Chapter 7.5, Problem 27PPS

(a)

To determine

To find:How much money she will save next year.

Answer to Problem 27PPS

Aida saves $\$530$ next year

Explanation of Solution

Given: In $2010$ Aida saves $\$500$ . She plans to save $6\%$ more than her previous years saving for the next several years.

Given: Aida saves $\$500$ in $2010$ .

Let she will save $x$ next years.

Then

  $\begin{array}{l}x=500+6\%of500\\ x=500+\frac{6}{100}\times 500\\ x=500+30\\ x=\$530\end{array}$

Hence, Aida saves $\$530$ next year.

(b)

To determine

To Make:The table for $2010-2015$

Answer to Problem 27PPS

  

Explanation of Solution

Given:In $2010$ Aida saves $\$500$ . She plans to save $6\%$ more than her previous years saving for the next several years.

Let $x$ represent the year and $y$ represent the amount of money she saved.

Let $x$ represent the year and $y$ represent the amount of money she saved.

Then

Table for $2010-2015$ is

  

(c)

To determine

To find:Does Aida’s saving increase by a constant amount each year? Explain.

Answer to Problem 27PPS

No

Explanation of Solution

Given: In $2010$ Aida saves $\$500$ . She plans to save $6\%$ more than her previous years saving for the next several years.

No,

The savings does not increase in the constant rate

Because the increment is depend on the previous year.