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pantera1 [17]
3 years ago
8

A company has two separate offices. One office has a color printer and the other office has a black and white

Mathematics
1 answer:
Vitek1552 [10]3 years ago
7 0

Answer:

the larger of two printers being used to print the payroll for a major corporation requres 40 min. to print the payroll.So the smaller printer's payroll printing rate is  

1 payroll per x min or  or  

>>...After both printers have been operating for 10 min, the larger printer malfunctions...<<

So the fraction of the payroll that the larger printer did in the 10 minutes

is  = payroll.

Step-by-step explanation:

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Suppose that the weights of passengers on a flight to Greenland on Frigid Aire Lines are normal with mean 175 pounds and standar
Natali5045456 [20]

Answer:

The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds

Step-by-step explanation:

Problems of normally distributed samples are 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 = 175, \sigma = 22

What is the minimum weight for a passenger who outweighs at least 90% of the other passengers?

90th percentile

The 90th percentile is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. So

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

1.28 = \frac{X - 175}{22}

X - 175 = 22*1.28

X = 203.16

The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds

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The question presented is not answerable because it just presented a mere statement. 
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