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evablogger [386]
3 years ago
11

A brick wall will be shaped like a rectangular prism. The wall needs to be 3ft tall and the builders have enough bricks for the

wall to have a volume of 330 cubic feet. What tool can be used to find possible dimensions for the base of the wall?
Mathematics
1 answer:
faltersainse [42]3 years ago
7 0

Answer:

For a rectangular prism that has a base area B, and a height H, the volume is:

V = B*H

This is the only "tool" that we need to use in this situation.

We know that the wall needs to be 3ft tall, then H = 3ft

And we also know that we have enough bricks for a wall of a volume:

V = 330 ft^3

If we replace these two in the volume equation, we get:

330ft^3 = B*3ft

Now we can solve this for B, and thus find the volume of the base of the wall.

To do it, we just need to divide both sides by 3ft

(330 ft^3)/3ft = B*3ft/3ft

110 ft^2 = B

This means that the area of the base is 110ft^2

As the base is a rectangle of width W and length L, we must have:

110ft^2 = B = L*W

Then the possible measures of the base are given by the linear relation:

L = 110ft^2/W

Where we also need to add some trivial restrictions, like:

L > 0 ft

W > 0ft

This only means that we can not have a length or width equal to or smaller than zero, as those do not have physical sense.

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A machine that puts corn flakes into boxes is adjusted to put an average of 15.6 ounces into each box, with standard deviation o
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Answer:

\chi^2 =\frac{15-1}{0.0676} 0.1521 =31.5

p_v =P(\chi^2_{14} >31.5)=0.0047

If we compare the p value and the significance level assumed of 0.05 we see that p_v so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly higher than 0.26^2=0.0676.

Step-by-step explanation:

Notation and previous concepts

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=15 represent the sample size

\alpha represent the confidence level  

s^2 =0.39^2 =0.1521 represent the sample variance obtained

\sigma^2_0 =0.26^2 =0.0676 represent the value that we want to test

Null and alternative hypothesis

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

Null Hypothesis: \sigma^2 \leq 0.0676

Alternative hypothesis: \sigma^2 >0.0676

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{15-1}{0.0676} 0.1521 =31.5

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case df=15-1=14. And since is a right tailed test the p value would be given by:

p_v =P(\chi^2_{14} >31.5)=0.0047

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

"=1-CHISQ.DIST(31.5,14,TRUE)"

Conclusion

If we compare the p value and the significance level assumed of 0.05 we see that p_v so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly higher than 0.26^2=0.0676.

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