Answer:
a. P = 190t -373,930 . . . . . t=actual year number
b. P = 6640
Explanation:
Given the two points (1991, 4360) and (1999, 5880) on a linear relation between years and moose population, you want a formula for the population (P) and an estimate of the population in 2003.
a. Formula
We can use the point-slope form of the equation for a line to write the formula for moose population. To do that, we need to know the slope of the line.
The slope is given by the formula ...
m = (y2 -y1)/(x2 -x1) . . . . . . . . line through points (x1, y1) and (x2, y2)
Using the given point values, we have ...
m = (5880 -4360)/(1999 -1991) = 1520/8 = 190
The point-slope equation for a line with slope m through point (h, k) is ...
y -k = m(x -h)
Using the first ordered pair, we can write the equation as ...
y -4360 = 190(x -1991)
y = 190x -373,930 . . . . . . . where x is the actual year number
Using P and t for the variables, the formula is ...
P = 190t -373,930
b. Population in 2003
Using t=2003, the above formula evaluates to ...
P = 190(2003) -373,930 = 380,570 -373,930
P = 6,640
The linear model predicts the 2003 population to be 6640 moose.
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Additional comment
The formula above uses actual year number. The year value can be translated any way you might want. For example, using t = years after 1991, the formula would be ...
P = 190(t +1991) -373,930
P = 190t +4360
Then t=12 for year 2003.