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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?

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2 Answers

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

Comments

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.

Comment

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

Comments

yep missed the "changes linearly"

 

Comment