Respuesta :
Answer:
The percent change in the population of Town A is of 2.45%.
Step-by-step explanation:
The exponential model of population growth is given by:
[tex]P(t) = P_{0}e^{rt}[/tex]
In which P(t) is the population after t years, [tex]P_{0}[/tex] is the initial population and r is the annual percent change.
The population of Town A increases by 31% every 11 years. What is the annual percent change in the population of Town A?
We have that:
[tex]P(11) = 1.31P_{0}[/tex]
Applying this to the equation, we find r.
[tex]P(t) = P_{0}e^{rt}[/tex]
[tex]1.31P_{0} = P_{0}e^{11r}[/tex]
[tex]e^{11r} = 1.31[/tex]
Applying ln to both sides:
[tex]\ln{e^{11r}} = \ln{1.31}[/tex]
[tex]11r = \ln{1.31}[/tex]
[tex]r = \frac{\ln{1.31}}{11}[/tex]
[tex]r = 0.0245[/tex]
The percent change in the population of Town A is of 2.45%.
Annual percent change in the population of Town A will be 2.485%.
Exponential growth model:
- If the population of certain species grows exponentially, final population will be modeled by,
[tex]P(t)=P_0(1+r)^t[/tex]
Here, [tex]P_0=[/tex] Initial population
[tex]r=[/tex] percentage growth per year
[tex]t=[/tex] Duration of growth
Given in the question,
- Initial population = P
- Population after 11 years = (P + 0.31P)
= 1.31P
- Duration = 11 years
Substitute these values in the expression for the population,
[tex]1.31P=P(1+r)^{11}[/tex]
[tex]1.31=(1+r)^{11}[/tex]
[tex]\text{log}(1.31)=\text{log}(1+r)^{11}[/tex]
[tex]\text{log}(1+r)=\frac{0.117271}{11}[/tex]
[tex](1+r)=10^{0.010661}[/tex]
[tex]r=1.02485-1[/tex]
[tex]r=0.02485[/tex]
[tex]r\approx 2.485[/tex]%
Therefore, annual percent change in the population of Town A will be 2.485%.
Learn more about the exponential growth here,
https://brainly.com/question/2193820?referrer=searchResults
