Chapter 4, Exponential and Logarithmic Functions - Section 4.2 - The Natural Exponential Function - 4.2 Exercises - Page 379: 33

$A(2) \approx 2145.02$ $A(4) \approx 2300.5$ $A(12) \approx 3043.9$

We are given $P=\$ 2000$ $r=0.035$ We use the formula for continously compounded interest with a. $t=2$ $A(2)=2000e^{(0.035)2}=2000e^{(0.07)} \approx 2145.02$ b. $t=4$ $A(4)=2000e^{(0.035)4}=2000e^{(0.14)} \approx 2300.5$ c. $t=12$ $A(12)=2000e^{(0.035)12}=2000e^{(0.42)} \approx 3043.9$