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ki77a [65]
3 years ago
10

A pair of 120 shoes was discounted 20 percent a costumer had a coupon for an additional 5 off what was the final price of the sh

oes ?
Mathematics
1 answer:
STatiana [176]3 years ago
8 0

Answer:

The final price of the shoes is $91.20.  

Step-by-step explanation:

In order to find the final price of the shoes, we need to calculate each discount separately.  Since the initial cost of the shoes is $120, we need to first find the cost after the 20% discount.  If the shoes are discounted 20%, then we can find the cost by multiplying the original amount by 80%, or 0.8:  $120 x 0.80 = $96.00.  Since the customer also has an additional 5% off with the coupon, we can then take the new total, $96.00, and multiply by 95%, or 0.95, to get the final cost:  $96.00 x 0.95 = $91.20.  

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The U.S. Bureau of Labor Statistics reports that of persons who usually work full-time, the average number of hours worked per w
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Answer:

The standard deviation of number of hours worked per week for these workers is 3.91.

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 we have that:

The average number of hours worked per week is 43.4, so \mu = 43.4.

Suppose 12% of these workers work more than 48 hours. Based on this percentage, what is the standard deviation of number of hours worked per week for these workers.

This means that the Z score of X = 48 has a pvalue of 0.88. This is Z between 1.17 and 1.18. So we use Z = 1.175.

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

1.175 = \frac{48 - 43.4}{\sigma}

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\sigma = \frac{4.6}{1.175}

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The standard deviation of number of hours worked per week for these workers is 3.91.

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