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Vera_Pavlovna [14]
3 years ago
5

Gabrielle buys jeans pants for $32 each and marks up

Mathematics
1 answer:
const2013 [10]3 years ago
6 0

Answer: $11.20

Work: She buys each Jean for $32, and the markup is 35%. After a little multiplication, the new price is $43.20. When someone buys the Jeans at that price, she makes $11.20 because $43.20 is $11.20 more then the price she bought it for.

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It’s option A, Choi’s option A

Step-by-step explanation:

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Plz help me with this
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So far soon-jin has sent out 48 invitations to her party. This is 4/5 a of all of the invitations she will send out.
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3 years ago
Assuming that the heights of college women are normally distributed with mean 64 inches and standard deviation 1.5 inches, what
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Answer:

15.74% of women are between 65.5 inches and 68.5 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}

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 64, \sigma = 1.5

What percentage of women are between 65.5 inches and 68.5 inches?

This percentage is the pvalue of Z when X = 68.5 subtracted by the pvalue of Z when X = 65.5.

X = 68.5

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

Z = \frac{68.5 - 64}{1.5}

Z = 3

Z = 3 has a pvalue of 0.9987

X = 65.5

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

Z = \frac{65.5 - 64}{1.5}

Z = 1

Z = 1 has a pvalue of 0.8413

So 0.9987 - 0.8413 = 0.1574 = 15.74% of women are between 65.5 inches and 68.5 inches.

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