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2 years ago

10

I need to solve for the indicated pieces of the right triangle and round sides and angles to the nearest tenth also find What x

= and m

1 answer:

Troyanec [42]2 years ago

3 0

Check the picture below.

Make sure your calculator is in Degree mode.

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The average height of a 13 year old male in the U.S. is 60 inches, with a standard deviation of 2 inches. The average weight of

Romashka [77]

Since the average height is 60 inches and its deviation is 2 inches, one deviation to the right (or higher) is 62 inches (60 + 2). Two deviations is 64 inches, three deviations is 66 inches, and four deviations is 68 inches.

Since the average weight is 100 pounds and its deviation is 5 inches, we repeat the process from finding heights to get to 115 pounds. That takes three deviations.

The MORE deviations away, the more unusual it is. So the height (4 deviations) is more unusual than the weight (3 deviations).

5 0

3 years ago

Read 2 more answers

How do you do this?

sammy [17]

its all about the angles just subtract the number you have from the total

5 0

3 years ago

If f(x)=4^x, find f(4)

lina2011 [118]

Answer:

$\huge\boxed{f(4) = 256}$

Step-by-step explanation:

<u>Given function is:</u>

$f(x) = 4^x$

Put x = 4

So,

$f(4) = 4^4\\\\f(4) = 256\\\\\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807</h3>

6 0

2 years ago

I need help asap

Nastasia [14]

Gggggggggggggggggggggggggggggggggggggg. It would be 7

8 0

3 years ago

In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there

Vedmedyk [2.9K]

Answer:

a) 151lb.

b) 6.25 lb

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

In this problem, we have that:

$\mu = 151, \sigma = 25, n = 16$

So

a) The expected value of the sample mean of the weights is 151 lb.

(b) What is the standard deviation of the sampling distribution of the sample mean weight?

This is $s = \frac{\sigma}{\sqrt{n}} = \frac{25}{\sqrt{16}} = 6.25$

8 0

3 years ago