In 1991, the moose population in a park was measured to be 3710. By 1998, the population was measured again to be 4130. 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)=
Since this is a linear (non-exponential) population problem you can just use the standard y=mx+b form of an equation. Where m = (change in population/change in years)
The numbers you were provided state that over the course of 7 years (1998-1991) the population increased by 420 people (4130-3710). So, (420/7) = 60 = m. Assuming that the growth rate for 1990 is the same as 1991. then you would have a starting population of (3710-60) or 3650, that would be your "b" value since at t=0 P(t) = 3650. This yields a final equation of P(t) = 60t +3650. Check the answer at t=1 and you get the population during 1991: 3710.
WyzAnt.com
Comments
1990=x=0
1991: (1, 3710) 1998: (8, 4130)
find slope: (4130-3710)/(8-1)=420/7=60
m=60
find y-intercept (b) by plugging in what you have for slope-intercept equation:
y=mx+b
choose a point: (1,3710)
3710=60(1)+ b
subtract 60 on both sides to isolate "b"
3650=b
final linear equation is: y=60x+3650
- Vilma M. 1/30/2013