To complete the input-output table based on the equation \( y = 4060(0.95)^t \), we will calculate the number of birds (y) for each given year (t).
Here’s the breakdown for each year listed in the table:
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For \( t = 0 \): \[ y = 4060(0.95)^0 = 4060(1) = 4060 \]
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For \( t = 5 \): \[ y = 4060(0.95)^5 \approx 4060 \times 0.77378 \approx 3138.5 \quad \text{(rounding to 3139)} \]
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For \( t = 10 \): \[ y = 4060(0.95)^{10} \approx 4060 \times 0.59049 \approx 2393.0 \quad \text{(rounding to 2393)} \]
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For \( t = 15 \): \[ y = 4060(0.95)^{15} \approx 4060 \times 0.46329 \approx 1882.0 \quad \text{(rounding to 1882)} \]
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For \( t = 20 \): \[ y = 4060(0.95)^{20} \approx 4060 \times 0.35849 \approx 1450.0 \quad \text{(rounding to 1450)} \]
Now we can fill in the input-output table:
\[ \begin{array}{|c|c|} \hline \text{Years (t)} & \text{Number of Birds (y)} \ \hline 0 & 4060 \ 5 & 3139 \ 10 & 2393 \ 15 & 1882 \ 20 & 1450 \ \hline \end{array} \]
Conclusion
The completed table shows the decay of the species of birds over time according to the given exponential decay model.
To determine which graph matches this sequence of values, the graph would need to show a decreasing trend that approaches zero but never actually reaches it, which is characteristic of exponential decay. Based on the values computed, the graph should start at 4060 at \( t=0 \) and gradually decline, ideally showing a smooth curve that gets shallower as time progresses.
3 answers