Mark M. answered 02/17/20
Mathematics Teacher - NCLB Highly Qualified
The population declines by 4800 over 6 years. The rate of change is -80
In 1990 the population would have been 4730 + 3(80) or 4970
P(t) = -80t + 4980
Erin R.
asked 02/16/20In 1993, the moose population in a park was measured to be 4730. By 1999, the population was measured again to be 4250. If the population continues to change linearly:
Find a formula for the moose population, P, in terms of t, the years since 1990.
What does your model predict the moose population to be in 2002?
Mark M. answered 02/17/20
Mathematics Teacher - NCLB Highly Qualified
The population declines by 4800 over 6 years. The rate of change is -80
In 1990 the population would have been 4730 + 3(80) or 4970
P(t) = -80t + 4980
Difference in y over difference in x.
(4730 - 4250)/(1993 - 1999) = 480/-6 = -80
Since the time from 1990 to 1993 is half the time from '93 to '99, we can save work and assume, because the decline is linear that the decline was half of the already figured loss.
Remembering that the population would have been greater in 1990, 4730 + 240 = 4970.
P(t) = -80t + 4970
Since 2002 - 1990 = 12...
P(12) = -80(12) + 4970
P(12) = -960 + 4970
P(12) = 4010
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