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Natali5045456 [20]
3 years ago
14

A manufacturing company plans to progressively increase its production capacity over the next few quarters. (A quarter is a peri

od of three months.) The increase in production can be modeled by the equation y = x6 − 25x4 + 199x2, where x is the number of quarters. What is the minimum duration required for the company to reach a production capacity of 4,975 units?

Mathematics
1 answer:
mihalych1998 [28]3 years ago
7 0

Answer:

  5 months

Step-by-step explanation:

We assume that y represents production capacity, rather than <em>increase</em> in production capacity. Then we want to solve the 6th-degree equation ...

  x^6 -25x^4 +199x^2 -4975 = 0

This can be factored in groups as ...

  x^4(x^2 -25) + 199(x^2 -25) = 0

  (x^4 +199)(x^2 -25) = 0

This has 4 complex solutions and 2 real solutions.

  x^2 = 25

  x = ±5

The duration required for capacity to reach 4975 units is 5 months.

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Using the binomial distribution, the probabilities are given as follows:

  • 0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.
  • 0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.
  • Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

P(X > 4) = 1 - P(X \leq 4)

In which:

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Then:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061

P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403

P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209

P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150

P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502

Hence:

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325

P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106

P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016

P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at brainly.com/question/24863377

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