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Questions: Algebra BusinessCalculus
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Suppose that $19911 is invested at an interest rate of 43% per year, compounded continuously. What is the account balance after 15 years?
Using \(\displaystyle P=P_0e^{kt}\) gives \(\displaystyle P=19911e^{\frac{43 t}{100}}\). Evaluating at \(\displaystyle t=15\) gives \(\displaystyle P=12597735\) dollars.
\begin{question}Suppose that \$19911 is invested at an interest rate of 43\% per year, compounded continuously. What is the account balance after 15 years? \soln{4.5cm}{Using $P=P_0e^{kt}$ gives $P=19911e^{\frac{43 t}{100}}$. Evaluating at $t=15$ gives $P=12597735$ dollars.} \end{question}
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<p> <p>Suppose that $19911 is invested at an interest rate of 43% per year, compounded continuously. What is the account balance after 15 years? </p> </p>
<p> <p>Using <img class="equation_image" title=" \displaystyle P=P_0e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20P%3DP_0e%5E%7Bkt%7D%20" alt="LaTeX: \displaystyle P=P_0e^{kt} " data-equation-content=" \displaystyle P=P_0e^{kt} " /> gives <img class="equation_image" title=" \displaystyle P=19911e^{\frac{43 t}{100}} " src="/equation_images/%20%5Cdisplaystyle%20P%3D19911e%5E%7B%5Cfrac%7B43%20t%7D%7B100%7D%7D%20" alt="LaTeX: \displaystyle P=19911e^{\frac{43 t}{100}} " data-equation-content=" \displaystyle P=19911e^{\frac{43 t}{100}} " /> . Evaluating at <img class="equation_image" title=" \displaystyle t=15 " src="/equation_images/%20%5Cdisplaystyle%20t%3D15%20" alt="LaTeX: \displaystyle t=15 " data-equation-content=" \displaystyle t=15 " /> gives <img class="equation_image" title=" \displaystyle P=12597735 " src="/equation_images/%20%5Cdisplaystyle%20P%3D12597735%20" alt="LaTeX: \displaystyle P=12597735 " data-equation-content=" \displaystyle P=12597735 " /> dollars.</p> </p>