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kelly has 47 model cars. Jamal has 58 model cars. Yesterday , jamal traded 8 model cars to kelly for 14 of her cars. How many ca

lana [24]

Answer:

33 cars please give Brainliest

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

What is the slope of the line? A. 0 B. 5 C. 11 D. Undefined

True [87]

Answer:

D. Undefined

Step-by-step explanation:

The slope is the ratio of rise to run. When the line is vertical, the run is zero, so calculating the slope involves division by zero. That makes the slope be undefined.

3 0

4 years ago

A solid object was sliced to form two new objects Each of the two new objects had a circular base Which shape could not have bee

SpyIntel [72]

The answer is a prism. If you think about it when you cut a cone, cylinder, or sphere in half you get a circle. If you cut a prism you'd get a quadrilateral.

5 0

3 years ago

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Suppose a normal distribution has a mean of 266 and a standard deviation of

Elis [28]

Answer:

I think the answer is P(x>178)

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers