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Furkat [3]
3 years ago
5

There are 5 boys in the eighth grade class and 9 girls . If mr. Jowers randomly selects one student to carry the Eagle folder ,

what is the probability it will be a boy ? Please help asap
Mathematics
1 answer:
Valentin [98]3 years ago
7 0

Answer: 36.78% chance

Step-by-step explanation: Add 9+5 and divide 5 to 14

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Please help me I'm not sure what the answer is
gogolik [260]

Step-by-step explanation:

yes yours answer is correct .

4 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
Write answer on correct order so i would not be confused + no need for explanation
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Answer:

Step-by-step explanation:

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Answer:

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