Question

19. In a small town named Danville, the population today is 6999. For the last sixyears, the population has grown by 20% every year. If the growth stay thesame, what is the expected population in Danville 4 years from today?O 14487O14496O14560OAll changes saveO 14513

Answer 1

Explanation

From the statement, we know that the population in Danville:

• today is P₀ = 6999,

,

• for the last six years, the population has grown by r = 20% = 0.2 every year.

Supposing the same growth rate for the following years, we must compute the expected population in Danville n = 4 years from today.

(1) The general formula for the population with constant growth is given:

$P_n=P_0\cdot(1+r)^n.$

Where:

• P₀ = 6999 is the initial population,

,

• r = 20% = 0.2 is the growth rate in decimals,

,

• n = the number of years from today.

(2) Replacing the data from above, we have the following formula:

$P_n=6999\cdot(1+0.2)^n=6999\cdot1.2^n.$

(3) Evaluating this formula for n = 4, we get:

$P_4=6999\cdot1.2^4\cong14513.$

So the expected population in Danville 4 years from today is about 14513 people.

Answer

14513