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zepelin [54]
3 years ago
15

A local nursery sells a large number of ornamental trees every year. The owners have determined the cost per tree C for buying a

nd caring for each tree before it is sold is C = 0.001n2 - 0.3n + 50. In this function, C is the cost per tree in dollars and n is the number of trees in stock.
a. How many trees will minimize the cost per tree?
b. What will the minimum cost per tree be?
Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0

The minimum cost is achieved when 150 trees are produced giving a minimum cost of $27.5

Polynomial is an expression that involves the <em>operations of addition, subtraction, multiplication of variables.</em>

Let C represent the cost for buying and caring for n trees. Given that:

C = 0.001n² - 0.3n + 50.

The minimum cost is at dC/dn = 0, hence:

dC/dn = 0.002n - 0.3

0.002n - 0.3 = 0

0.002n = 0.3

n = 150

C(150) = 0.001(150)² - 0.3(150) + 50 = 27.5

The minimum cost is achieved when 150 trees are produced giving a minimum cost of $27.5

Find out more at: brainly.com/question/25836610

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artcher [175]

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Step-by-step explanation:

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Answer:

2.79

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3 years ago
Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points. Rachel scored 78 p
Artemon [7]

Answer:

Keenan's z-score was of 0.61.

Rachel's z-score was of 0.81.

Step-by-step explanation:

Z-score:

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.

Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points.

This means that X = 80, \mu = 77, \sigma = 4.9

So

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

Z = \frac{80 - 77}{4.9}

Z = 0.61

Keenan's z-score was of 0.61.

Rachel scored 78 points on an exam that had a mean score of 75 points and a standard deviation of 3.7 points.

This means that X = 78, \mu = 75, \sigma = 3.7. So

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

Z = \frac{75 - 78}{3.7}

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