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When the KJ Reservoir (KJR) was built, a single, well-known species of fish was introduced for its role in the new ecosystem. KJR is a popular destination for a public fishing season that extends from June through August each year. Each May 1, the area's wildlife management team generates an estimate of the number of these fish in the reservoir. The team has determined that the fish's population should be no less than 1,500 fish at the start of the fishing season—less than 1,500, and the team acquires and releases enough fish to make up the difference.
The fish of this species reproduce once a year in mid-September. Due to environmental factors, of the species's total population at the end of September, 80% will survive until the following May. It is estimated that KJR cannot support a healthy, reproductive fish population greater than 2,000.
Based on the information provided, which one of the following was the team's estimate of KJR's fish population?
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| Information from Dataset | Analysis |
|---|---|
| "a single, well-known species of fish was introduced for its role in the new ecosystem" |
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| "public fishing season that extends from June through August each year" |
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| "Each May 1, the area's wildlife management team generates an estimate" |
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| "fish's population should be no less than 1,500 fish at the start of the fishing season" |
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| "less than 1,500, and the team acquires and releases enough fish to make up the difference" |
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| "reproduce once a year in mid-September" |
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| "80% will survive until the following May" |
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| "KJR cannot support a healthy, reproductive fish population greater than 2,000" |
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Summary: KJR manages a single fish species with annual assessments every May 1, maintaining the population between 1,500-2,000 fish through restocking when needed.
| Information from Dataset | Analysis |
|---|---|
| "capture a number (M) of individual animals...Each is marked, released" |
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| "A number (N) of individual animals are then captured in a second group" |
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| "some of which will have been captured in both groups, provided both...groups are sufficiently large" |
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| "number of fish common to both groups/N = M/P" |
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Summary: Source B describes the capture-recapture method for estimating animal populations, which is the technique used by KJR's wildlife team for their annual May 1 fish count.
| Information from Dataset | Analysis |
|---|---|
| "predict the annual increase in fish population in KJR due to reproduction" |
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| "fish population at the start of September (P)" |
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| "maximum supported healthy fish population (C)" |
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| "population increase = rP(1 - P/C)" |
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| "May 1 last year...marked the 60 fish comprising the capture group" |
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| "recapture group had 100 fish, including exactly 5 from the capture group" |
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Summary: Source C provides both the mathematical model for predicting fish reproduction and real data showing how the capture-recapture method was used last year, with the chart confirming that KJR's management thresholds keep the fish population in a healthy growth range.
The three sources reveal a complete fish management system at KJ Reservoir:
The logistic growth model demonstrates that maximum reproduction occurs at 1,000 fish (50% of capacity), confirming that the management strategy maintains the population well within sustainable limits.
The question asks for the wildlife management team's calculated estimate of the total number of fish in KJR based on their capture-recapture study. This requires:
Our analysis provides all necessary components for this calculation. Source C contains the specific capture-recapture data (60 marked fish, 100 recaptured fish, 5 found in common), while Source B provides the standard capture-recapture formula. These sources work together to enable a complete calculation, and the analysis confirms we can answer this question using the available data.
The capture-recapture method uses the fundamental relationship expressed in the formula: \(\frac{\mathrm{common}}{\mathrm{N}} = \frac{\mathrm{M}}{\mathrm{P}}\), where M represents marked fish (60), N represents recaptured fish (100), common represents fish found in both groups (5), and P represents the total population estimate we need to calculate.
Using our data: M = 60, N = 100, and 5 common fish. We can solve for P by rearranging the formula to: \(\mathrm{P} = \frac{\mathrm{M} \times \mathrm{N}}{\mathrm{common}} = \frac{60 \times 100}{5} = 1200\)
Verification confirms our calculation and contextual understanding:
1200
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