Suppose a rumor is going around a group of 191 people. Initially, only 38 members of the group have heard the rumor, but 3 days later 68 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.)

Respuesta :

Answer:

106 people.

Step-by-step explanation:

Logistic equation:

The logistic equation is given by:

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

In which

[tex]A = \frac{K - P_0}{P_0}[/tex]

K is the carrying capacity, k is the growth/decay rate, t is the time and P_0 is the initial value.

Suppose a rumor is going around a group of 191 people. Initially, only 38 members of the group have heard the rumor.

This means that [tex]K = 191, P_0 = 38[/tex], so:

[tex]A = \frac{191 - 38}{38} = 4.03[/tex]

Then

[tex]P(t) = \frac{191}{1+4.03e^{-kt}}[/tex]

3 days later 68 people have heard it.

This means that [tex]P(3) = 68[/tex]. We use this to find k.

[tex]P(t) = \frac{191}{1+4.03e^{-kt}}[/tex]

[tex]68 = \frac{191}{1+4.03e^{-3k}}[/tex]

[tex]68 + 274.04e^{-3k} = 191[/tex]

[tex]e^{-3k} = \frac{191-68}{274.04}[/tex]

[tex]e^{-3k} = 0.4484[/tex]

[tex]\ln{e^{-3k}} = \ln{0.4484}[/tex]

[tex]-3k = \ln{0.4484}[/tex]

[tex]k = -\frac{\ln{0.4484}}{3}[/tex]

[tex]k = 0.2674[/tex]

Then

[tex]P(t) = \frac{191}{1+4.03e^{-0.2674t}}[/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{191}{1+4.03e^{-0.2674*6}} = 105.52[/tex]

Rounding to the nearest whole number, 106 people.