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miskamm [114]
2 years ago
6

Jack knows the surface area of a cylinder and its radius. He wants to find the cylinder's height. He needs to rewrite the formul

a to find the cylinder's height (h) in terms of the cylinder’s surface area (A) and its radius (r). Which is the correct formula?
Mathematics
2 answers:
Sindrei [870]2 years ago
4 0
This formula should help you.

bagirrra123 [75]2 years ago
3 0

Answer:

h=\frac{A}{2\pi r} -r

Step-by-step explanation:

We know the formula for the surface area of a cylinder is:

A=2\pi r(r+h)

where;

A is the surface area of the cylinder

r is the radius of the cylinder

h is the height of the cylinder

Now, we have to rewrite the formula to find the cylinder's height (h)

So, we will isolate the variable 'h'

Dividing both sides by 2\pi r , we get

h=\frac{A}{2\pi r} -r

Thus, the formula for 'h' with respect to A and 'r' is h=\frac{A}{2\pi r} -r

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Answer:

We conclude that the mean Ohio score is below the national average.

Step-by-step explanation:

We are given that a random survey of 1000 students nationwide showed a mean ACT score of 21.1. Ohio was not used.

A survey of 500 randomly selected Ohio scores showed a mean of 20.8. The population standard deviation is 3.

<u><em>Let </em></u>\mu<u><em> = mean Ohio scores.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 21.1      {means that the mean Ohio score is above or equal the national average}

Alternate Hypothesis, H_A : \mu < 21.1      {means that the mean Ohio score is below the national average}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about population standard deviation;

                    T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n}}}  ~ N(0,1)

where, \bar X = sample mean Ohio score = 20.8

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So, <u><em>test statistics</em></u>  =  \frac{20.8-21.1}{\frac{3}{\sqrt{500}}}  

                              =  -2.24

The value of z test statistics is -2.24.

<em>Now, at 0.1 significance level the z table gives critical value of -1.2816 for left-tailed test.</em><em> Since our test statistics is less than the critical values of z as -2.24 < 1.2816, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the mean Ohio score is below the national average.

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