Answer:
Step-by-step explanation:
Letting 2001 equate to t=0 for our starting point
We started at (0, 4840) and 8 years later we are at (8, 6200)
4840 + 8t = 6200 The population went from 4840 to 6200 over 8 years.
Subtract 4840 from both sides
8t = 6200 - 4840
8t = 1360
divide both sides by 8 to isolate t
t = 1360/8
t = 170
The population rises by 170 moose per year
P(t) = 170t + 4840
b) In 2020... t = 2020 - 2001 = 19 years
P(t) = 170t + 4840
P(t) = 170(19) + 4840
P(t) = 8070
By our linear model there should be approximately 8070 moose by 2020.
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You could also take the two initial points (0, 4840), (8, 6200)
and find the slope m to set up the equation in
slope intercept form y = mt + b
where m is the slope and the point (0,b) = (0, 4840) is y-intercept.
m = (rise/run) = (6200-4840)/(8-0) = 170
y(t) = mt + b = 170t + 4840
There is another, probably easier, way to find the "245" for this
problem. We have two points (0, 4240) and (8, 6200)
We can use that to find the slope of our equation to set up the
equation in slope intercept form of y = mx + b... or, in this case
y = mt + b since y is a function of t
m = slope of the line, and the point (0,b) is the y-intercept = (0, 4240)
m = (change in y)/(change in t) = (6200-4240)/(8-0) = 1960/8 = 245
We can then set up our equation y = mt + b
using 245 for m, and 4240 for b
y = 245t + 4240