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Sagot :
Sure, let's complete the table and find the balances for each compounding frequency.
1. Principal (P): The initial amount invested is [tex]$3500. 2. Annual interest rate (r): 3% or 0.03 as a decimal. 3. Time (t): 10 years. The formula for compound interest is given by: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) is the annual interest rate (in decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for, in years. Let's calculate the balances for different values of \( n \): ### When \( n = 1 \) (Compounded Annually) \[ A = 3500 \left(1 + \frac{0.03}{1}\right)^{1 \cdot 10} \] \[ A = 3500 \left(1 + 0.03\right)^{10} \] \[ A = 3500 \left(1.03\right)^{10} \] \[ A \approx 4703.71 \] ### When \( n = 2 \) (Compounded Semiannually) \[ A = 3500 \left(1 + \frac{0.03}{2}\right)^{2 \cdot 10} \] \[ A = 3500 \left(1 + 0.015\right)^{20} \] \[ A = 3500 \left(1.015\right)^{20} \] \[ A \approx 4713.99 \] ### When \( n = 4 \) (Compounded Quarterly) \[ A = 3500 \left(1 + \frac{0.03}{4}\right)^{4 \cdot 10} \] \[ A = 3500 \left(1 + 0.0075\right)^{40} \] \[ A = 3500 \left(1.0075\right)^{40} \] \[ A \approx 4719.22 \] ### When \( n = 12 \) (Compounded Monthly) \[ A = 3500 \left(1 + \frac{0.03}{12}\right)^{12 \cdot 10} \] \[ A = 3500 \left(1 + 0.0025\right)^{120} \] \[ A = 3500 \left(1.0025\right)^{120} \] \[ A \approx 4722.74 \] ### When \( n = 365 \) (Compounded Daily) \[ A = 3500 \left(1 + \frac{0.03}{365}\right)^{365 \cdot 10} \] \[ A = 3500 \left(1 + 0.00008219\right)^{3650} \] \[ A = 3500 \left(1.00008219\right)^{3650} \] \[ A \approx 4724.45 \] Now, let's fill in the table with these balances: \[ \begin{tabular}{c|l|l|l|l} \hline n & 1 & 2 & 4 & 12 & 365 \\ \hline A & 4703.71 & 4713.99 & 4719.22 & 4722.74 & 4724.45 \\ \hline \end{tabular} \] So, the balance \( A \) after 10 years for each compounding frequency \( n \) is: - For annual compounding (\( n = 1 \)): $[/tex]4703.71
- For semiannual compounding ([tex]\( n = 2 \)[/tex]): [tex]$4713.99 - For quarterly compounding (\( n = 4 \)): $[/tex]4719.22
- For monthly compounding ([tex]\( n = 12 \)[/tex]): [tex]$4722.74 - For daily compounding (\( n = 365 \)): $[/tex]4724.45
1. Principal (P): The initial amount invested is [tex]$3500. 2. Annual interest rate (r): 3% or 0.03 as a decimal. 3. Time (t): 10 years. The formula for compound interest is given by: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) is the annual interest rate (in decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for, in years. Let's calculate the balances for different values of \( n \): ### When \( n = 1 \) (Compounded Annually) \[ A = 3500 \left(1 + \frac{0.03}{1}\right)^{1 \cdot 10} \] \[ A = 3500 \left(1 + 0.03\right)^{10} \] \[ A = 3500 \left(1.03\right)^{10} \] \[ A \approx 4703.71 \] ### When \( n = 2 \) (Compounded Semiannually) \[ A = 3500 \left(1 + \frac{0.03}{2}\right)^{2 \cdot 10} \] \[ A = 3500 \left(1 + 0.015\right)^{20} \] \[ A = 3500 \left(1.015\right)^{20} \] \[ A \approx 4713.99 \] ### When \( n = 4 \) (Compounded Quarterly) \[ A = 3500 \left(1 + \frac{0.03}{4}\right)^{4 \cdot 10} \] \[ A = 3500 \left(1 + 0.0075\right)^{40} \] \[ A = 3500 \left(1.0075\right)^{40} \] \[ A \approx 4719.22 \] ### When \( n = 12 \) (Compounded Monthly) \[ A = 3500 \left(1 + \frac{0.03}{12}\right)^{12 \cdot 10} \] \[ A = 3500 \left(1 + 0.0025\right)^{120} \] \[ A = 3500 \left(1.0025\right)^{120} \] \[ A \approx 4722.74 \] ### When \( n = 365 \) (Compounded Daily) \[ A = 3500 \left(1 + \frac{0.03}{365}\right)^{365 \cdot 10} \] \[ A = 3500 \left(1 + 0.00008219\right)^{3650} \] \[ A = 3500 \left(1.00008219\right)^{3650} \] \[ A \approx 4724.45 \] Now, let's fill in the table with these balances: \[ \begin{tabular}{c|l|l|l|l} \hline n & 1 & 2 & 4 & 12 & 365 \\ \hline A & 4703.71 & 4713.99 & 4719.22 & 4722.74 & 4724.45 \\ \hline \end{tabular} \] So, the balance \( A \) after 10 years for each compounding frequency \( n \) is: - For annual compounding (\( n = 1 \)): $[/tex]4703.71
- For semiannual compounding ([tex]\( n = 2 \)[/tex]): [tex]$4713.99 - For quarterly compounding (\( n = 4 \)): $[/tex]4719.22
- For monthly compounding ([tex]\( n = 12 \)[/tex]): [tex]$4722.74 - For daily compounding (\( n = 365 \)): $[/tex]4724.45
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