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N76 [4]
3 years ago
9

Solve for x. Enter your answer in the box. x =____°

Mathematics
2 answers:
irga5000 [103]3 years ago
7 0
The sum of all interior angles is (n-2)x180 while n=number of sides. So (6-2)x180=720. The angles you have add up to 588. All you need to do now is 720-588. x=132
Vlad1618 [11]3 years ago
6 0

we know that

The sum of all interior angles in the polygon is equal to the formula

(n-2)*180\°

where

n is the number of sides of the polygon

In this problem we have

n=6

substitute the value in the formula

(6-2)*180\°=720\°

<u>Find the value of x</u>

720\°=x\°+108\°+113\°+132\°+75\°+160\°\\720\°=x\°+588\°\\x\°=720\°-588\°\\x=132\°

therefore

<u>the answer is</u>

the measure of x is 132\°


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4 0
3 years ago
Suppose a certain airline uses passenger seats that are 16.2 inches wide. Assume that adult men have hip breadths that are norma
Pachacha [2.7K]

Answer:

Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.

There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem

Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that \mu = 14.4, \sigma = 1.1.

What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?

For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:

Z = \frac{X - \mu}{\sigma}

Z = \frac{16.2 - 14.4}{1.1}

Z = 1.64

Z = 1.64 has a pvalue of 0.9495.

This means that each male has a 1-0.9495 = 0.0505 = 5.05% probability of having a hip width greater than 16.2 inches.

For the average of the sample

What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches?

Now, we need to find the standard deviation of the sample before using the zscore formula. That is:

s = \frac{\sigma}{\sqrt{110}} = 0.1.

Now

Z = \frac{X - \mu}{\sigma}

Z = \frac{16.2 - 14.4}{0.1}

Z = 18

Z = 18 has a pvalue of 0.9999.

This means that there is a 1-0.9999 = 0.0001 = 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

7 0
3 years ago
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