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1 year ago

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Mathematics

1 answer:

igor_vitrenko [27]1 year ago

5 0

The marginal cost function is C'(x) = 4 + 0.06x + 0.0009x². The cost of 150th pair of jeans is $13.135. The actual cost of 151st pairs jeans is $13.1359.

The cost of the pair of jeans is defined by the function,

C(x) = 2000 + 4x + 0.03x² + 0.0003x³

a. Marginal cost function is,

Differentiating with respect to x,

C'(x) = 4 + 0.06x + 0.0009x²

b. The value of C'(150),

Putting x = 150,

C'(150) = 4 + 0.06(150) + 0.0009(150)

C'(150) = 4 +9 + 0.135

C'(150) = $13.135.

It predicts the cost of 150th pair of jeans.

c. The cost of 151st pair is,

C(151) = 4 + 0.06(151) + 0.0009(151)

C(151) = 4 + 9.06 + 0.1359

C(151) = $13.1359.

To know more about differentiation, visit,

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3 years ago

Five individuals from an animal population thought to be near extinction in a certain region have been caught, tagged, and relea

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Answer:

a) For this case the random variable X follows a hypergometric distribution.

b) $E(X)= n\frac{M}{N}=10 \frac{5}{25}=2$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1$

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Step-by-step explanation:

The hypergometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:

$P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}$

Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes

The expected value and variance for this distribution are given by:

$E(X)= n\frac{M}{N}$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}$

a. What is the distribution of X?

For this case the random variable X follows a hypergometric distribution.

b. Compute the values for E(X) and Var(X)

For this case n=10, M=5, N=25, so then we can replace into the formulas like this:

$E(X)= n\frac{M}{N}=10 \frac{5}{25}=2$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1$

c. What is the probability that none of the animals in the second sample are tagged?

So for this case we want this probability:

$P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565$

d. What is the probability that all of the animals in the second sample are tagged?

So for this case we want this probability:

$P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474$

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Step-by-step explanation:

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