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3 years ago

Suppose a family has four children. What is the probability that at least two are female?

1 answer:

4 0

P(all 4 are male) = 1/2^4 = 1/16 ( assuming probability of 1 male = 1/2)

P( 3 are male) = 1/2^4*4 = 1/4

The rest will have at least 2 females

So probability of at least 2 females) = 1 - 1/16 - 1/4 = 1 - 5/16 = 11/16 answer

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If x+1/x= 3, then prove that m^5+1/m^5= 123

alina1380 [7]

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Explanation:

We can start with the relations ...

$\displaystyle\left(x+\frac{1}{x}\right)^3=\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)\\\\\left(x+\frac{1}{x}\right)^5=\left(x^5+\frac{1}{x^5}\right)+5\left(x^3+\frac{1}{x^3}\right)+10\left(x+\frac{1}{x}\right)\\\\\textsf{From these, we can derive ...}\\\\x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)-10\left(x+\frac{1}{x}\right)$

$\displaystyle x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(x+\frac{1}{x}\right)^3+5\left(x+\frac{1}{x}\right)\right)\\\\x^5+\frac{1}{x^5}=3^5 -5(3^3)+5(3)\\\\=((3^2-5)3^2+5)\cdot3=(4\cdot9+5)\cdot3=(41)(3)\\\\=\boxed{123}$

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2 years ago

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3 years ago

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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