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## In 1994, the moose population in a park was measured to be 4290

In 1994, the moose population in a park was measured to be 4290. By 1998, the population was measured again to be 3850. If the population continues to change linearly:

Find a formula for the moose population, P, in terms of t, the years since 1990.

P(t)=

What does your model predict the moose population to be in 2006?

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# 1 Answer

We will consider x on the graph to be years since 1990 and y on the graph to be the number of Moose.

We have two points on the graph:  (4, 4290) and (8, 3850)

Using the slope formula (m = [y2 - y1]/[x2 - x1]), we can find the slope of this line:

m = (y2 - y1)/(x2 - x1)                 Slope formula

m = (3850 - 4290)/(8 - 4)          Substitution

m = -440/4                               Solve the subtraction problems in the parenthesis

m = -110                                  Reduce the fraction

Now that we have the slope of the line, we can use the point-slope formula (y - y1 = m[x - x1]) to find the equation for the line.  I will use the first point, but either point will produce the same answer.

y - y1 = m(x - x1)                                     Point-slop e formula

y - 4290 = -110(x - 4)                              Substitution

y - 4290 = -110x + 440                            Distribute the -110

y - 4290 + 4290 = -110x + 440 + 4290      Add 4290 to each side

y = -110x + 4730                                     Simplify

Since the formula that is being sought is supposed to be interms of P and t, I will replace y with P and x with t.

P(t) = -110t + 4730

To predict the population in 2006, we need to determine the years from 1990 to 2006.

2006 - 1990 = 16

P(16) = -110 (16) + 4730

P(16) = -1760 + 4730

P(16) = 2970