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Darya [45]
3 years ago
7

In a class of 25 students, 15 of them have a cat,

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
7 0

Answer: 9/25

Step-by-step explanation:

Since 3 of them have neither a cat or a dog, that means 22 (25-3) of them have a dog, a cat, or both.

16 + 15 = 31, which is 9 more than 22. That means 9 of them have both a dog and a cat.

Out of the 25 people, 9 have both a cat and dog, which means the probablity a student chosen at random has both is 9/25.

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Bovine deposits $7,000 into an investment that pays 4% interest compounded quarterly. if she leaves it in the account for 9 year
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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
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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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Look at image for question
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