Answer :
To determine which function best models the population of butterflies [tex]\( x \)[/tex] years from the first measurement, we start by considering the given conditions:
- Initial population ([tex]\( P_{0} \)[/tex]): 2000 butterflies
- Final population ([tex]\( P_{f} \)[/tex]): 2800 butterflies
- Time span ([tex]\( t \)[/tex]): 7 years
We need to evaluate each function and see which one produces values that reasonably match the population growth from 2000 to 2800 over the 7-year period.
### Option A: [tex]\( f(x) = 2000 \left( 1 + 0.4 \right)^{\frac{x}{7}} \)[/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2000 \left( 1 + 0.4 \right)^{\frac{1}{7}} \][/tex]
[tex]\[ f(1) \approx 2000 \times 1.049 \][/tex]
[tex]\[ f(1) \approx 2098.483 \][/tex]
### Option B: [tex]\( f(x) = 2000 \left( 1 + 0.4^7 \right)^x \)[/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2000 \left( 1 + 0.4^7 \right)^1 \][/tex]
[tex]\[ f(1) \approx 2000 \times 1.00164 \][/tex]
[tex]\[ f(1) \approx 2003.2768 \][/tex]
### Option C: [tex]\( f(x) = 2000 \left( 1 + \frac{0.4}{7} \right)^x \)[/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2000 \left( 1 + \frac{0.4}{7} \right)^1 \][/tex]
[tex]\[ f(1) \approx 2000 \times 1.05714 \][/tex]
[tex]\[ f(1) \approx 2114.286 \][/tex]
### Option D: [tex]\( f(x) = 2000 \left( 1 + \frac{0.4}{7} \right)^{7x} \)[/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2000 \left( 1 + \frac{0.4}{7} \right)^7 \][/tex]
[tex]\[ f(1) \approx 2000 \times 1.475 \][/tex]
[tex]\[ f(1) \approx 2950.977 \][/tex]
### Conclusion:
Given that the butterfly population grows from 2000 to 2800 over 7 years, we are looking for a model that captures a significant increase over time. From our evaluations:
- Option A: [tex]\( \approx 2098.483 \)[/tex]
- Option B: [tex]\( \approx 2003.2768 \)[/tex]
- Option C: [tex]\( \approx 2114.286 \)[/tex]
- Option D: [tex]\( \approx 2950.977 \)[/tex]
Model [tex]\( D \)[/tex] estimates a final population of approximately 2950.977, which is the closest to the actual final population of 2800. Therefore, the best function that models the population is:
[tex]\[ f(x) = 2000 \left( 1 + \frac{0.4}{7} \right)^{7x} \][/tex]
- Initial population ([tex]\( P_{0} \)[/tex]): 2000 butterflies
- Final population ([tex]\( P_{f} \)[/tex]): 2800 butterflies
- Time span ([tex]\( t \)[/tex]): 7 years
We need to evaluate each function and see which one produces values that reasonably match the population growth from 2000 to 2800 over the 7-year period.
### Option A: [tex]\( f(x) = 2000 \left( 1 + 0.4 \right)^{\frac{x}{7}} \)[/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2000 \left( 1 + 0.4 \right)^{\frac{1}{7}} \][/tex]
[tex]\[ f(1) \approx 2000 \times 1.049 \][/tex]
[tex]\[ f(1) \approx 2098.483 \][/tex]
### Option B: [tex]\( f(x) = 2000 \left( 1 + 0.4^7 \right)^x \)[/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2000 \left( 1 + 0.4^7 \right)^1 \][/tex]
[tex]\[ f(1) \approx 2000 \times 1.00164 \][/tex]
[tex]\[ f(1) \approx 2003.2768 \][/tex]
### Option C: [tex]\( f(x) = 2000 \left( 1 + \frac{0.4}{7} \right)^x \)[/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2000 \left( 1 + \frac{0.4}{7} \right)^1 \][/tex]
[tex]\[ f(1) \approx 2000 \times 1.05714 \][/tex]
[tex]\[ f(1) \approx 2114.286 \][/tex]
### Option D: [tex]\( f(x) = 2000 \left( 1 + \frac{0.4}{7} \right)^{7x} \)[/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2000 \left( 1 + \frac{0.4}{7} \right)^7 \][/tex]
[tex]\[ f(1) \approx 2000 \times 1.475 \][/tex]
[tex]\[ f(1) \approx 2950.977 \][/tex]
### Conclusion:
Given that the butterfly population grows from 2000 to 2800 over 7 years, we are looking for a model that captures a significant increase over time. From our evaluations:
- Option A: [tex]\( \approx 2098.483 \)[/tex]
- Option B: [tex]\( \approx 2003.2768 \)[/tex]
- Option C: [tex]\( \approx 2114.286 \)[/tex]
- Option D: [tex]\( \approx 2950.977 \)[/tex]
Model [tex]\( D \)[/tex] estimates a final population of approximately 2950.977, which is the closest to the actual final population of 2800. Therefore, the best function that models the population is:
[tex]\[ f(x) = 2000 \left( 1 + \frac{0.4}{7} \right)^{7x} \][/tex]
Answer:
A. Â [tex] f(x) = 2000(1 + 0.4)^{\frac{1}{7}x} [/tex]
Step-by-step explanation:
The number of years we know about is 7, so x = 7.
[tex] f(x) = 2000(1 + 0.4)^{\frac{1}{7}x} [/tex]
[tex] f(x) = 2000(1.4)^{\frac{1}{7} \times 7} [/tex]
[tex] f(x) = 2000(1.4)^{1} [/tex]
[tex] f(x) = 2000(1.4) [/tex]
[tex] f(x) = 2800 [/tex]
Answer: Choice A.
