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Suppose a rumor is going around a group of 210 people. Initially, only 34 members of the group have heard the rumor, but 3 days later 69 people have heard it. Using a logistic growth model, how many people are expected to have heard the rumor after 6 days total have passed since it was initially spread? (Round your answer to the nearest whole person.)

Answer :

Answer:

116 people are expected to have heard the rumor after 6 days total have passed since it was initially spread.

Step-by-step explanation:

Logistic function:

The logistic function is given by:

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

In which:

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

Considering that K is the carrying capacity, k is the growth/decay rate and P(0) is the initial population.

Suppose a rumor is going around a group of 210 people.

This means that [tex]K = 210[/tex]

Initially, only 34 members of the group have heard the rumor:

This means that [tex]P(0) = 34[/tex] and:

[tex]A = \frac{210 - 34}{34} = 5.1765[/tex]

So

[tex]P(t) = \frac{210}{1 + 5.1765e^{-kt}}[/tex]

3 days later 69 people have heard it.

This means that [tex]P(3) = 69[/tex], and we use this to find k.

[tex]69 = \frac{210}{1 + 5.1765e^{-3k}}[/tex]

[tex]69 + 357.1785 e^{-3k} = 210[/tex]

[tex]357.1785 e^{-3k} = 141[/tex]

[tex]e^{-3k} = \frac{141}{357.1785}[/tex]

[tex]\ln{e^{-3k}} = \ln{\frac{141}{357.1785}}[/tex]

[tex]-3k = \ln{\frac{141}{357.1785}}[/tex]

[tex]k = -\frac{\ln{\frac{141}{357.1785}}}{3}[/tex]

[tex]k = 0.3098[/tex]

So

[tex]P(t) = \frac{210}{1 + 5.1765e^{-0.3098t}}[/tex]

How many people are expected to have heard the rumor after 6 days total have passed since it was initially spread?

This is P(6), so:

[tex]P(6) = \frac{210}{1 + 5.1765e^{-0.3098*6}} = 116[/tex]

116 people are expected to have heard the rumor after 6 days total have passed since it was initially spread.

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