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Thepotemich [5.8K]
3 years ago
7

Find the area of A cylinder has a volume of 175 cubic units and a height of 7 units. The diameter of the cylinder is

Mathematics
1 answer:
damaskus [11]3 years ago
5 0
To find the area of the cylinder we need to find its volume first. Remember that the formula for the volume of a cylinder is V= \pi r^{2} h
where:
V is the volume 
r is the radius 
h is the height 
From the question we know that A=175 and h=7. Lets replace those values in our volume formula:
175= \pi r^{2} 7
Now we can solve for r to find our radius:
r^{2} = \frac{175}{7 \pi }
r^{2} = \frac{25}{ \pi }
r= \sqrt{ \frac{25}{ \pi } }
r= \frac{5}{ \sqrt{ \pi } }

Now that we know the radius, we can use the formula for the area of a cylinder A=2 \pi rh+2 \pi r^{2}
where:
A is the area
r is the radius 
h is the height
We know now that r= \frac{5}{ \sqrt{ \pi } } and h=7, so lets replace those values in our area formula:
A=2 \pi ( \frac{5}{ \sqrt{ \pi } } )(7)+2 \pi ( \frac{5}{ \sqrt{ \pi } })^{2}
A= \frac{70 \pi }{ \sqrt{ \pi } } +50
A=174.07

We can conclude that the area of a cylinder that has a volume of 175 cubic units and a height of 7 units is 174.07 square units.

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df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday  Tuesday  Wednesday Thursday Friday Saturday    Total

 12             9                 11                 10           9            9              60

The level of significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{60}{6}= 10 and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0
3 years ago
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