To model the deer population with an exponential function, we can use the general formula for exponential growth:
\[ P(t) = P_0 \times (1 + r)^t \]
where:
- \( P(t) \) is the population at time \( t \),
- \( P_0 \) is the initial population,
- \( r \) is the growth rate (as a decimal), and
- \( t \) is the time in years.
In this case:
- The initial population \( P_0 = 100 \),
- The growth rate \( r = 0.09 \) (since 9% expressed as a decimal is 0.09).
Plugging these values into the formula gives:
\[ P(t) = 100 \times (1 + 0.09)^t \]
which simplifies to:
\[ P(t) = 100 \times (1.09)^t \]
Now, to find the population for each year from 0 to 10, we can evaluate \( P(t) \) for \( t = 0, 1, 2, \ldots, 10 \).
For the values:
- \( P(0) = 100 \times (1.09)^0 = 100 \)
- \( P(1) = 100 \times (1.09)^1 \approx 109 \)
- \( P(2) = 100 \times (1.09)^2 \approx 118.81 \)
- \( P(3) = 100 \times (1.09)^3 \approx 128.50 \)
- \( P(4) = 100 \times (1.09)^4 \approx 139.22 \)
- \( P(5) = 100 \times (1.09)^5 \approx 150.15 \)
- \( P(6) = 100 \times (1.09)^6 \approx 161.23 \)
- \( P(7) = 100 \times (1.09)^7 \approx 172.59 \)
- \( P(8) = 100 \times (1.09)^8 \approx 184.16 \)
- \( P(9) = 100 \times (1.09)^9 \approx 196.01 \)
- \( P(10) = 100 \times (1.09)^{10} \approx 208.19 \)
Thus, the population after 10 years will be approximately 208 deer.
The exponential function describing the deer population is:
\[ P(t) = 100 \times (1.09)^t \]
for \( t = 0, 1, 2, ..., 10 \).
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