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2 years ago

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A Norman window is rectangular on three sides with a semi-circular top. You decide to install a Norman window in your family roo

m with the dimensions indicated in the diagram. What is its perimeter?
a.
15.71 feet
b.
14.23 feet
c.
18.45 feet
d.
14 feet

2 answers:

sergiy2304 [10]2 years ago

8 0

Given that the window is rectangular on three sides with semi-circular top and with the following dimensions: 4 as the length and 3 as the width. We can calculate the perimeter using the given values, and it gives us the perimeter of 15.71 feet.

Alexxandr [17]2 years ago

5 0

Answer:

Option A. Perimeter = 15.71 feet

Step-by-step explanation:

A Norman window is rectangular on three sides with semicircular top.

We have to find the perimeter of the window.

Perimeter of Norman window = perimeter of semicircular top + perimeter of rectangle at the bottom

Perimeter of semicircular top = $\frac{2\pi r}{2}=\pi r$

= $(\pi)(\frac{3}{2})=(3.14)(1.5)$

= 4.71 feet

Perimeter of the rectangular base = (Length) + 2(width)

= 3 + (2×4) = 3 + 8

= 11 feet

Total perimeter = 4.71 + 11 = 15.71 feet

Answer is Option A.15.71 feet will be the perimeter.

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∠abc measures 115° and ∠dbc measures 70°. What is the measurement of ∠abd?

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Answer:

45 °

Step-by-step explanation:

m< ABC = m< DBC + m< ABD

= 115 _ 70

= 45 °

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2 years ago

The answer to the problem

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She puts 8% of her salary in, so multiply her pay by 8% to get her yearly amount she saves:

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3 years ago

Suppose that a random sample of 10 newborns had an average weight of 7.25 pounds and sample standard deviation of 2 pounds. a. T

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Answer:

$z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565$

$p_v =P(Z$

a) If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.

b) $\chi^2 =\frac{10-1}{1.96} 4 =18.367$

$p_v =P(\chi^2 >18.367)=0.0311$

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

Step-by-step explanation:

Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"

1) Data given and notation

$\bar X=7.25$ represent the sample mean

$s=1.2$ represent the sample standard deviation

$\sigma=1.4$ represent the population standard deviation

$n=10$ sample size

$\mu_o =7.5$ represent the value that we want to test

$\alpha=0.05,0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is less than 7.5, the system of hypothesis would be:

Null hypothesis: $\mu \geq 7.5$

Alternative hypothesis: $\mu < 7.5$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565$

4)P-value

Since is a left tailed test the p value would be:

$p_v =P(Z$

5) Conclusion

Part a

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.

Part b

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=10$ represent the sample size

$\alpha=0.01$ represent the confidence level

$s^2 =4$ represent the sample variance obtained

$\sigma^2_0 =1.96$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance increase, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \leq 1.96$

Alternative hypothesis: $\sigma^2 >1.96$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{10-1}{1.96} 4 =18.367$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 9. And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2 >18.367)=0.0311$

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(18.367,9,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

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<span>

</span>

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