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# In 1992, the moose population in a park was measured to be 4280

In 1992, the moose population in a park was measured to be 4280. By 1999, the population was measured again to be 6170. 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)= syntax error

What does your model predict the moose population to be in 2007?

So this is an exponential growth problem of the form: P(t) = P*ekt     where P = original population

k = constant we need to find  ;  t = the time in years[Remember the first year, 1992, is the beginning.

1992 = first started( t = 0). So we just plug in everything and solve.

The Start:  In 1992  t = 0(We are starting here), the popultion is 4280(P = 4280)

P(t) = 4280*ekt , when t = 0 the population is 4280

Next Interval: 1999(now 7 years passed so now t = 7) and the population is 6170

P(t)  =  4280*ekt   <==>  when t = 7 we know the population is 6170

P(7) =  4280*e7 k  = 6170   ;  Now the only variable is k , to solve for k

e7k  =  6170/4280

ln (e7k) =  ln (6170/4280)    <===>    7k =  ln (6170/4280)

k  =  [ln (6170/4280)]/7  ≈  .05224

Equation of Populace==> P(t) = 4280*e.05224t ; and in 2007 it will have been 15 yrs( so t = 15)

P(t) =  4280*e.05224*15   =  10679.80  =  approx. 10,680

This is not an exponential growth problem. It is clearly stated 'the population continues to change linearly'. So the answer is 8330 as solved using linear equations. 2/13/2013

Find the equation of the line:

Since we have 2 points, we will use the point slope form.

Points (1999, 6170) and (1992, 4280)

slope m = change in y / change in x = (6170-4280) / (1999-1992) = 1890/7 = 270

Therefore the equation of the line is

(Y - Y1) = m (X - X1)

(Y - 6170) = 270 (X - 1999)

To determine the population in 2007, use 2007 for X and slove for Y

(Y - 6170) = 270 (2007 - 1999)

Y - 6170 = 270 (8)

Y -6170 = 2160

Y = 8330

yep missed the "changes linearly"

2/23/2013

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