In 2000, the population of Sheboygan, WI was about 113,000 and was growing at a rate of 0.85% per year. Find an exponential function that describes the population P in terms of t, the number of Years after 2000.
A) $P ( t ) = 113,000 + 1.85 ^ { t }$
B) $P ( t ) = 113,000 ( 1.0085 ) ^ { t }$
C) $P ( t ) = 113,000 ( 1.85 ) ^ { t }$
D) $P ( t ) = 0.85 ( 113,000 )$

Question

In 2000, the population of Sheboygan, WI was about 113,000 and was growing at a rate of 0.85% per year. Find an exponential function that describes the population P in terms of t, the number of Years after 2000. 
A) $P ( t ) = 113,000 + 1.85 ^ { t }$
B) $P ( t ) = 113,000 ( 1.0085 ) ^ { t }$
C) $P ( t ) = 113,000 ( 1.85 ) ^ { t }$
D) $P ( t ) = 0.85 ( 113,000 )$

Quizplus · Accepted Answer

The correct formula for exponential growth is $P(t) = P_0(1 + r)^t$, where $P_0$ is the initial population, $r$ is the growth rate as a decimal, and $t$ is the time in years. Therefore, the correct formula is $P(t) = 113,000(1.0085)^t$, which matches choice B.

In 2000, the Population of Sheboygan, WI Was About 113,000 P(t)=113,000+1.85tP ( t ) = 113,000 + 1.85 ^ { t }P(t)=113,000+1.85t

In 2000, the Population of Sheboygan, WI Was About 113,000 $P ( t ) = 113,000 + 1.85 ^ { t }$