To complete the input-output table using the equation \( y = 4,060(0.95)^t \), we will substitute the values of \( t \) (0, 5, 10, 15, and 20) into the equation to find the corresponding values of \( y \).
Calculating the values:
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For \( t = 0 \): \[ y = 4,060(0.95)^0 = 4,060 \times 1 = 4,060 \]
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For \( t = 5 \): \[ y = 4,060(0.95)^5 \approx 4,060 \times 0.77378 \approx 3,140.81 \]
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For \( t = 10 \): \[ y = 4,060(0.95)^{10} \approx 4,060 \times 0.59874 \approx 2,429.00 \]
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For \( t = 15 \): \[ y = 4,060(0.95)^{15} \approx 4,060 \times 0.46329 \approx 1,883.72 \]
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For \( t = 20 \): \[ y = 4,060(0.95)^{20} \approx 4,060 \times 0.34868 \approx 1,413.73 \]
Completed Input-Output Table:
\[ \begin{array}{|c|c|} \hline \text{Years (t)} & \text{Number of Birds (y)} \ \hline 0 & 4,060 \ 5 & 3,141 \ 10 & 2,429 \ 15 & 1,884 \ 20 & 1,414 \ \hline \end{array} \]
Summary of Values:
- \( t = 0 \): \( y = 4,060 \)
- \( t = 5 \): \( y \approx 3,141 \)
- \( t = 10 \): \( y \approx 2,429 \)
- \( t = 15 \): \( y \approx 1,884 \)
- \( t = 20 \): \( y \approx 1,414 \)
Identifying the Correct Graph:
To determine which graph corresponds to this decay model, look for a graph that depicts an exponential decay with a gradually decreasing rate of the bird population. Such a graph would start at approximately 4,060 and approach 0 but never quite reach it, displaying a gentle slope.
If you can visualize or reference specific graphs, match it to this behavior to find the correct representation of the data calculated from the equation. If you need further assistance in identifying the graph, please describe the available graphs or their key features.
4 answers