Question

The population of a pack of wolves is 88. the population is expected to grow at a rate of 2.5% each year. what function equation represents the population of the pack of wolves after t years?
a.f(t)=88(0.025)^t
b.f(t)=88(1.25^)t
c.f(t)=88(2.5)^t
d.f(t)=88(1.025)^t

Answer 1

D

First year: 88 + 88 * 2.5% = 88 + 88 * 0.025 = 88 * 1.025
Second year: 88 * 1.025 + 88 * 1.025 * 2.5% = 88 * 1.025 + 88 * 1.025 * 0.025 = 88 * 1.025 * 1.025  = 88 * 1.025^2

Answer 2

Answer:

The population model of wolves after t years is given by

$f(t)=88(1.025)^t$

D is the correct option.

Step-by-step explanation:

The exponential function can be represented as

$f(t)=a(1+r)^t$

a = initial amount

r = rate

t = time

Now, we have been given that

r = 2.5% = 0.025

a = 88

On substituting these values in the above exponential model

$f(t)=88(1+0.025)^t\\f(t)=88(1.025)^t$

The population model of wolves after t years is given by

$f(t)=88(1.025)^t$