In 1992, the moose population in a park was measured to be 4260. By 1996, the population was measured again to be 3660. If the population continues to change linearly: A.) Find a formula for the moose population, P , in terms of t , the years since 1990. P ( t ) = B.) What does your model predict the moose population to be in 2008?
Answer:
Model predict the moose population to be in 2008 is 1860
Solution:
Let us consider this problem on graph, taking the x axis to be years since 1990 and y on the graph be the number of moose
We have two points on the graph: (2, 4260) and (6, 3660)
Using the slope formula we can find the slope :
[tex]\mathrm{m}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
[tex]m=\frac{(3660-4260)}{(6-2)}[/tex]
m = -150
Now that we have the slope of the line, we can use the point-slope formula to find the equation for the line as follows:-
[tex]\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right)} \\\\ {y-4260=-150(x-2)} \\\\ {y-4260=-150 x+300} \\\\ {y=-150 x+4560}\end{array}[/tex]
Since the formula that is being sought is supposed to be interms of P and t, we will replace y with P and x with t
P(t) = -150t + 4560
Number of years from 1990 to 2008 is
2008 - 1990 = 18
So, population in 2008 will be
P(18) = -150 (18) + 4560
P(18) = -2700 + 4560
P(18) = 1860
Thus the model to predict the moose population in 2008 is found
