Question

Population in a Small Town, USA was 15,587 in 2010. The populations increase by 2% annually. Show and explain all work. a. Write the exponential growth function to represent this function. b. What will the population be in 2028? Round your answer to the nearest person

Answer 1

the other answer is wrong, you have to use the function a=Pe^(rt). from then you plug in 0.02 for r and 18 for t and intial value of 15587 for p to become 15587e^(0.02(18)) = ^0.36 and then solve from there to get 22341. also e not just a random variable but a set value of 2.718282.

Answer 2

a) Y = 15,587(1.02)^t

where Y is the population at a certain year t after 2010

b) 22,262 persons

Explanation:

a) we want to write an exponential equation;

Generally, we have this as;

Y = P(1 + r)^t

where Y is the value at a time t

P is the initial value

r is the rate of change

t is the time

From the question;

P is 15,587

r is 2% = 2/100 = 0.02

So, we have the equation as;

Y = 15,587(1 + 0.02)^t

Y = 15,587(1.02)^t

b) The population in 2028

To get this, we need the value of t

What we have to do here is to subtract 2010 from 2028

We have this as; 2028 - 2010 = 18

So substituting this value into the exponential equation above, we have

Y = 15,587(1 + 0.02)^18

Y = 15,587(1.02)^18

= 22,262