To solve the given problems based on the population model [tex]\( f(t) = 151000 \cdot (0.962)^t \)[/tex], let's go through each part step-by-step.
### Part 1: Determining if the population is increasing or decreasing
The population function [tex]\( f(t) = 151000 \cdot (0.962)^t \)[/tex] involves an exponential term [tex]\( (0.962)^t \)[/tex]. Since the base of the exponent (0.962) is less than 1, the exponent is decreasing as [tex]\( t \)[/tex] increases.
So, the population is decreasing.
### Part 2: Calculating the annual percent decrease
The annual decrease factor in the model is 0.962, which means every year the population reaches 96.2% of its value from the previous year.
To find the annual percent decrease:
[tex]\[
\text{Annual percent decrease} = 1 - 0.962 = 0.038 \text{ or } 3.8\%
\][/tex]
By what percent? The population is decreasing by 3.8% per year.
### Part 3: Finding the population in 2032
To determine the population in 2032, we need to find [tex]\( t \)[/tex] by calculating the number of years from 2002 to 2032:
[tex]\[
t_{2032} = 2032 - 2002 = 30
\][/tex]
Now we substitute 30 into the population model:
[tex]\[
f(30) = 151000 \cdot (0.962)^{30}
\][/tex]
Calculating this, we get:
[tex]\[
f(30) \approx 47231
\][/tex]
So, the population in 2032 is approximately 47,231.
### Part 4: Finding the year when the population will be 142,453
To find when the population will be 142,453, we need to solve the equation:
[tex]\[
151000 \cdot (0.962)^t = 142453
\][/tex]
Rewriting this equation to solve for [tex]\( t \)[/tex]:
[tex]\[
(0.962)^t = \frac{142453}{151000}
\][/tex]
Taking the natural logarithm on both sides:
[tex]\[
t \cdot \ln(0.962) = \ln \left(\frac{142453}{151000}\right)
\][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[
t = \frac{\ln \left(\frac{142453}{151000}\right)}{\ln (0.962)}
\][/tex]
Using a numerical solver, we find:
[tex]\[
t \approx 2
\][/tex]
So the year when the population will be 142,453 is:
[tex]\[
2002 + 2 = 2004
\][/tex]
Therefore, the population will be 142,453 in the year 2004.
