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Sagot :
To solve this problem, we start with the formula for continuous compounding interest, which is given by:
[tex]\[ A = P \cdot e^{rt} \][/tex]
where:
- [tex]\( A \)[/tex] is the amount to which the investment grows,
- [tex]\( P \)[/tex] is the principal or initial investment,
- [tex]\( r \)[/tex] is the interest rate per year (expressed as a decimal),
- [tex]\( t \)[/tex] is the time in years,
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately 2.71828).
For the given problem:
- The initial investment [tex]\( P \)[/tex] is [tex]$3000, - The time \( t \) is 13 years, - The interest rate \( r \) needs to be expressed as a decimal (e.g., 1% = 0.01). Using the formula \( A = 3000 \cdot e^{13r} \), we can calculate the amounts for various interest rates \( r \). Let's fill in the amounts for each of the given interest rates: 1. For an interest rate of 1% (0.01): \[ A = 3000 \cdot e^{13 \times 0.01} = 3000 \cdot e^{0.13} \] This computation gives an amount of \( \$[/tex]3416.49 \).
2. For an interest rate of 2% (0.02):
[tex]\[ A = 3000 \cdot e^{13 \times 0.02} = 3000 \cdot e^{0.26} \][/tex]
This computation gives an amount of [tex]\( \$3890.79 \)[/tex].
3. For an interest rate of 3% (0.03):
[tex]\[ A = 3000 \cdot e^{13 \times 0.03} = 3000 \cdot e^{0.39} \][/tex]
This computation gives an amount of [tex]\( \$4430.94 \)[/tex].
4. For an interest rate of 4% (0.04):
[tex]\[ A = 3000 \cdot e^{13 \times 0.04} = 3000 \cdot e^{0.52} \][/tex]
This computation gives an amount of [tex]\( \$5046.08 \)[/tex].
5. For an interest rate of 5% (0.05):
[tex]\[ A = 3000 \cdot e^{13 \times 0.05} = 3000 \cdot e^{0.65} \][/tex]
This computation gives an amount of [tex]\( \$5746.62 \)[/tex].
6. For an interest rate of 6% (0.06):
[tex]\[ A = 3000 \cdot e^{13 \times 0.06} = 3000 \cdot e^{0.78} \][/tex]
This computation gives an amount of [tex]\( \$6544.42 \)[/tex].
So the completed table should look like this:
[tex]\[ \begin{tabular}{|c|c|} \hline \begin{tabular}{l} Rate \\ per \\ Year \end{tabular} & Amount \\ \hline 1\% & \$3416.49 \\ \hline 2\% & \$3890.79 \\ \hline 3\% & \$4430.94 \\ \hline 4\% & \$5046.08 \\ \hline 5\% & \$5746.62 \\ \hline 6\% & \$6544.42 \\ \hline \end{tabular} \][/tex]
[tex]\[ A = P \cdot e^{rt} \][/tex]
where:
- [tex]\( A \)[/tex] is the amount to which the investment grows,
- [tex]\( P \)[/tex] is the principal or initial investment,
- [tex]\( r \)[/tex] is the interest rate per year (expressed as a decimal),
- [tex]\( t \)[/tex] is the time in years,
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately 2.71828).
For the given problem:
- The initial investment [tex]\( P \)[/tex] is [tex]$3000, - The time \( t \) is 13 years, - The interest rate \( r \) needs to be expressed as a decimal (e.g., 1% = 0.01). Using the formula \( A = 3000 \cdot e^{13r} \), we can calculate the amounts for various interest rates \( r \). Let's fill in the amounts for each of the given interest rates: 1. For an interest rate of 1% (0.01): \[ A = 3000 \cdot e^{13 \times 0.01} = 3000 \cdot e^{0.13} \] This computation gives an amount of \( \$[/tex]3416.49 \).
2. For an interest rate of 2% (0.02):
[tex]\[ A = 3000 \cdot e^{13 \times 0.02} = 3000 \cdot e^{0.26} \][/tex]
This computation gives an amount of [tex]\( \$3890.79 \)[/tex].
3. For an interest rate of 3% (0.03):
[tex]\[ A = 3000 \cdot e^{13 \times 0.03} = 3000 \cdot e^{0.39} \][/tex]
This computation gives an amount of [tex]\( \$4430.94 \)[/tex].
4. For an interest rate of 4% (0.04):
[tex]\[ A = 3000 \cdot e^{13 \times 0.04} = 3000 \cdot e^{0.52} \][/tex]
This computation gives an amount of [tex]\( \$5046.08 \)[/tex].
5. For an interest rate of 5% (0.05):
[tex]\[ A = 3000 \cdot e^{13 \times 0.05} = 3000 \cdot e^{0.65} \][/tex]
This computation gives an amount of [tex]\( \$5746.62 \)[/tex].
6. For an interest rate of 6% (0.06):
[tex]\[ A = 3000 \cdot e^{13 \times 0.06} = 3000 \cdot e^{0.78} \][/tex]
This computation gives an amount of [tex]\( \$6544.42 \)[/tex].
So the completed table should look like this:
[tex]\[ \begin{tabular}{|c|c|} \hline \begin{tabular}{l} Rate \\ per \\ Year \end{tabular} & Amount \\ \hline 1\% & \$3416.49 \\ \hline 2\% & \$3890.79 \\ \hline 3\% & \$4430.94 \\ \hline 4\% & \$5046.08 \\ \hline 5\% & \$5746.62 \\ \hline 6\% & \$6544.42 \\ \hline \end{tabular} \][/tex]
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