For part a, a linear model has a constant rate of change. Linear models usually take the form

y = mx + b

with m being the rate of change of our dependent variable y, as our independent variable x changes. The value b is the initial value y has (when x = 0). For this setup, we're dealing with swan populations (our y; dependent variable) as a function of time (our x; independent variable). If we think of 2002 as our initial year (when x = 0) then we know b will be 41313 swans. We want the number of swans to be 3000 (y = 3000) by year 2013 (11 years after 2002; x = 11). Plugging everything into our linear equation, we have

y = m*x + b

3000 = (m)(11) + 14313

Simplifying and solving for m, we get

(3000 - 14313) / 11 = m

-1028.5 = m

Since m is our rate of change (how much the number of swans (y) changes per year (x)), we have our answer.

For part b, we have an exponential equation of the form

y = A*e^{k*x}

with A being the initial value (when x = 0) of our dependent variable y (swans), k being the decay/growth factor, and x being our independent variable (time). Again, taking 2002 to be the initial year, (year zero; x = 0), we know A = 14313. In 2013, 11 years after 2002 (x = 11), we want our y value to be 3000. So let's plug and chug, and solve for our decay factor k.

y = A*e^{k*x}

3000 = 14313*e^{k*11}

3000 / 14313 = e^{k*11}

Now we take the natural log of both sides to get

ln(3000/14313) = k*11

ln(3000/14313) / 11 = k

-.1 = k

And there's your decay factor! Hope this was a help. :)

- Jon