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Doss [256]
3 years ago
7

Two seventh grade math classes are competing in an academic challenge. Each class will randomly select one class member to compe

te each week. Each class has 26 students, and half of the students in each class are boys.
The classes want to design a simulation that will give the probability of both class members participating in the academic challenge being girls. Two possible simulations are given below.

Simulation 1: Flip a coin 26 times. If it lands on heads, count the flip as a girl. If it lands on tails, count the flip as a boy. Count how many times the coin lands on heads.

Simulation 2: Use two standard decks of cards. Randomly draw one card from each deck. Each diamond or heart drawn will represent a girl. Each spade or club drawn will represent a boy. After the two cards are drawn, they are replaced and the decks are shuffled. Count how many times a heart or diamond is drawn from both decks for 100 trials.
(Simuation 1 or Simulation 2) would be best to use to simulate a girl being randomly selected from each class to participate in the academic challenge. Based on this simulation, the probability that two girls will be selected is about (15%, 50%, 25%, or 75%).
Mathematics
1 answer:
erastovalidia [21]3 years ago
7 0
Probability of a coin flip would be 50%
probability f a red card = 50%

50% x 50% = 25%
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Solve using Fourier series.
Olin [163]
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\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.
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Answer:

1/2 or 0.5 (same value, different form)

Step-by-step explanation:

Ok...since I can't actually graph it, we can make a table out of it....

x      y

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From -1 to 0, we add 1 (this is for the y section)

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Now all we have to do is divide the y with the x

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alukav5142 [94]

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Plug x = -3 and y = 0 into the 2 inequalities and see if they fit:

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8 0
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