Question

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.

Answer 1

Tewksbury Population In 2011

=
$25000 {( {1.022)}^{2011 - 2006} }$
= 27,874

Answer 2

Answer:

The growth factor is 1.022

y = (25000) (1.022)^x

y =27874 people in 2011

Step-by-step explanation:

Exponential growth is modeled by the equation

y= a b^x

where a is the initial amount and b is 1 + the growth percent  and x is the time in years

a =25000

b = 1+.022 = 1.022

y = (25000) (1.022)^x

We need to find the population ins 2011

2011-2006 is 5 years  so x=5

y = 25000 (1.022)^5

y = 27873.7 people

Rounding because we don't have parts of people

y =27874 people