High School

A town has a population of 126,000 and shrinks at a rate of [tex]4.3\%[/tex] every year. Which equation represents the town's population after 2 years?

A. [tex]P=126,000(1.043)^2[/tex]
B. [tex]P=126,000(1-0.043)(1-0.043)[/tex]
C. [tex]P=126,000(0.957)(0.957)[/tex]
D. [tex]P=126,000(0.043)^2[/tex]

Answer :

Let's start by understanding how population shrinkage works year over year. The town's population decreases by 4.3% each year.

1. Initial population is [tex]\( P_0 = 126,000 \)[/tex].
2. The shrinkage rate is 4.3%, which can be written as [tex]\(\text{shrink rate} = 0.043\)[/tex].

To find the population after one year, we calculate:
[tex]\[
P_1 = P_0 \times (1 - \text{shrink rate})
\][/tex]
[tex]\[
P_1 = 126,000 \times (1 - 0.043)
\][/tex]
[tex]\[
P_1 = 126,000 \times 0.957
\][/tex]
[tex]\[
P_1 = 120,582.0
\][/tex]

To find the population after the second year, we apply the shrinkage rate again to [tex]\( P_1 \)[/tex]:
[tex]\[
P_2 = P_1 \times (1 - \text{shrink rate})
\][/tex]
[tex]\[
P_2 = 120,582 \times 0.957
\][/tex]
[tex]\[
P_2 = 115,396.974
\][/tex]

Alternatively, we can also find the population after 2 years using the formula:
[tex]\[
P = P_0 \times (1 - \text{shrink rate})^2
\][/tex]
Plugging in the values:
[tex]\[
P = 126,000 \times (1 - 0.043)^2
\][/tex]
[tex]\[
P = 126,000 \times 0.957^2
\][/tex]

When we calculate [tex]\( 0.957^2 \)[/tex]:
[tex]\[
0.957^2 = 0.915849
\][/tex]

Thus, the population after 2 years is:
[tex]\[
P = 126,000 \times 0.915849
\][/tex]
[tex]\[
P = 115,396.97399999999
\][/tex]

Given the options, the correct equation representing the town's population after 2 years is

[tex]\[
P = 126,000 \times 0.957 \times 0.957
\][/tex]

This equation matches:
[tex]\[
P = 126,000 \times (0.957)^2
\][/tex]
The equivalent choice is:

[tex]\[
P = 126,000 (0.957)(0.957)
\][/tex]