## Precalculus (6th Edition) Blitzer

The investment with $4\%$ compounded is better by $179$ dollars.
Step 1. For the first account, we have $P=6000, t=5, r=0.04, n=12$ and the end balance is $A_1=P(1+\frac{r}{n})^{nt}=6000(1+\frac{0.04}{12})^{12(5)}\approx7326$ dollars. Step 2. For the second account, we have $P=6000, t=5, r=0.035$ and the end balance is $A_2=Pe^{rt}=6000(e)^{0.035(5)}\approx7147$ dollars. Step 3. Since $A_1\gt A_2$, the first investment is better by $A_1-A_2=179$ dollars.