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.

a)

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.

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Item a:

The exponential model is:

In which:

Thus, the model is:

Item b:

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

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

Item c:

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

The fish population will reach 14 million after 14 years.

Item d:

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

A similar problem is given at link

(a) n(t) = P(0)*e^(0.010939940t)

(b) 12,674,681 (nearest unit)

(c) 14 years (nearest year)

Step-by-step explanation:

rate = 1.1% / year = 1.011

(a)

P(0) = 12,000,000  = population in 2010

In compound interest format, after t years

P(t) = P(0)* (1.011)^t

Given format = P(0)* e^(rt)

therefore

e^(rt) = 1.011^t       use law of exponents

(e^r)^t = 1.011^t

e^r = 1.011

r = log_e(1.011) = 0.010939940   (to 9 decimal places)

required formula is

n(t) = P(0)*e^(0.010939940t)

(b)

in 2015,

P(0)=12000000, n = 5 (years after 2010)

n(5) = 12000000*e^( 0.010939940 * 5 ) = 12,674,680.6 = 12,674,681 (nearest unit)

(c)

to reach 14 million, we equate

n(t) = 14,000,000

12,000,000 *e^(0.010939940*t) = 14,000,000

e^(0.010939940*t) = 14000000/12000000 = 7/6

take log on both sides

0.010939940*t = log(7/6)

t = log(7/6) / 0.010939940 = 14.091 years = 14 years to the nearest year.

See graph attached.  Y-axis is in millions, x-axis is in years.

answer: -1

step-by-step explanation: