Chapter 6.4, Problem 62E

(a)

To determine

The functions that represent the balances of the accounts after t years.

Answer to Problem 62E

The model is:

  $y=300{\left(1.08\right)}^t$

Explanation of Solution

Given information:

Deposit $300

Annual interest 12%

The given savings graph as shown below,

  

Formula used:

  $y=P{\left(1+\frac{r}{n}\right)}^{nt}$

Calculation:

FOR INVESTMENT ACCOUNT:

The compound interest formula as shown below,

  $y=P{\left(1+\frac{r}{n}\right)}^{nt}$

Substitute the given equation as shown below,

  $P=300,r=0.12\&n=4\left(quartely\right)$

Simplify the equation as shown below,

  $\begin{array}{l}y=300{\left(1+\frac{0.12}{4}\right)}^{4t}\\ y=300{\left(1.03\right)}^{4t}\end{array}$

FOR SAVINGS ACCOUNT:

Make a table of approximated values:

  

From the table, a = 300 (initial value) & (1 + r) = 1.08

Therefore, the model is:

  $y=300{\left(1.08\right)}^t$

Conclusion:

The model is:

  $y=300{\left(1.08\right)}^t$

(b)

To determine

The graph of the function in the same coordinate plane.

Answer to Problem 62E

The investment account grows at a faster rate at 12% quarterly than the savings account which grows at a rate of 8% per year.

Explanation of Solution

Given information:

Deposit $300

Annual interest 12%

The given savings graph as shown below,

  

Formula used:

The x axis represent the year and the y axis represent the Balance in dollars.

Calculation:

  

The investment account grows at a faster rate at 12% quarterly than the savings account which grows at a rate of 8% per year.

Conclusion:

The investment account grows at a faster rate at 12% quarterly than the savings account which grows at a rate of 8% per year.