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In this task, you will calculate the balance of an account using simple, compound, and continuously compounding interest.

Question 1

Brianna wants to save the [tex]\$ 435[/tex] she earned from being a lifeguard over the summer. She's trying to decide which type of savings account would be best for her. All of the accounts she's considering offer the same interest rate of [tex]3.9 \%[/tex], however each account calculates interest differently:
- Account 1: The balance is calculated using simple interest.
- Account 2: The balance is calculated using interest compounded annually.
- Account 3: The balance is calculated using interest compounded quarterly.
- Account 4: The balance is calculated using interest compounded continuously.

Part A

Write an expression to model the interest earned on each account, then find the balance of each option after 3 years.

Complete the table. Round your responses to the nearest whole cent, if necessary.



Answer :

GinnyAnswer
To help Brianna decide which savings account is the best for her, we can calculate the balance of each account using the different methods of interest calculation. Here are the detailed steps and formulas we will use:

### Account 1: Simple Interest
Formula:
[tex]\[ A = P (1 + rt) \][/tex]

where:
- [tex]\( P \)[/tex] is the principal amount (initial investment) = \[tex]$435 - \( r \) is the annual interest rate = 3.9% = 0.039 - \( t \) is the time the money is invested for in years = 3 Calculation: \[ A = 435 \times (1 + 0.039 \times 3) \] \[ A = 435 \times (1 + 0.117) \] \[ A = 435 \times 1.117 \] \[ A = 485.895 \] So, the balance after 3 years with simple interest is approximately \$[/tex]485.89.

### Account 2: Interest Compounded Annually
Formula:
[tex]\[ A = P (1 + \frac{r}{n})^{nt} \][/tex]

where:
- [tex]\( P \)[/tex] is the principal amount = \[tex]$435 - \( r \) is the annual interest rate = 0.039 - \( n \) is the number of times interest is compounded per year = 1 (annually) - \( t \) is the time in years = 3 Calculation: \[ A = 435 \times \left(1 + \frac{0.039}{1}\right)^{1 \times 3} \] \[ A = 435 \times (1 + 0.039)^{3} \] \[ A = 435 \times 1.039^{3} \] \[ A = 435 \times 1.122838919 \] \[ A = 487.912928 \] So, the balance after 3 years with interest compounded annually is approximately \$[/tex]487.91.

### Account 3: Interest Compounded Quarterly
Formula:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]

where:
- [tex]\( P \)[/tex] is the principal amount = \[tex]$435 - \( r \) is the annual interest rate = 0.039 - \( n \) is the number of times interest is compounded per year = 4 (quarterly) - \( t \) is the time in years = 3 Calculation: \[ A = 435 \times \left(1 + \frac{0.039}{4}\right)^{4 \times 3} \] \[ A = 435 \times \left(1 + 0.00975\right)^{12} \] \[ A = 435 \times 1.00975^{12} \] \[ A = 435 \times 1.123489359 \] \[ A = 488.705 \] So, the balance after 3 years with interest compounded quarterly is approximately \$[/tex]488.71.

### Account 4: Interest Compounded Continuously
Formula:
[tex]\[ A = P e^{rt} \][/tex]

where:
- [tex]\( P \)[/tex] is the principal amount = \[tex]$435 - \( r \) is the annual interest rate = 0.039 - \( t \) is the time in years = 3 - \( e \) is the mathematical constant approximately equal to 2.71828 Calculation: \[ A = 435 \times e^{0.039 \times 3} \] \[ A = 435 \times e^{0.117} \] \[ A = 435 \times 1.124298 \] \[ A = 488.989 \] So, the balance after 3 years with interest compounded continuously is approximately \$[/tex]488.99.

### Summary Table
| Account Type | Balance after 3 years |
|-----------------------------|-----------------------|
| Simple Interest | \[tex]$485.89 | | Compounded Annually | \$[/tex]487.91 |
| Compounded Quarterly | \[tex]$488.71 | | Compounded Continuously | \$[/tex]488.99 |

Brianna can now compare the different accounts and decide which one maximizes her savings.

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