Problem 2 Quiz navigation 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 ? \[ y=113,505 \] What number will you fill in for $\boldsymbol{x}$ to solve the equation? 5 (Hint: For this one, how many years after 2007 is $2012 ?$ \[ y=15 \] (Round to the nearest whole person.)

Step 1 ：The problem is asking for the population of the town in 2012, given that it is increasing at a rate of 1.2% per year from the base year of 2007. The population in 2007 is given as 113,505.

Step 2 ：The number of years from 2007 to 2012 is 5 years.

Step 3 ：We can use the formula for compound interest to calculate the population in 2012. The formula is: \[A = P(1 + r/n)^{nt}\] where: - A is the amount of money accumulated after n years, including interest. - P is the principal amount (the initial amount of money) - r is the annual interest rate (in decimal) - n is the number of times that interest is compounded per year - t is the time the money is invested for in years

Step 4 ：In this case, the principal amount P is the initial population 113,505, the annual interest rate r is 1.2% or 0.012 in decimal, n is 1 since the population increases once per year, and t is 5 years.

Step 5 ：We can substitute these values into the formula and calculate the population in 2012.

Step 6 ：Final Answer: The population of the town in 2012 will be \(\boxed{120481}\).