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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

Sagot :

mhanifa

Answer:

  • Option A

Step-by-step explanation:

Decrease rate is 4%

  • 1 - 4% =
  • 1 - 0.04 =
  • 0.96

The equation for this situation is

  • [tex]y = 1150*0.96^t[/tex], where t is the number of years passed since 2004

The fish population in 2009

  • [tex]y = 1150*0.96^{2009-2004}=1150*0.96^5 = 938 (rounded)[/tex]

Correct choice is A

semsee45

Answer:

[tex]\textsf{A:} \quad y = 1150(0.96)^t[/tex]

      2009 = 938

Step-by-step explanation:

Exponential Function

General form of an exponential function:

  [tex]y=ab^x[/tex]

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:

[tex]\implies y=1150(0.96)^t[/tex]

If the initial year was 2004 then:

⇒ t = 2009 - 2004 = 5

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

[tex]\implies y=1150(0.96)^5[/tex]

[tex]\implies y=937.6786022...[/tex]

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

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