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Masteriza [31]
3 years ago
8

The pentagon below has an area of 8.3 square centimeters and a side of 2.2. What is the approximate length of the apothem? A.0.4

cm B.0.8 cm C.1.5 cm D.1.9 cm
Mathematics
2 answers:
RideAnS [48]3 years ago
7 0

Answer:

C-1.5 cm

Step-by-step explanation:

nalin [4]3 years ago
3 0

Answer:

C. 1.5 cm

Step-by-step explanation:

A pentagon is a polygon with 5 sides.  

The formula for the area of a regular polygon is

A=\frac{1}{2}ap

where a is the apothem and p is the perimeter.  The perimeter is found by multiplying the number of sides of the figure by the length of 1 of those sides:

5(2.2) = 11

Now we have everything we need to solve for the length of the apothem.  

8.3=\frac{1}{2}a(11)

Begin by multiplying both sides by 2 to get rid of the fraction, so

16.6 = 11a  Divide both sides by 11 to get

a = 1.5090909 or

C. 1.5 cm

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A researcher wishes to see if the five ways (drinking decaffeinated beverages, taking a nap, going for a walk, eating a sugary s
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Answer:

\chi^2 = \frac{(21-12)^2}{12}+\frac{(16-12)^2}{12}+\frac{(10-12)^2}{12}+\frac{(8-12)^2}{12}+\frac{(5-12)^2}{12}=13.833

p_v = P(\chi^2_{4} >13.833)=0.00785

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(13.833,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the  drowsiness are NOT equally distributed among office workers .

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".

Assume the following dataset:

Method   Beverage   Nap   Walk   Snack   Other

Number       21             16       10         8         5

The total on this case is 60

We need to conduct a chi square test in order to check the following hypothesis:

H0: Drowsiness are equally distributed among office workers

H1: Drowsiness IS NOT equally distributed among office workers

The level of significance assumed for this case is \alpha=0.1

The statistic to check the hypothesis is given by:

\chi^2 = \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{total}{5}

And the calculations are given by:

E_{Beverage} =\frac{60}{5}=12

E_{Nap} =\frac{60}{5}=12

E_{Walk} =\frac{60}{5}=12

E_{Snack} =\frac{60}{5}=12

E_{Other} =\frac{60}{5}=12

And the expected values are given by:

Method   Beverage   Nap   Walk   Snack   Other

Number       12             12       12         12         12

And now we can calculate the statistic:

\chi^2 = \frac{(21-12)^2}{12}+\frac{(16-12)^2}{12}+\frac{(10-12)^2}{12}+\frac{(8-12)^2}{12}+\frac{(5-12)^2}{12}=13.833

Now we can calculate the degrees of freedom for the statistic given by:

df=(catgories-1)=(5-1)=4

And we can calculate the p value given by:

p_v = P(\chi^2_{4} >13.833)=0.00785

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(13.833,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the  drowsiness are NOT equally distributed among office workers .

7 0
3 years ago
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