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2 years ago

5

The area of a circle is 36

class="latex-formula">
cm?.
What is the radius of this circle?

2 answers:

dmitriy555 [2]2 years ago

8 0

6 ??????????????????????????????????????

Anettt [7]2 years ago

4 0

I’m pretty sure it’s 6 bc to calculate the volume it’s pie•radius^2

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2 years ago

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

Testing the hypothesis, it is found that:

a)

The null hypothesis is: $H_0: \mu \leq 10$

The alternative hypothesis is: $H_1: \mu > 10$

b)

The critical value is: $t_c = 1.74$

The decision rule is:

If t < 1.74, we <u>do not reject</u> the null hypothesis.
If t > 1.74, we <u>reject</u> the null hypothesis.

c)

Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of <u>at most 10 pounds</u>, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of <u>more than 10 pounds</u>, that is:

$H_1: \mu > 10$

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a <u>significance level of 0.05</u> and 18 - 1 = <u>17 df.</u>

Hence, using a calculator for the t-distribution, the critical value is: $t_c = 1.74$ .

Hence, the decision rule is:

If t < 1.74, we <u>do not reject</u> the null hypothesis.
If t > 1.74, we <u>reject</u> the null hypothesis.

Item c:

We have the <u>standard deviation for the sample</u>, hence the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

For this problem, we have that:

$\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18$

Thus, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}$

$t = 1.41$

Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

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2 years ago