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The town of Sickville, with a population of 2603 is exposed to the Blue Moon Virus, against which there is no immunity. The number of people infected when the virus is detected is 95. Suppose the number of infections grows logistically, with k=0.59

Respuesta :

Using the logistic equation, it is found that the number of people infected after t days is given by:

[tex]P(t) = \frac{2603}{1 + 26.4e^{-0.59t}}[/tex]

What is the logistic equation?

It is given by:

[tex]P(t) = \frac{K}{1 + Ae^{-kt}}[/tex]

[tex]A = \frac{K - P(0)}{P(0)}[/tex]

In which:

  • K is the carrying capacity.
  • P(0) is the initial amount.
  • k is the rate.

In this problem, the parameters are as follows: K = 2603, P(0) = 95, k = 0.59.

Hence:

[tex]A = \frac{K - P(0)}{P(0)} = \frac{2603 - 95}{95} = 26.4[/tex]

Then:

[tex]P(t) = \frac{K}{1 + Ae^{-kt}}[/tex]

[tex]P(t) = \frac{2603}{1 + 26.4e^{-0.59t}}[/tex]

More can be learned about the logistic equation at https://brainly.com/question/25697660

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