Question

The fish population of Lake Collins is decreasing at a rate of 4% per year. In 2004 there were about 1,150 fish. Write an exponential decay function to model this situation. Then find the population in 2009.

A: y = 1,150(0.96)^t
2009=938
B: y = 1,150(0.96)^t
2009=942.67
C: y = 1,150(0.6)^5
2009=89
D: y = 1,150(0.04)^t

Answer 1

`\textsf{A:} \quad y = 1150(0.96)^t`

2009 = 938

Explanation:

Exponential Function

General form of an exponential function:

`y=ab^x`

where:

• a is the initial value (y-intercept)
• b is the base (growth/decay factor) in decimal form
• x is the independent variable
• y is the dependent variable

If b > 1 then it is an increasing function

If 0 < b < 1 then it is a decreasing function

If the rate of decrease is 4% per year, that means that each year the number of fish is 96% of the previous year, since 100% - 4% = 96%

Given:

• a = 1150
• b = 96% = 0.96
• x = t = time in years
• y = total of fish

Substitute the given values into the exponential equation:

`\implies y=1150(0.96)^t`

If the initial year was 2004 then:

⇒ t = 2009 - 2004 = 5

Substitute t = 5 into the equation and solve for y:

`\implies y=1150(0.96)^5`

`\implies y=937.6786022...`

Therefore, the population in 2009 was 938 (to the nearest whole number).

Answer 2

• Option A

Explanation:

Decrease rate is 4%

• 1 - 4% =
• 1 - 0.04 =
• 0.96

The equation for this situation is

• 
  `y = 1150*0.96^t`, where t is the number of years passed since 2004

The fish population in 2009

• 
  `y = 1150*0.96^(2009-2004)=1150*0.96^5 = 938 (rounded)`

Correct choice is A