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A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

tankabanditka [31]

The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

 $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$

P(all 6 girls chosen)=35/286=0.12

B.
"The probability that a randomly chosen team has 3 girls and 7 boys."

with the same logic as in A, the number of groups were all 7 boys are in, is

 $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$

so the probability is 20/286=0.07

C.
"The probability that a randomly chosen team has either 4 or 6 boys."

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

 $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$

the probability is 105/286=0.367

since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in

 $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

6 0

3 years ago

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A sine function has the following key features: Period = 4 Amplitude = 3 Midline: y=−1 y-intercept: (0,-1) The function is not a

mezya [45]

Answer:

See the graphs attached and the explanation below

Explanation:

The most simple sine function, considered the parent function, is:

$y=sin(x)$

That function has:

Midline, also known as rest or equilibrium position: y = 0
Minimum: - 1
Maximum: 1
Amplitude: the distance between a minimum or a maximum and the midline = 1
period: the interval of repetition of the function = 2π

The more general sine function is:

$y=Asin(Bx+C)+D$

That function has:

Midline: y = D (it is a vertical shift from the parent function)
Minimum: - A + D
Maximum: A + D
Amplitude: A
period: 2π/B
phase shift: C (it is a horizontal shift of the from the parent function)

Now, you have to draw the sine function with the given key features:

Period = 4 ⇒ 2π/B = 4 ⇒ B = π/2

Amplitude, A = 3

midline y = - 1 ⇒ D = - 1

y-intercept = (0, -1)

Substitute the know values and use the y-intercept to find C:

$y=3sin(2x/\pi+C)-1$

Substitute (0, -1)

$-1=3sin(0+C)-1\\ \\ 3sin(C)=0\\ \\ sin(C)=0\\ \\ C=0$

Hence, the function to graph is:

$y=3sin(\pi x/2)-1$

To draw that function use this:

Maxima: 3(1) - 1 = 3 - 1 = 2, at x = 1 ± 4n (n = 0, 1, 2, 3, ...)
Minima: 3(-1) - 1 = - 3 - 1 = -4
y-intercept: (0, - 1)
x-intercepts: the solutions to 0 = 3sin(πx/2) = - 1
first point of the midline: (0, -1) it is the same y-intercept

With that you can understand the graphs attached.

4 0

3 years ago

The ratio of the length to the width of a picture is 5:2. The picture is enlarged so that its length is 15 inches now. What is t

Maru [420]

The answer is B.
The ratio is multiplied by three so 5 times 3 is 15 and 2 times 3 is 6.

8 0

3 years ago

Select the correct answer.

Ratling [72]

I think it’s D but not completely sure!

5 0

2 years ago

Please help me asap! I will give brainly!

MariettaO [177]

Answer:

Divide the number of contestants by 4 and when there is 120 contestants there is 30 winners

Step-by-step explanation:

Looking back at past sequences we can see that when we had 8 contestants we had 2 winners 8/4=2 and we also saw 16/4=4 so that was the formula now we just had to do it on 120 contestants and now we get 30 winners....Hope this helps! :)

3 0

2 years ago

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