Question

The table shows population data for a community

a.) What is the percent change from 2007 to 2013? (The answer for 'a' is 6.9; I already know A and I am looking for B)
b.) Use this percent change to predict the population in 2009

In 2007, the population was 118,000
In 2013, the population was 138,000

Can anyone help me with letter b?

Answer 1

It might have been 125,537.Hope that this help's you!!!!

Answer 2

Answer: a) 16.9491525424 %

b) 124322 ( approx )

Explanation:

a) The population in 2009 = 118000

While the population in 2013 = 138,000

Thus, the total percentage change in the population from 2009 to 2013,

`=\frac{\text{ Population in 2013 - Population in 2009 }}{\text{Population in 2009}}* 100`

`=(138000-118000)/(118000)* 100`

`=(2000000)/(118000)`

`=16.9491525424\% \approx 16.94\%`

b) Let the function that shows the population after x years since 2007,

`f(x)=ab^(x)`

Where a and b are any unknown number,

Since, for x = 0, f(0) = 118000 ( given)

⇒
`118000 = ab^0`

⇒
`a=118000`

Now, For x = 6, f(6) = 138,000 ( given )

⇒
`138000 = ab^6`

⇒
`138000= 118000 b^6`

⇒
`b=1.0264382948`

Hence, the function that shows the population after x years since 2007,

`f(x)=118000(1.0264382948)^(x)`

For, x = 2,

`f(2)=118000(1.0264382948)^(2)=124321.917618\approx 124322`

⇒ The predicted population in 2009 is approximately 124322.