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3 years ago

Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!

Mathematics

2 answers:

Novosadov [1.4K]3 years ago

6 0

Answer:

101

Step-by-step explanation:

h(t)= -20+11t

h(11)= -20+11(11)

h(11)= -20+121

h(11) = 101

Kitty [74]3 years ago

3 0

Answer & Step-by-step explanation:

In the problem, we are given an equation. h(t) = -20 + 11t

In the equation, t represents an unknown value. So, if we are given a number that is in the replacement of t, then we can plug that number in to where t is at.

t = 11

h(11) = -20 + 11(11)

h(11) = -20 + 121

h(11) = 101

So, h(11) is equal to 101

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Answer:

A. mean=14.29

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C. option 3

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Step-by-step explanation:

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3 years ago

The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.

Rainbow [258]

Answer:

(a) 0.2061

(b) 0.2514

Step-by-step explanation:

Let <em>X</em> denote the heights of women in the USA.

It is provided that <em>X</em> follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

$P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z$

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

$P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z$

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

$P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0$

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

8 0

3 years ago