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Gwar [14]
3 years ago
12

What if the sample size used was some unspecified value larger than 30. Describe how this change in sample size would have affec

ted the following, if at all: The standard deviation of the distribution for the sample mean rent price for a one-bedroom apartment in this large city would ___________.
Mathematics
1 answer:
arlik [135]3 years ago
6 0

Answer:

Option D is correct.

Decreases.

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Before an election 23% said that would vote for A 9% said they would vote for B 20% said they would not vote these all votes as
Bad White [126]
A got the most votes because the ration works with the percentage very well
7 0
3 years ago
2. In the following figure, triangle ABC and triangle DEF are similar. What is the value of x ? Triangle ABC and triangle DEF ar
gavmur [86]
FD/CA = EF/BC
x/(8.5 mm) = (12 mm)/(4 mm)
x = (8.5 mm)*(12/4)
x = 25.5 mm . . . . . . . . . . . the 3rd selection
5 0
3 years ago
Solve using Fourier series.
Olin [163]
With 2L=\pi, the Fourier series expansion of f(x) is

\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L
\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx

where the coefficients are obtained by computing

\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx
\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx

\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx
\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx

\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx
\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx

You should end up with

a_0=0
a_n=0
(both due to the fact that f(x) is odd)
b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)

Now the problem is that this expansion does not match the given one. As a matter of fact, since f(x) is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of f(x), which is to say we're actually considering the function

\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3

and enforcing a period of 2L=2\pi. Now, you should find that

\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)

The value of the sum can then be verified by choosing x=0, which gives

\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)
\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots

as required.
5 0
3 years ago
Please helppppppp ❤️
liberstina [14]

Answer:

Inverse of f exists.

Step-by-step explanation:

From the graph attached,

If we do the horizontal line test for the function graphed,

We find the function as one to one function.

In other words for every input value (x-value) there is a different output value.

Since, for one-to-one functions, inverse of the functions exist.

Therefore, the answer will be,

The inverse of 'f' exists.

7 0
3 years ago
Which expression is equivalent to 5(w+4) when w=9
erica [24]

Answer: Should be 5(9)+4

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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