what annual interest rates r yield a balance of more than $3,000 on a 20 - year investment of $1,000? (enter…

Explanation:

Step1: Use compound - interest formula

The compound - interest formula is $A = P(1 + r)^t$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate, and $t$ is the number of years. Here, $P=$1000$, $A > 3000$, and $t = 20$. So, $3000<1000(1 + r)^{20}$.

Step2: Simplify the inequality

Divide both sides of the inequality $3000<1000(1 + r)^{20}$ by $1000$. We get $3<(1 + r)^{20}$.

Step3: Solve for $r$

Take the 20 - th root of both sides. Since the function $y = x^{\frac{1}{20}}$ is an increasing function, we have $3^{\frac{1}{20}}<1 + r$. Then subtract 1 from both sides to isolate $r$: $r>3^{\frac{1}{20}}-1$.

Answer:

$r > 3^{\frac{1}{20}}-1$