M. Guthis Classroom
Pretest: Functions

Select the correct answer.

A population of frogs in a pond increases at an annual rate of [tex]22\%[/tex]. If there are 100 frogs in the pond, which equation gives you a model for how many frogs there will be in time [tex]t[/tex] years?

A. [tex]P_t = 100 e^{0.22 t}[/tex]
B. [tex]P_t = 100 e^{\frac{9 n}{1}}[/tex]
C. [tex]P_t = 100 e^{\frac{-2 x}{t}}[/tex]
D. [tex]P_t = 100 e^{-0.22 t}[/tex]



Answer :

To determine the model for the population of frogs in the pond over time, we need to use the exponential growth formula since the population increases at a constant annual rate.

The general formula for exponential growth is:
[tex]\[ P_t = P_0 \cdot e^{rt} \][/tex]
where:
- [tex]\( P_t \)[/tex] is the population at time [tex]\( t \)[/tex],
- [tex]\( P_0 \)[/tex] is the initial population,
- [tex]\( r \)[/tex] is the growth rate (as a decimal), and
- [tex]\( t \)[/tex] is the time in years.

Given:
- An initial population [tex]\( P_0 \)[/tex] of 100 frogs,
- An annual growth rate [tex]\( r \)[/tex] of 22%, which needs to be converted to a decimal, so [tex]\( r = 0.22 \)[/tex].

Substituting these values into the exponential growth formula provides:
[tex]\[ P_t = 100 \cdot e^{0.22t} \][/tex]

Now, we compare this to the options provided:
A. [tex]\( P_t = 100 e^{0.22 t} \)[/tex]
B. [tex]\( P_t = 100 e^{\frac{9 n}{1}} \)[/tex]
C. [tex]\( P_t = 100 e^{\frac{-2 x}{t}} \)[/tex]
D. [tex]\( P_t = 100 e^{-0.22 t} \)[/tex]

The correct equation that matches our derived model is option:
A. [tex]\( P_t = 100 e^{0.22 t} \)[/tex]

So, the answer is:
1. A