Question

In 2006, the population of Tewksbury, Rhode Island was 25,000, and it was growing at an annual rate of 2.2%.Part AWhat is the growth factor for the town?Part BWrite an equation to model the cars value.Part CUse your equation to estimate the population of Tewksbury in 2011. round your response to the nearest whole number

Answer 1

Part A. To find the growth factor of the town, you can proceed like this

$\begin{gathered} 2.2\text{ \% }=\frac{2.2}{100}=0.022 \\ \text{Then,} \\ 25000\cdot0.022=550\Rightarrow\text{ Population increase in one year} \\ 25000+550=25550\Rightarrow\text{ Population in 2007},\text{ adding the population in 2006 plus the increase} \\ \end{gathered}$

So, the growth factor for the town will be

$\frac{25550}{25000}=1.022$

For part B, exponential growth is modeled by the equation

$\begin{gathered} y=ab^x \\ \text{Where} \\ a\text{ is the initial amount} \\ b\text{ is the growth factor} \\ x\text{ is the time in years} \end{gathered}$

So, you have

$\begin{gathered} a=25000 \\ b=1.022 \\ \text{ Then,} \\ y=(25000)(1.022)^x \end{gathered}$

For part C, you need to find the population in 2011, so

$2011-2006=5\text{ years}$

Then, plug in the equation found x = 5

$\begin{gathered} y=(25000)(1.022)^x \\ y=(25000)(1.022)^5 \\ y=27873.7 \\ \text{Rounding} \\ y=27874 \end{gathered}$

Therefore, the population of Tewksbury in 2011 will be 27874 people.