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ohaa [14]
3 years ago
7

21. The Johnsons framed a family picture to hang on the wall. The perimeter of the frame is 72 inches. Use the formula P = 2 l +

2 w to find the length of the frame if the width is 14 inches. 21. The Johnsons framed a family picture to hang on the wall. The perimeter of the frame is 72 inches. Use the formula P = 2 l + 2 w to find the length of the frame if the width is 14 inches.
Mathematics
2 answers:
Arlecino [84]3 years ago
8 0
So here's the solution to the given problem above.
Given that the perimeter of the frame is 72 inches and the width is 14 inches, we are going to look for the length. The formula for perimeter is 2L + 2W.
So let us substitute the values.
P = 2L + 2w
72 = 2L + 2(14)
72 = 2L + 28
72 -28 = 2L
44 = 2L <<<divide both sides by 2
22 = L
So the length of the picture frame is 22 inches.
To check, 22 in x 2 is 44 inches + 14 x 2 is 28 inches. 28 + 44 is 72 inches.
Hope this answer helps. Thanks for posting your question.
alex41 [277]3 years ago
5 0
P = 2l + 2w
72 = 2l + 2(14) = 2l + 28
2l = 72 - 28 = 44
l = 44/2 = 22 inches.
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Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the online platform was accessed significantly more so than before the stay at home order.

Then, the null and alternative hypothesis are:

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The significance level is 0.05.

The sample has a size n=108000.

The sample mean is M=1378.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=489.

The estimated standard error of the mean is computed using the formula:

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Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1378-1000}{1.488}=\dfrac{378}{1.488}=254.036

The degrees of freedom for this sample size are:

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This test is a right-tailed test, with 107999 degrees of freedom and t=254.036, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t>254.036)=0

As the P-value (0) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the online platform was accessed significantly more so than before the stay at home order.

The effect size can be estimated with the Cohen's d.

This can be calculated as:

d=\dfrac{M-\mu}{\sigma}=\dfrac{1378-1000}{489}=\dfrac{378}{489}=0.77

The values of Cohen's d between 0.2 and 0.8 are considered "Medium", so in this case, the effect size d=0.77 is medium.

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