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 :


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:-

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