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4 years ago

The blue segment below is a diameter of 0. What is the length of the radius of the circle

Mathematics

2 answers:

WINSTONCH [101]4 years ago

7 0

Answer:

<h2>B. 5.1 units.</h2>

Step-by-step explanation:

Givens

The diameter is

$d=10.2u$

Now, by definition, the radius is

$r=\frac{d}{2}$

Replacing given values, we have

$r=\frac{10.2u}{2}\\ r=5.1 u$

Therefore, the right answer is B.

Norma-Jean [14]4 years ago

3 0

Answer:

Option (B) 5.1 units

Step-by-step explanation:

Since the radius of the circle is half of its diameter and given the diameter of the circle is 10.2 units then the radius would be $\frac{10.2}{2} = 5.1 units$

Thus the answer.

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What is the measure of <R?

Verdich [7]

Answer:

<R=23°

Step-by-step explanation:

The sum of angles in a triangle is 180°

So, <R + <S + <T = 180°

<R+90°+67°=180°

<R+157°=180°

<R=23°

7 0

4 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:

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

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

The critical value is: $t_c = 1.74$

The decision rule is:

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

Since t = 1.41 < 1.74, we do not reject the null hypothesis, 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 at most 10 pounds, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of more than 10 pounds, 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 significance level of 0.05 and 18 - 1 = 17 df.

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

Item c:

We have the standard deviation for the sample, 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 do not reject the null hypothesis, 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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