Answer :
To solve this problem, we first need to determine the linear relationship between the moose population and time. We're told that the moose population changes linearly over time, and we have two data points:
- In 1992, the moose population was 3700.
- In 1997, the population was 1300.
Let's express the population as a linear function of time, [tex]P(t) = mt + b[/tex], where:
- [tex]P(t)[/tex] is the population at time [tex]t[/tex].
- [tex]m[/tex] is the rate of change of the population per year.
- [tex]b[/tex] is the population at [tex]t = 0[/tex], which corresponds to the year 1990 since [tex]t[/tex] represents the years since 1990.
We need to find [tex]m[/tex] and [tex]b[/tex]. First, calculate [tex]m[/tex], the rate of change:
The change in population from 1992 to 1997 is:
[tex]\Delta P = 1300 - 3700 = -2400[/tex]
The change in time from 1992 to 1997 is:
[tex]\Delta t = 1997 - 1992 = 5 \, \text{years}[/tex]
Thus, the slope [tex]m[/tex] is:
[tex]m = \frac{\Delta P}{\Delta t} = \frac{-2400}{5} = -480[/tex]
This means the moose population decreases by 480 moose per year.
Next, find [tex]b[/tex], the population at [tex]t = 0[/tex]:
Since [tex]t = 2[/tex] in 1992, you can write:
[tex]P(2) = -480 \times 2 + b = 3700[/tex]
[tex]-960 + b = 3700[/tex]
[tex]b = 3700 + 960 = 4660[/tex]
Thus, the equation for the population is:
[tex]P(t) = -480t + 4660[/tex]
Prediction for 2008:
Find the population for [tex]t = 18[/tex] (since 2008 is 18 years after 1990):
[tex]P(18) = -480 \times 18 + 4660[/tex]
[tex]P(18) = -8640 + 4660 = -3980[/tex]
This calculation suggests a prediction of the population to be negative, which mathematically indicates that the moose population would have reached zero and means the population model is no longer valid beyond this point (indicating extinction). In a real-world context, this implies the population would become extinct before 2008 under this linear model.
