Respuesta :
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]n(t) = 120 e^{(0.056t)}[/tex]
b
[tex]n(36) = 901 \ rats[/tex]
c
[tex]t = 50 \ months[/tex]
Step-by-step explanation:
From the question we are told that
The original number of rats that swim from the wreckage to a deserted island is
The population on the island grows exponentially according to the model
[tex]n(t) = n_o e^{rt}[/tex]
The number of rats after t = 15 months is 280
So
[tex]n(15) = 120 e^{15 * r } = 280[/tex]
=> [tex]120 e^{15 * r } = 280[/tex]
=> [tex]e^{15r} = 2.33[/tex]
=> [tex]15r = 0.8459[/tex]
=> [tex]r = 0.056[/tex]
Therefore the population t months after the arrival of the rats is mathematically represented as
[tex]n(t) = 120 e^{(0.056t)}[/tex]
Here t is in months
Considering question b
For t = 3 years = 12 * 3 = 36 months
Then the number of rats that will be present is mathematically represented as
[tex]n(36) = 120 e^{0.056 * 36 }[/tex]
[tex]n(36) = 901 \ rats[/tex]
Considering question c
Now the number of rats considered is n(t) = 2000
So
[tex]n(t) = 2000 = 120e^{0.056t}[/tex]
=> [tex]2000 = 120e^{0.056t}[/tex]
=> [tex]16.67 = e^{0.056t}[/tex]
Taking natural log of both sides
=> [tex]0.056 t = 2.81[/tex]
=> [tex]t = 50 \ months[/tex]
