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Helga [31]
3 years ago
9

10 points someone help me please!?

Mathematics
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

\large\boxed{b=\sqrt{95}}

Step-by-step explanation:

Use the Pyhagorean theorem:

leg^2+leg^2=hypotenuse^2

We have

leg=b,\ leg=7,\ hypotenuse=12

Substitute:

b^2+7^2=12^2

b^2+49=144       <em>subtract 49 from both sides</em>

b^2=95\to b=\sqrt{95}

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Assume that​ women's heights are normally distributed with a mean given by mu equals 62.5 in​,and a standard deviation given by
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Answer:

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Let X be the random variable that represents the height of a woman. Then, X is normally distributed with  

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(a) P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx = 0.5899

   (in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking P(\bar{X} < 63) where n = 37. \bar{X} is normally distributed with mean 62.5 in and standard deviation 2.2/\sqrt{37}. So, the probability density function is given by

g(x) = \frac{1}{\sqrt{2\pi}\frac{2.2}{\sqrt{37}}}\exp{-\frac{(x-62.5)^{2}}{2(2.2/\sqrt{37})^{2}}}, and

P(\bar{X} < 63) = \int\limits_{-\infty}^{63}g(x)dx = 0.9166

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2/sqrt(37))

You can use a table from a book to find the probabilities or a programming language like the R statistical programming language.

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3 years ago
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