Find the balance for $4000 invested at 9.3%
for t 13 years and compounded:
a) yearly.
Answer in units of dollars. Your answer
must be within ± 0.001%
Your answer must be within ±0.001 %.
part 2 of 6
b) biannually.
Answer in units of dollars. Your answer
must be within ±0.001%
Your answer must be within ±0.001 %.



Answer :

Sure, let's solve these questions step by step using the compound interest formula.

The compound interest formula is given by:

[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]

where:
- [tex]\( A \)[/tex] is the amount of money accumulated after n years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial sum of money).
- [tex]\( r \)[/tex] is the annual interest rate (decimal).
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year.
- [tex]\( t \)[/tex] is the number of years the money is invested for.

### Part a: Compounded Yearly

For yearly compounding:
- Principal ([tex]\( P \)[/tex]) = [tex]$4000 - Annual Interest Rate (\( r \)) = 9.3% = 0.093 - Time (\( t \)) = 13 years - Number of Times Compounded Per Year (\( n \)) = 1 Plugging these values into the formula: \[ A = 4000 \left(1 + \frac{0.093}{1}\right)^{1 \cdot 13} \] \[ A = 4000 \left(1 + 0.093\right)^{13} \] \[ A = 4000 (1.093)^{13} \] We'll calculate \( (1.093)^{13} \): \[ (1.093)^{13} \approx 3.146 \] Now, multiplying by the principal: \[ A \approx 4000 \times 3.146 \] \[ A \approx 12584 \] So, the balance for $[/tex]4000 invested at 9.3% for 13 years and compounded yearly is approximately [tex]$12,584.00. ### Part b: Compounded Biannually For biannual compounding: - Principal (\( P \)) = $[/tex]4000
- Annual Interest Rate ([tex]\( r \)[/tex]) = 9.3% = 0.093
- Time ([tex]\( t \)[/tex]) = 13 years
- Number of Times Compounded Per Year ([tex]\( n \)[/tex]) = 2

Plugging these values into the formula:

[tex]\[ A = 4000 \left(1 + \frac{0.093}{2}\right)^{2 \cdot 13} \][/tex]
[tex]\[ A = 4000 \left(1 + 0.0465\right)^{26} \][/tex]
[tex]\[ A = 4000 (1.0465)^{26} \][/tex]

We'll calculate [tex]\( (1.0465)^{26} \)[/tex]:

[tex]\[ (1.0465)^{26} \approx 3.283 \][/tex]

Now, multiplying by the principal:

[tex]\[ A \approx 4000 \times 3.283 \][/tex]
[tex]\[ A \approx 13132 \][/tex]

So, the balance for [tex]$4000 invested at 9.3% for 13 years and compounded biannually is approximately $[/tex]13,132.00.

Hence, the final answers are:
- Compounded yearly: [tex]$12,584.00 - Compounded biannually: $[/tex]13,132.00

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