A city with a population of 1,000,000 is growing at a rate of 15% per decade. The population of the city can be predicted using the exponential growth formula p(d)=1,000,000(1+0.15)d

Question

, where d is the number of decades past the current year. Which of the following formulas would more specifically predict the population based on 2-year increments rather than a 10-year increment?(1 point)
Responses

p(d)=1,000,000(5+0.15)d
p left parenthesis d right parenthesis equals 1,000,000 left parenthesis 5 plus 0.15 right parenthesis superscript d baseline

p(d)=1,000,000(1+0.60)d
p left parenthesis d right parenthesis equals 1,000,000 left parenthesis 1 plus 0.60 right parenthesis superscript d baseline

p(d)=1,000,000(1+0.03)d
p left parenthesis d right parenthesis equals 1,000,000 left parenthesis 1 plus 0.03 right parenthesis superscript d baseline

p(d)=1,000,000(1+0.05)d

Answer 1

To adjust the exponential growth formula from a 10-year increment to a 2-year increment, we need to understand the growth rate in terms of 2 years instead of decades (10 years).

The growth rate for 10 years is 15%, equivalent to a growth factor of:

$$1+0.15=1.15$$

To find the growth rate for 2 years, we need to find the 2-year growth factor. This can be calculated by taking the 10-year growth factor and raising it to the power of $\frac{2}{10}$ (because we want the growth over 2 years):

$$(1.15{)}^{\frac{2}{10}}=(1.15{)}^{0.2}$$

Calculating this:

$$(1.15{)}^{0.2}\approx 1.0287$$

The growth in 2 years is approximately 2.87%, which as a decimal is approximately 0.0287. This gives us a growth factor of:

$$1+0.0287\approx 1.03$$

Thus, the new population model based on 2-year increments can be represented as:

$$p(d)=1,000,000(1+0.03{)}^d$$

where $d$ is the number of 2-year increments.

From the given options, the correct choice is:

$${\text{p(d) = 1,000,000(1 + 0.03)}}^d$$