Question

In the year 2005, a total of 750 fish were introduced into a manmade lake. The fish population was expected to grow at a rate of 8.3% each year. What will the fish population be in 2050? Round to the nearest whole number. Enter your answer in the box.please help!!!

Answer 1

(100+8.3%=108.3\100=1.083)
difference in years=45
=750 x 1.083 raised exponent45
=27123.25528

Answer 2

Exponential change can be modeled by:
N = N(0) x λⁿ; where λ is the rate of change. It is greater than 1 when there is growth and less than 1 when there is decay. n is the number of time periods passed.
λ is 1 + 0.083 = 1.083
N(0) is the initial value of 750
n is 2050 - 2005 = 45 years
N = 750 x (1.083)⁴⁵
= 27,000 fish