To convert the growth rate from a decade (10 years) to a 2-year increment, we first need to find the effective growth rate for 2 years based on the given rate of 15% per decade.
The formula for the population is: \[ p(d) = 1,000,000(1 + 0.15)^d \] where \(d\) is the number of decades.
For 2-year increments, we have:
- 10 years (1 decade) = 15% growth
- 2 years = \(\frac{2}{10} = \frac{1}{5}\) of a decade.
We need to find the equivalent growth for this portion of a decade. The growth factor over a full decade (10 years) is \(1 + 0.15 = 1.15\).
To find the growth factor for 2 years, we take the 10-year growth factor to the power of \( \frac{1}{5} \): \[ (1.15)^{\frac{1}{5}} \]
Now, we can calculate \( (1.15)^{0.2} \) (the same as taking the fifth root of 1.15).
Using a calculator or exponentiation: \[ (1.15)^{0.2} \approx 1.02899 \approx 1 + 0.03 \]
This indicates that the growth rate for each 2-year increment is approximately 3%.
Therefore, the population model for 2-year increments becomes: \[ p(d) = 1,000,000(1 + 0.03)^d \]
where \(d\) is the number of 2-year periods.
Among your options, the correct formula to predict the population based on 2-year increments is: \[ p(d) = 1,000,000(1 + 0.03)^d \]
4 answers