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Lunna [17]
3 years ago
15

Find the unknown side length of the rectangle is 3 2/3 and if the area is 11m2

Mathematics
1 answer:
patriot [66]3 years ago
7 0

Answer:

The width of the rectangle given is equal to 3m.

Step-by-step explanation:

In order to solve this question we need to know that the areas of the rectangle is equal to A_{rectangle} = (l)(w), where "l" is length and "w" is width. So know in order to solve this question we would have to find out the unknown width by using area and the length of the triangle. Based on the formula I wrote down earlier we can understand the following.....

A_{rectangle} = (l)(w)

\frac{A_{rectangle} }{l}  = \frac{(l)(w)}{l}

w = \frac{A_{rectangle} }{l}

Just now we found out the formula that we can use to solve for the width when we know the area and the length of the rectangle. Now in order to get the answer we just have to plug in the values we know, and we get........

w = \frac{A_{rectangle} }{l}

w = \frac{11m^{2} }{(3\frac{2}{3}m )}

w = 3m

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Answer:

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

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Step-by-step explanation:

Data given

\bar X_{1}=810 represent the mean for sample 1  

\bar X_{2}=770 represent the mean for sample 2  

s_{1}=67 represent the sample standard deviation for 1  

s_{2}=56 represent the sample standard deviation for 2  

n_{1}=13 sample size for the group 2  

n_{2}=19 sample size for the group 2  

\alpha=0.05 Significance level provided

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

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

And the degrees of freedom are given by df=n_1 +n_2 -2=13+19-2=30  

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.  

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

t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771  

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

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Comparing the p value with the significance level \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

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