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IrinaVladis [17]

3 years ago

10

10 men and 5 women are meeting in a conference room. Four people are chosen at random from the 15 to form a committee. (a) What

is the probability that the committee consists of 2 men and 2 women? (b) Among the 15 is a couple, Bob and Jane. What is the probability that Bob and Jane both end up on the committee? (c) What is the probability that Bob ends up on the committee but Jane doesn’t?

1 answer:

nata0808 [166]3 years ago

7 0

Answer:

Step-by-step explanation:

Given

There are 10 men and 5 women in a conference room

total People=15

(a)Probability that the committee consists of 2 men and 2 women

we have to select from 10 men and 5 women

$P=\frac{^5C_2\times ^{10}C_2}{^{15}C_4}$

(b)Bob and Jane end up on the same committee

as bob and Jane are necessary to include so there are 4 women and 9 men left to select remaining 2 person

$P=\frac{^{13}C_2}{^{15}C_4}$

(c)Bob is in the committee but Jane is not

9 men and 4 women left to choose other 3 person

$P=\frac{^{13}C_3}{^{15C_4}}$

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Step-by-step explanation:

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Answer:

А.The system has two solutions, but only one is viable because the other results in a negative width.

Step-by-step explanation:

Given

Let:

$L_A \to$ length of play area A

$W_A \to$ width of play area A

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$W_B \to$ width of play area B

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From the question, we have the following:

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$W_B = 2 + W_A$

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$x = y$

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This gives:

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Open bracket

$x = W_A + 4W_A^2$

The area of B is:

$y = L_B * W_B$

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Substitute: $W_B = 2 + W_A$

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Open brackets

$y = (2 + 6 + 3W_A) * ( 2 + W_A)$

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Expand

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Using quadratic calculator, we have:

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