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Juliette [100K]
3 years ago
8

I will give you brainliest!!!! Which statement is true about the values x=3 and y=-1 ?

Mathematics
1 answer:
GenaCL600 [577]3 years ago
6 0

Answer:

D

Step-by-step explanation:

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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
Please help ill give whoever answers the best the brainliest answer
GalinKa [24]
The third chart i think
6 0
2 years ago
Frank invested $30,000 at 4% simple interest. How much interest will he earn each year?
Oduvanchick [21]

Answer:

$1200 interest

Step-by-step explanation:

$30,000 divided by 100 = 300 , 300 multipled by 4 = 1200 , thats how to find what 4% would be

3 0
3 years ago
If a tires k=0.85, what does this tell you about the tire installed on the car as opposed to the factory-installed tire?
Gala2k [10]

Answer: since I don’t understand this question I’m choosing D

Step-by-step explanation:

8 0
3 years ago
5. Complete the following equation using &lt; &gt;, or =<br> 7___24/4
Zepler [3.9K]

Answer:

7 > 24/4

Step-by-step explanation:

7 __ 24/4

simplify

7 __ 6

7  > 6

3 0
3 years ago
Read 2 more answers
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