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Fynjy0 [20]
3 years ago
14

Assume that a student attends a four-year college with tuition costs of $20,000 per year, room and board costs of $5,000 per yea

r, and books/entertainment costs of $1,000 per year. If the student did not go to college, she would work at a job that pays $25,000 per year but still face the same room and board and entertainment expenses. The opportunity cost of attending college for this student for four years is:
Mathematics
1 answer:
Kamila [148]3 years ago
7 0

Answer:

$180,000.

Step-by-step explanation:

We have been given that a student attends a four-year college with tuition costs of $20,000 per year, room and board costs of $5,000 per year, and books/entertainment costs of $1,000 per year. If the student did not go to college, she would work at a job that pays $25,000 per year but still face the same room and board and entertainment expenses.

We know that opportunity cost is the profit lost, when one alternative is selected over another.

Opportunity cost of attending college for 4 years would be amount lost from job in 4 years plus amount spent on studies because the student lost $25,000 for four years for not doing the job and spent $20,000 for 4 years on studies.

4(\$25000+\$20,000)=4(\$45,000)=\$180,000

Therefore, the opportunity cost of attending college for this student for four years is $180,000.

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Answer:

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Step-by-step explanation:

5(2x + 1) < 10

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Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

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

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

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

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

5 0
2 years ago
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