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3 years ago

Circumference of 150.7in. what is the area

1 answer:

3 0

   Circumference of any circle = (pi) (diameter)

      150.7 inches = (pi) (diameter)

Divide each side by (pi): 150.7/pi inches = diameter

Radius = (1/2) (diameter) = 150.7/ 2pi inches

   Area of any circle = (pi) (radius)²

      Area of THIS circle = (pi) (150.7in / 2pi)²

   = (pi) (150.7in)² / 4 pi²

   =     (150.7in)² / 4 pi

   = 1,807.2 inch² .

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What is the first step in constructing an inscribed circle inside triangle XYZ?

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Answer:

Construct the angle bisector of angle Y

Step-by-step explanation:

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3 years ago

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3 years ago

An insurance company states that it settles 85% of all life insurance claims within 30 days. A consumer group asks the state ins

BlackZzzverrR [31]

Using the z-distribution, it is found that since the p-value of the test is less than 0.05, we reject the null hypothesis and find that there is enough evidence to conclude that the true proportion of all life insurance claims made to this company that are settled within 30 days is less than 85%.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is not enough evidence that the proportion is below 85%, that is:

$H_0: p \geq 0.85$

At the alternative hypothesis, it is tested if there is enough evidence that the proportion is below 85%, that is:

$H_0: p < 0.85$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

$\overline{p}$ is the sample proportion.

p is the proportion tested at the null hypothesis.

n is the sample size.

For this problem, the parameters are:

$p = 0.85, n = 250, \overline{p} = \frac{203}{250} = 0.812$

Hence the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.812 - 0.85}{\sqrt{\frac{0.85(0.15)}{250}}}$

z = -1.68

<h3>What is the p-value and the conclusion</h3>

Using a z-distribution calculator, with a left-tailed test, as we are testing if the proportion is less than a value, and z = -1.68, the p-value is of 0.0465.

Since the p-value of the test is less than 0.05, we reject the null hypothesis and find that there is enough evidence to conclude that the true proportion of all life insurance claims made to this company that are settled within 30 days is less than 85%.

More can be learned about the z-distribution at brainly.com/question/16313918

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2 years ago

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Answer:

A. 26

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4 years ago