In 2006, the population of tewksbury, rhode island was 25,000, and it was growing at an annual rate of 2.2%.
what is the growth factor for the town?
write an equation to model tewksbury’s growth.
use your equation to estimate the population of tewksbury in 2011. round your response to the nearest whole number.

Assuming the growth is of 1st order, we can start using the formula for rate of 1st order reaction:

dN / dt = k * N

Rearranging,

dN / N = k dt

Where N = amount of sample, k = growth factor, t = time

Integrating the equation from N = Ni to Nf and t = ti to tf will result in:

ln (Nf / Ni) = k (tf – ti)

Finding for the growth factor k:

k = ln (Nf / Ni) / (tf – ti)

k = ln (1.022 Ni / Ni) / 1 year

k = 0.02176 / year

The population in 2011 is:

ln (Nf / Ni) = k (tf – ti)

ln (Nf / 25000) = (0.02176 / year) * (2011 – 2006)

Nf = 27,873.7 = 27,874

i just took the test, the correct answer is .. 31.5 in.

step-by-step explanation:

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when the line is dashed:

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this > inequality sign means:

this < inequality sign means:

about the problem:

srry i'm not done ! i'll