Respuesta :
Answer:
(a)[tex]y=80e^{\dfrac{ln(2.7)t}{12}[/tex]
(b)301
Step-by-step explanation:
By the exponential population model,the population P(t) at any time t is given as: [tex]P(t)=P_{0}e^{rt}[/tex] where [tex]P_{0}[/tex] is the initial population.
P(12)=y=216 at t=12 years
[tex]P_{0}=80[/tex]
[tex]P(t)=P_{0}e^{rt}[/tex]
[tex]216=80e^{12r}\\\frac{216}{80} =e^{12r}[/tex]
Taking the natural logarithm of both sides
12r=ln(2.7)
[tex]r=\frac{ln (2.7)}{12}[/tex]
[tex]y=80e^{\dfrac{ln(2.7)t}{12}[/tex]
(b) After 16 years
At t=16
[tex]y=80e^{\dfrac{ln(2.7)t}{12}}\\y=80e^{\dfrac{ln(2.7)X16}{12}}\\y=300.77[/tex]
After 16 years, there are approximately 301 fishes in the lake.
Given question is incomplete; find the complete question,
An initial population of 80 fish is introduced into a lake. This fish population grows according to a continuous exponential growth model. There are 216 the lake after 12 years.
(a) Let t be the time (in years) since the initial population is introduced and let y be the number of fish at time t. Write a formula relating y to t. Use exact expressions to fill in the missing parts of the formula. Do not use approximations.
(b) How many fish are there 3 years after the initial population is......? Do not round any intermediate computations, and round your answer to the nearest whole number.
Expression for the population growth will be,[tex]y=80e^{0.08277t}[/tex] and there will be 103 fishes in the pond after 3 years.
(a). Given in the question,
- Initial population of fish = 80
- Final population of fish = 216
- Time taken to reach the final population = 12 years
Expression for the population after time 't' is given by,
[tex]y=P_0e^{rt}[/tex]
Here, [tex]P_0=[/tex] Initial population
[tex]t=[/tex] Duration
Expression for the population,
[tex]y=P_0e^{rt}[/tex]
[tex]216=80.e^{12r}[/tex]
[tex]e^{12r}=\frac{216}{80}[/tex]
[tex]12r[\text{ln(e)}]=\text{ln}(2.7)[/tex]
[tex]r=\frac{0.99325}{12}[/tex]
[tex]r=0.08277[/tex]
Expression for the population growth will be,
[tex]y=80e^{0.08277t}[/tex]
(b). For t = 3 years,
[tex]y=80.e^{0.08277\times 3}[/tex]
[tex]=80\times 1.28186[/tex]
[tex]=102.55[/tex]
[tex]\approx103[/tex]
Therefore, expression for the population growth will be [tex]y=80.e^{0.08277t}[/tex] and population of fish after 3 years will be nearly 103.
Learn more,
https://brainly.com/question/13674608
