Answer:
I think the answer is C ,The sample proportion is the fraction of samples which were successes, so. (1) For large , has an approximately normal distribution.
Step-by-step explanation:
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Here you'll want to add 5/8 with 5/8. Since they have the same denominator already then 5/8+5/8 = 10/8. Simplifying that down to 5/4 then changing to a mixed number is 1 1/4 inches.
Answer:

Step-by-step explanation:
![if \: the \: question \: is \: f[g(4)] \\ then \: at \: first \: solve \: for \: g(4) \\ g(4) = {4}^{2} \\ f[g(4)] = 4( {4}^{2} ) + 2 \\ f[g(4)] = 4(16) + 2 \\ f[g(4)] =64 + 2 \\ f[g(4)] = \boxed{66}](https://tex.z-dn.net/?f=%20if%20%5C%3A%20the%20%5C%3A%20question%20%5C%3A%20is%20%5C%3A%20f%5Bg%284%29%5D%20%5C%5C%20then%20%5C%3A%20at%20%5C%3A%20first%20%5C%3A%20solve%20%5C%3A%20for%20%5C%3A%20g%284%29%20%5C%5C%20g%284%29%20%3D%20%20%7B4%7D%5E%7B2%7D%20%20%5C%5C%20f%5Bg%284%29%5D%20%20%3D%204%28%20%7B4%7D%5E%7B2%7D%20%29%20%2B%202%20%5C%5C%20f%5Bg%284%29%5D%20%20%3D%204%2816%29%20%2B%202%20%5C%5C%20f%5Bg%284%29%5D%20%20%3D64%20%2B%202%20%5C%5C%20%20%20f%5Bg%284%29%5D%20%20%3D%20%20%5Cboxed%7B66%7D)
Answer: The z-scores for a woman 6 feet tall is 2.96 and the z-scores for a a man 5'10" tall is 0.25.
Step-by-step explanation:
Let x and y area the random variable that represents the heights of women and men.
Given : The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches.
i.e.

Since , 
Then, z-score corresponds to a woman 6 feet tall (i.e. x=72 inches).
[∵ 1 foot = 12 inches , 6 feet = 6(12)=72 inches]

Men the same age have mean height 69.3 inches with standard deviation 2.8 inches.
i.e.

Then, z-score corresponds to a man 5'10" tall (i.e. y =70 inches).
[∵ 1 foot = 12 inches , 5 feet 10 inches= 5(12)+10=70 inches]

∴ The z-scores for a woman 6 feet tall is 2.96 and the z-scores for a a man 5'10" tall is 0.25.
Answer:
b i think
Step-by-step explanation: