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KIM [24]
3 years ago
7

Slope????????????????​

Mathematics
1 answer:
MAXImum [283]3 years ago
6 0
M = Undefined (because the line is perfectly vertical)
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$12 meal; 4.5% tax find total cost to the nearest cent
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3 years ago
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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3 years ago
Use the distributive property to simplify the following expression: 3y(2 - y) + 8y
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3y(2 - y) + 8y
6y - 3y^2 + 8y
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