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Margarita [4]
3 years ago
11

The perimeter of a rectangle is 34 units. Its width is 6.5 point, 5 units.

Mathematics
1 answer:
12345 [234]3 years ago
6 0
<h3><u>Question:</u></h3>

The perimeter of a rectangle is 34 units. Its width W is 6.5 units.

Write an equation to represent the perimeter in terms of the length L, and find the value of L

<h3><u>Answer:</u></h3>

The length of rectangle is 10.5 units

<h3><u>Solution:</u></h3>

Given that,

Perimeter of rectangle = 34 units

Width of rectangle = 6.5 units

Let "L" be the length of rectangle

<em><u>The perimeter of rectangle is given by formula:</u></em>

Perimeter = 2(length + width)

<em><u>Substituting the values we get,</u></em>

34 = 2(L + 6.5)

Thus the equation is found

<em><u>Solve for "L"</u></em>

L + 6.5 = \frac{34}{2}\\\\L + 6.5 = 17\\\\L = 17 - 6.5\\\\L =10.5

Thus length of rectangle is 10.5 units

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Null hypothesis:\mu_{1}\leq \mu_{2}

Alternative hypothesis:\mu_{1} > \mu_{2}

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t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value.

Based on the significance level\alpha=0.05 and \alpha/2=0.025 we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

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Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874

What is the p-value for this hypothesis test?

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

p_v =P(t_{70}>1.874)=0.033

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.

The best option would be:

Yes, since the test statistic value is greater than the critical value.

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= 9.6

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