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3 years ago

Suppose that the dollar cost of producing x appliances is c(x)=1000+ 90x- 0.2x 2

Mathematics

1 answer:

Aleksandr [31]3 years ago

6 0

Answer:

Step-by-step explanation:

Given that the dollar cost of producing x appliances is

$c(x)=1000+ 90x- 0.2x^2$

A) Average cost can be obtained by dividing c(x) by units = x

i.e. average cost = $\frac{1000}{x} +90-0.2x$

B) Marginal cost = dC/dx = $90-0.4x$

When x=140 marginal cost $= 90-0.4(140)\\=34$

C) We calculate

c(140) = $=1000+90(140)-0.2(140^2)\\\\=9680$

c(141) = $=1000+90(141)-0.2(141^2)\\\\\\=9713.80$

Marginal cost = difference = 33.80

Same as 34 shown in marginal cost.

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sergeinik [125]

Answer:

C. 1:27

Step-by-step explanation:

a²/b²=1/9 => a=1 and b=3

a³/b³=1³/3³=1/27

C. 1:27

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3 years ago

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vladimir1956 [14]

Answer:

neither

Step-by-step explanation:

Take out 5 as a common factor. It will be easier to look at.

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3 years ago

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ludmilkaskok [199]

Answer:

Step-by-step explanation:

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2 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$