In Texas, the population of a species of bats is decreasing at a rate of -0.021 per year. The population was 22,000 in 2010. According to the function [tex]A(t)=A_0 e^{(n t)}[/tex], what is the predicted population in 2015?

A. 19,807
B. 21,543
C. 4,130
D. 14,130



Answer :

Certainly! To determine the predicted population of a species of bats in Texas in 2015, we use the exponential decay function given by [tex]\( A(t) = A_0 e^{(nt)} \)[/tex]. Here's the step-by-step solution:

1. Identify the given values:
- The initial population in 2010, [tex]\( A_0 \)[/tex], is 22,000.
- The rate of decrease per year, [tex]\( n \)[/tex], is -0.021.
- The time period, [tex]\( t \)[/tex], is from 2010 to 2015. Therefore, [tex]\( t = 2015 - 2010 = 5 \)[/tex] years.

2. Substitute the given values into the exponential decay formula:
[tex]\[ A(t) = 22000 \cdot e^{(-0.021 \cdot 5)} \][/tex]

3. Calculate the exponent:
[tex]\[ -0.021 \cdot 5 = -0.105 \][/tex]

4. Compute the exponential term:
[tex]\[ e^{-0.105} \][/tex]

5. Multiply the initial population by the exponential term to get the predicted population:
[tex]\[ A(t) = 22000 \cdot e^{-0.105} \approx 22000 \cdot 0.9005 \approx 19807.139496897842 \][/tex]

6. Round the result to a whole number (since population counts are typically whole numbers):
[tex]\[ A(5) \approx 19807 \][/tex]

Therefore, the predicted population of the species of bats in Texas in the year 2015 is approximately 19,807.

Among the given options, the correct answer is:
- 19,807