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Sagot :
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
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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