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4 years ago

Please help me and show your work!!

1 answer:

4 0

$\bf \cfrac{\sqrt{250x^{16}}}{\sqrt{2x}}\qquad 
\begin{cases}
250=2\cdot 5\cdot 5\cdot 5\\
\qquad 2\cdot 5\cdot 5^2\\
x^{16}=x^{8\cdot 2}\\
\qquad (x^8)^2
\end{cases}\implies \cfrac{\sqrt{2\cdot 5\cdot 5^2\cdot (x^8)^2}}{\sqrt{2\cdot x}}$

$\bf \cfrac{5x^8\sqrt{2\cdot 5}}{\sqrt{2}\cdot \sqrt{x}}\implies \cfrac{5x^8\underline{\sqrt{2}}\cdot \sqrt{5}}{\underline{\sqrt{2}}\cdot \sqrt{x}}\implies \cfrac{5x^8\sqrt{5}}{\sqrt{x}}\impliedby 
\begin{array}{llll}
\textit{and rationalizing the}\\
\textit{denominator}
\end{array}$

$\bf \cfrac{5x^8\sqrt{5}}{\sqrt{x}}\cdot \cfrac{\sqrt{x}}{\sqrt{x}}\implies \cfrac{5x^8\sqrt{5}\cdot \sqrt{x}}{(\sqrt{x})^2}\implies \cfrac{5x^8\sqrt{5x}}{x}\implies 5x^8x^{-1}\sqrt{5x}
\\\\\\
5x^{8-1}\sqrt{5x}\implies 5x^7\sqrt{5x}$

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Answer:

P(X = 0) = 0.1897

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Step-by-step explanation:

For each owner, there are only two possible outcomes. Either they are insured against an earthquake, or they are not. The probability of a homeowner being insured against an earthquake is independent of other homeowners. So the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Suppose that in one metropolitan area, the chance a homeowner is insured against an earthquake is 0.34.

This means that $p = 0.34$

A sample of four homeowners are to be selected at random.

This means that $n = 4$

(a) Find the probability mass function of X. (Round your answers to four decimal places.)

The pmf is the probability of each outcome.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.34)^{0}.(0.66)^{4} = 0.1897$

$P(X = 1) = C_{4,1}.(0.34)^{1}.(0.66)^{3} = 0.3910$

$P(X = 2) = C_{4,2}.(0.34)^{2}.(0.66)^{2} = 0.3021$

$P(X = 3) = C_{4,3}.(0.34)^{3}.(0.66)^{1} = 0.1038$

$P(X = 4) = C_{4,4}.(0.34)^{4}.(0.66)^{0} = 0.0134$

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