Respuesta :
Answer:
The correct answer is E. none of the above. The population will drops below 100 when t ≥ 38.
Explanation:
Given A= A0 e^kt. The population 10 years ago is A0, the population today is A(10), and we have to find the value of "k" and then the time when population drops below 100.
So, A(t) = 1700 e^kt ⇒ A(10) = 1700 e^k(10) ⇒ 800 = 1700 e^k(10) ⇒
800/1700 = e^k(10) ⇒ln (800/1700) = k(10) ln e ⇒ -0.754/10 = k ⇒
k = -0.0754.
Now you have all the parameters, so you can find the time at which the population drops below 100.
A(t) = 1700 e^kt ⇒ 100 = 1700 e^(-0.0754)t ⇒100/1700 = e^(-0.0754)t ⇒
ln(100/1700) = (-0.0754)t ln e ⇒ [ln(100/1700)]/(-0.0754) = t ⇒
t = 38.
So, the population will drops below 100 when t ≥ 38.
The population of the given species can be calculated by the given exponential growth formula:
A[tex]_{\text t}[/tex] = A₀ x [tex]\text e^{\text {{ -kt}}[/tex].
Such that,
- A₀ = population 10 years ago = 1700.
- k = ?
- A[tex]_{\text t}[/tex] = 800
Substituting the values in the fomula:
A[tex]_{\text t}[/tex] = A₀ x [tex]\text e^{\text {{ -kt}}[/tex].
800 = 1700 x [tex]\text e^{\text {{ -kt}}[/tex].
[tex]\dfrac {800}{1700} = \text e^{\text{-kt}}[/tex]
Taking log:
ln ([tex]\dfrac {800}{1700}[/tex]) = k(10) ln e
k = [tex]\dfrac{0.754}{10}[/tex]
k = -0.0754
Now,
The time when population drops below 100:
A[tex]_{\text t}[/tex] = A₀ x [tex]\text e^{\text {{ -kt}}[/tex].
100 = 1700 x [tex]\text e^{\text {{ -kt}}[/tex].
[tex]\dfrac {100}{1700} = \text e^{\text{-kt}}[/tex]
Taking log:
ln ([tex]\dfrac {100}{1700}[/tex]) = [tex]\text e^{\text{-0.0754 x t}}[/tex]
t = [tex]\text{ln}\dfrac{\dfrac{100}{1700}}{(-0.0754)}[/tex]
t = 38
Therefore, the population will drop below a hundred when t> 38.
To know more about the exponential growth formula, refer to the following link:
https://brainly.com/question/3480875
