Question

The fish population of Lake Collins is decreasing at a rate of 3% per year. In 2004 there were about 1,300 fish. Write an exponential decay function to model this situation. Then find the population in 2010.

Answer 1

Answer:

1083 fish

Step-by-step explanation:

The fish population of Lake Collins is decreasing at a rate of 3% per year.

If starting population were about 1,300 in 2004 so the sequence will be

1,300, (1300 - 3% of 1300)..............

or 1300, 1261,.........

We know exponential function is modeled by

                   $A_{n} =A_{0} (r)^{n}$

Where        $A_{n}$ = population at time t.

                  $A_{0}$ = initial population

                   r = common ratio

                   n = time in years.

Common ratio (r) = $\frac{\text{second term}}{\text{first term}}$

                            = $\frac{1261}{1300}=(0.97)$

Then the function will be $A_{n}=1300(0.97)^n$

Then population in year 2010 (after 6 years) will be

$A_{6}=1300(0.97)^6$

   = 1300(0.8329)

  = 1082.86 ≈ 1083 fishes

Answer 2

For this case we have a function of the form:
 $y = A * (b) ^ x$ Where,
 A: initial amount
 b: decrease rate
 x: time in years
 Substituting values we have:
 $y = 1300 * (0.97) ^ x$
 For 2010 we have:
 $y = 1300 * (0.97) ^ 6 y = 1083$
 Answer:
 an exponential decay function to model this situation is:
 y = 1300 * (0.97) ^ x
 The population in 2010 is:
 y = 1083