To predict the population based on 2-year increments rather than a 10-year increment, we need to find the equivalent growth rate for 2 years based on the given growth rate for 10 years.
The original formula is:
\[ p(d) = 1,000,000(1 + 0.15)^d \]
Where \(d\) is the number of decades. Since 1 decade is equal to 10 years, we need to find the growth factor for a period of 2 years, which can be viewed as \(d = 0.2\) decades (since 2 years is 1/5 of a decade).
- Calculate the growth factor over 10 years:
\[ 1 + 0.15 = 1.15 \]
- For 2 years (which is 0.2 decades), we take the growth factor over 10 years to the power of 0.2:
\[ (1.15)^{0.2} \]
- To calculate this value:
\[ (1.15)^{0.2} \approx 1.02899 , (\text{approximately}) \]
Now, the approximate rate of increase for 2 years is:
\[ 1.02899 - 1 \approx 0.02899 \text{ or approximately } 0.03 \]
So, using this growth factor over each 2-year period, the new formula for the population every 2 years will be:
\[ p(d) = 1,000,000(1 + 0.03)^{d} \]
Where \(d\) is the number of 2-year increments.
Therefore, the correct option is:
A. \(p(d) = 1,000,000(1 + 0.03)^{d}\)
This predicts the population in 2-year increments based on the original growth rate.