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Vinvika [58]
3 years ago
6

kenny made pillows to sell at the fair. He paid $175.96 for supplies. He charged $8.75 for each pillow sold. Kenny sold 27 pillo

ws at the fair. What was kenny's profit in dollars
Mathematics
2 answers:
kolbaska11 [484]3 years ago
8 0

Answer:

Profit is the difference between the amount earned and the amount spent. So you multiply the number of items sold(27 pillows) by the amount they were sold for($8.75) and you get the amount earned($236.25). Then you find the difference(subtract) of the amount earned($236.25) and the amount spent($175.96) on the supplies and you come to the profit($60.29).

$8.75 × 27 = $236.25

$236.25 - $175.96 = $60.29

Alex777 [14]3 years ago
6 0

Answer:

$236.25

Step-by-step explanation:

MULTIPLY $8.75 AND 27

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1)5/6 of 3/4÷7/8×2/2<br>2)3/2of3/4÷8/2​
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7 0
3 years ago
PVC pipe is manufactured with a mean diameter of 1.01 inch and a standard deviation of 0.003 inch. The diameters are known to be
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Answer:

0.8185

Step-by-step explanation:

To solve this question, we have to understand the normal probability ditribution and the central limit theorem.

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 1.01, \sigma = 0.003, n = 9, s = \frac{0.003}{\sqrt{9}} = 0.001

Find the probability that a random sample of n = 9 sections of pipe will have a sample mean diameter greater than 1.009 inch and less than 1.012 inch.

This is the pvalue of Z when X = 1.012 subtracted by the pvalue of Z when X = 1.009. So

X = 1.012

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

By the Central limit theorem

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

Z = \frac{1.012 - 1.01}{0.001}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 1.009

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

Z = \frac{1.009 - 1.01}{0.001}

Z = -1

Z = -1 has a pvalue of 0.1587

0.9772 - 0.1587 = 0.8185

0.8185 = 81.85% probability that a random sample of n = 9 sections of pipe will have a sample mean diameter greater than 1.009 inch and less than 1.012 inch.

8 0
3 years ago
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