Question

The population of a town in 2007 is 113, 505

and is increasing at a rate of 1.2% per year.
What will the population be in 2012?

What number will you fill in for a to solve the
equation?
(Hint: For this one, how many years
after 2007 is 2012?

Answer 1

The population of the town in 2012 will be approximately 120,315.

Step-by-step explanation:

To find the population in 2012, we need to calculate the number of years from 2007 to 2012. This is done by subtracting the initial year (2007) from the final year (2012):

2012 - 2007 = 5 years

Next, we need to calculate the increase in population over the 5-year period. The population is increasing at a rate of 1.2% per year, so we can use the formula:

New Population = Initial Population + (Rate * Initial Population * Number of Years)

Substituting the given values:

New Population = 113,505 + (0.012 * 113,505 * 5)

Simplifying the equation:

New Population = 113,505 + (0.012 * 567,525)

New Population = 113,505 + 6,810.3

New Population = 120,315.3

Therefore, the population of the town in 2012 will be approximately 120,315.

Answer 2

from 2007 to 2012 is only 5 years, so we can see this as a compound interest with a rate of 1.2% per annum for 5 years, so

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$113505\\ r=rate\to 1.2\%\to (1.2)/(100)\dotfill &0.012\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per year, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases} \\\\\\ A=113505\left(1+(0.012)/(1)\right)^(1\cdot 5)\implies A=113505(1.012)^5\implies A\approx 120481`