Answer :
Certainly! Let's solve the given problem step-by-step using the compound interest formula:
[tex]\[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A(t) \)[/tex] is the amount of money accumulated after time [tex]\( t \)[/tex] (including interest).
- [tex]\( P \)[/tex] is the principal amount (the initial amount 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 time the money is invested for in years.
### Compounding Semi-Annually
1. Principal (P): \[tex]$6500 2. Annual Interest Rate (r): 3.6% or 0.036 3. Number of times interest applied per year (n): 2 (since it is compounded semi-annually) 4. Time (t): 20 years Plugging these values into the formula: \[ A_{semi-annually} = 6500 \left(1 + \frac{0.036}{2}\right)^{2 \cdot 20} \] Calculate inside the parentheses first: \[ 1 + \frac{0.036}{2} = 1.018 \] Next, calculate the exponent: \[ 2 \cdot 20 = 40 \] So, we have: \[ A_{semi-annually} = 6500 \cdot (1.018)^{40} \] Evaluating this expression, we get: \[ A_{semi-annually} \approx 13268.58 \] So, the account will be worth \$[/tex]13,268.58 after 20 years if compounded semi-annually.
### Compounding Weekly
1. Principal (P): \[tex]$6500 2. Annual Interest Rate (r): 3.6% or 0.036 3. Number of times interest applied per year (n): 52 (since it is compounded weekly) 4. Time (t): 20 years Plugging these values into the formula: \[ A_{weekly} = 6500 \left(1 + \frac{0.036}{52}\right)^{52 \cdot 20} \] Calculate inside the parentheses first: \[ 1 + \frac{0.036}{52} \approx 1.0006923 \] Next, calculate the exponent: \[ 52 \cdot 20 = 1040 \] So, we have: \[ A_{weekly} = 6500 \cdot (1.0006923)^{1040} \] Evaluating this expression, we get: \[ A_{weekly} \approx 13350.49 \] So, the account will be worth \$[/tex]13,350.49 after 20 years if compounded weekly.
In summary:
- The account balance compounded semi-annually after 20 years is approximately \[tex]$13,268.58. - The account balance compounded weekly after 20 years is approximately \$[/tex]13,350.49.
[tex]\[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A(t) \)[/tex] is the amount of money accumulated after time [tex]\( t \)[/tex] (including interest).
- [tex]\( P \)[/tex] is the principal amount (the initial amount 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 time the money is invested for in years.
### Compounding Semi-Annually
1. Principal (P): \[tex]$6500 2. Annual Interest Rate (r): 3.6% or 0.036 3. Number of times interest applied per year (n): 2 (since it is compounded semi-annually) 4. Time (t): 20 years Plugging these values into the formula: \[ A_{semi-annually} = 6500 \left(1 + \frac{0.036}{2}\right)^{2 \cdot 20} \] Calculate inside the parentheses first: \[ 1 + \frac{0.036}{2} = 1.018 \] Next, calculate the exponent: \[ 2 \cdot 20 = 40 \] So, we have: \[ A_{semi-annually} = 6500 \cdot (1.018)^{40} \] Evaluating this expression, we get: \[ A_{semi-annually} \approx 13268.58 \] So, the account will be worth \$[/tex]13,268.58 after 20 years if compounded semi-annually.
### Compounding Weekly
1. Principal (P): \[tex]$6500 2. Annual Interest Rate (r): 3.6% or 0.036 3. Number of times interest applied per year (n): 52 (since it is compounded weekly) 4. Time (t): 20 years Plugging these values into the formula: \[ A_{weekly} = 6500 \left(1 + \frac{0.036}{52}\right)^{52 \cdot 20} \] Calculate inside the parentheses first: \[ 1 + \frac{0.036}{52} \approx 1.0006923 \] Next, calculate the exponent: \[ 52 \cdot 20 = 1040 \] So, we have: \[ A_{weekly} = 6500 \cdot (1.0006923)^{1040} \] Evaluating this expression, we get: \[ A_{weekly} \approx 13350.49 \] So, the account will be worth \$[/tex]13,350.49 after 20 years if compounded weekly.
In summary:
- The account balance compounded semi-annually after 20 years is approximately \[tex]$13,268.58. - The account balance compounded weekly after 20 years is approximately \$[/tex]13,350.49.