Question

complete the table for a savings account in which interest is compounded continuously. (round any dollar amount to the nearest cent and any other value to 4 decimal places.) initial investment $1,000 annual % rate % time to double 11 yr amount after 12 years $

Explanation:

Step1: Find Annual Rate

For continuous compounding, the formula for doubling time \( t \) is \( t=\frac{\ln(2)}{r} \), where \( r \) is the annual rate (in decimal). Given \( t = 11 \) years, solve for \( r \):
\( r=\frac{\ln(2)}{t}=\frac{\ln(2)}{11}\approx\frac{0.6931}{11}\approx0.0630 \), so \( r\approx6.30\% \).

Step2: Find Amount After 12 Years

The formula for continuous compounding is \( A = Pe^{rt} \), where \( P = 1000 \), \( r = 0.0630 \), \( t = 12 \).
\( A = 1000e^{0.0630\times12}=1000e^{0.756}\approx1000\times2.131\approx2131.00 \).

Answer:

Annual Rate: \( \boldsymbol{6.30\%} \), Amount After 12 Years: \( \boldsymbol{\$2131.00} \)