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.

Answer 1

A = 750, the number of fish in the year 2005.
8.3% = 0.083, the growth rate
Let n =  the number of years, counted from 2005.
Let a(n) = the number of fish after n years.

When n=1,   a(1) = 750*1.083
When n=2,  a(2) = 750*1.083²
When n=3,  a(3) = 750*1.083³

Therefore, after n years, a(n) = 750*1.083ⁿ

The number of years between 2005 and 2050 is 2050-2005 = 45
Therefore in the year 2050, the number of fish will be
a(45) = 750*1.083⁴⁵ = 27123.26
We cannot have fractional fish, so a(45) =27123.

Answer: 27123

Answer 2

Using the given information, we can create the following exponential function.
Let t be the number of years after 2005.
Let F be the fish population.
F = 750(1.083)^t

In this case, t = 45 (2050 - 2005 = 45)

F = 750(1.083)^(45)
= 27123.255...
Round to the nearest whole number.

F = 27123
That's the expected population in 2050.

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