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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

Verified Answer :

Budek

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

SociometricStar

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

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

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