4.1 Homework Score: $42 / 100 \quad 6 / 15$ answered Question 12 Statesville's population in 2010 was about 24,400 , and was growing by about $1.1 \%$ each year. If this continues, what will Statesville's population be in 2015? [Round to the nearest person.] people Question Help: Video Submit Question

Step 1 ：The problem is asking for the population of Statesville in 2015 given that it was 24,400 in 2010 and it grows by 1.1% each year. This is a compound interest problem where the principal is the initial population, the rate is the growth rate, and the time is the number of years.

Step 2 ：The formula for compound interest 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 3 ：In this case, \(n\) is 1 because the population grows once a year, \(r\) is 1.1% or 0.011 in decimal, \(P\) is 24,400, and \(t\) is 5 because we want to find the population in 2015 and the initial population is given for 2010.

Step 4 ：Let's plug these values into the formula and calculate the population. After executing this, we will get the population of Statesville in 2015.

Step 5 ：Final Answer: The population of Statesville in 2015 will be approximately \(\boxed{25772}\) people.