Question

This exercise uses the population growth modet.

Answer 1

The following represents the value :

a) n(t) = $12e^{0.011t}$<u />

b) The estimate for the population in 2015 is of 12.7 million.

c) The fish population will reach 14 million after 14 years.

d) The sketch is given at the end of this answer.

How to calculate how many years will the fish population reach 14 million?

Item a:

The exponential model is:

n(t) = n(0)$e^{rt}$

In which:

n(0) is the population is 2010.

r is the growth rate, as a decimal.

Population of 12 million, thus n(0)  =12

Growth rate of 1.1%, thus r = 0.011

Thus, the model is:

n(t) = $12e^{0.011t}$

Item b:

2015 is 2015 - 2010 = 5 years after 2010, thus this is n(5).

n(5) = $12e^{0.0115}$ = 12.7

The estimate for the population in 2015 is of 12.7 million.

Item c:

This is t for which n(t) = 14, thus:

n(t) = $12e^{0.011t}$<u />

14 = $12e^{0.011t}$

$e^{0.011t}$ = 14/12

In $e^{0.011t}$ = In 14/12

t = In 14/12/0.011

t = 14

The fish population will reach 14 million after 14 years.

Item d:

At the end of this answer, the sketch is given.

The complete question is : The population of a certain species of fish has a relative growth rate of 1.1% per year. It is estimated that the population in 2010 was 12 million. a. Find an exponential model n(t)= no e^rt for the population t years after 2010. b. Estimate the fish population in the year 2015. c. After how many years will the fish population reach 14 million? d. Sketch a graph of the fish population.

To learn more about fish population refer to :

brainly.com/question/29490771

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