.warc.gz
| 559,810,257
| 27,773
|
# Convergence of this integral [duplicate]
Possible Duplicate:
Some questions about the gamma function
My statistics text book prescribed by my school states that the integral $$\Gamma(n)=\int_{0}^{\infty}e^{-x}x^{n-1}dx$$ is convergent for $n>0$.It does not prove the statement.So can anyone please help me prove it?Thanks again!
• Thank you.I flagged this question as well.Looks like I need some background in analysis before I enter deeper water. – Amitabh Udayiman May 6 '12 at 15:05
• I think that this question is at a lower level than the question Some questions about the gamma function. – Américo Tavares May 6 '12 at 17:33
I assume that $n$ is a real number. Split the gamma improper integral $$\Gamma(n)=\int_{0}^{\infty}e^{-x}x^{n-1}dx\tag{0}$$ into $I_1+I_2$, where $$I_1=\int_{0}^{1}e^{-x}x^{n-1}dx\tag{1}$$ and $$I_2=\int_{1}^{\infty}e^{-x}x^{n-1}dx\tag{2}$$
1. To prove that the integral $I_2$ is always convergent use the fact that for any real number $\alpha$ the integral $$\int_{1}^{\infty }e^{-x}x^{\alpha }dx\tag{3}$$ is convergent, by the limit comparison test $$\lim_{x\rightarrow \infty }\frac{e^{-x}x^{\alpha }}{x^{-2}}=0\tag{4}$$ with the convergent integral $$\displaystyle\int_{1}^{\infty }\dfrac{dx}{x^{2}}\tag{5}.$$
2. As for $I_1$ consider two cases. (a) If $n\geq 1$ observe that $\lim_{x\rightarrow 0}e^{-x}x^{n-1}=0$, so $I_1$ is a proper integral. (b) If $0<n<1$, the integrand $e^{-x}x^{n-1}$ behaves like $x^{n-1}$ near $n=0$, because $e^{-x}\rightarrow 1$ as $x\rightarrow 0$. Since $$\displaystyle\int_{0}^{1}\dfrac{dx}{x^{1-n}}\tag{6}$$ is convergent if and only if $1-n<1$, i.e. $n>0$, so is $I_1$.
It follows that $\Gamma(n)=I_1+I_2$ is convergent for $n>0.$
| 622
| 1,705
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.75
| 4
|
CC-MAIN-2020-05
|
latest
|
en
| 0.71411
|
https://www.physicsforums.com/threads/acceleration-of-the-block.534788/#post-3528969
| 1,638,585,092,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00626.warc.gz
| 1,013,456,784
| 15,394
|
# Acceleration of the block
1. What is the acceleration of the 20 kg block if the block is being pulled with a force of 150 N at 60 degrees above the X+ axis and there is a frictional force of 20 N acting on the block?
2. F=ma, Fcos(theta).
3. I did (150)cos(60)=(20)a ---> 75=20a ---> a=3.75 m/s
Just wondering if this was correct and if I did it the right way. Thank you
You have to remember that the friction force also applies here.
$\Sigma$F = m*a
$\Sigma$F = Fcos(theta) - friction = ma
Correct?
I don't think you used the frictional force of 20N in your equation.
You have to remember that the friction force also applies here.
$\Sigma$F = m*a
$\Sigma$F = Fcos(theta) - friction = ma
Correct?
Wouldn't it be $\Sigma$F = Fnetcos(theta) = ma
I don't think you used the frictional force of 20N in your equation.
Wouldn't it be $\Sigma$F = Fnetcos(theta) = ma
You wouldn't multiply the frictional force by cos(theta) though.
You have a friction force going straight backwards with a magnitude of 20 N. Therefore, it has an angle of 180 with direction of motion, so cos(180) = -1. The 60.0 degrees ONLY affects the diagonal force.
$\Sigma$F = Fcos(theta) - friction = ma
You wouldn't multiply the frictional force by cos(theta) though.
You have a friction force going straight backwards with a magnitude of 20 N. Therefore, it has an angle of 180 with direction of motion, so cos(180) = -1. The 60.0 degrees ONLY affects the diagonal force.
$\Sigma$F = Fcos(theta) - friction = ma
How do you know that the angle of the friction is 180? Wouldn't the angle of friction be 240 degrees, because of the 60 degree incline?
How did you draw your FBD?
How do you know that the angle of the friction is 180? Wouldn't the angle of friction be 240 degrees, because of the 60 degree incline?
Consider that the Force being applied is someone pulling on a string. That string is 60.0 degrees to the flat surface the block is on. Ergo, the Force is at 60.0 degrees to the horizontal. And the friction is acting on the flat surface, keeping the angle at 180.
Even if the block was on a 60 degree incline, the friction force's angle would still be 180 degrees. Since the direction of motion would still be going up the ramp, the direction of friction would be going the exact opposite of the direction of motion, making a 180 degree angle (straight line).
I drew my FBD:
Weight force going down.
Normal force going up.
Friction force going to the right.
Some force going up and to the left, forming a 60.0 angle with the horizontal.
I assume the block is on a flat surface.
Last edited:
Oh, I assumed the block was on a ramp. That's why I was getting confused.
| 695
| 2,674
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.953125
| 4
|
CC-MAIN-2021-49
|
latest
|
en
| 0.929185
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.