suppose a deposit of $2,000 in a savings account that paid an annual interest rate $r$ (compounded yearly)…

suppose a deposit of $2,000 in a savings account that paid an annual interest rate $r$ (compounded yearly) is worth $2,222 after 2 years. using the formula $a = p(1 + r)^t$, we have\n$2,222 = 2,000(1 + r)^2$\nsolve for $r$ to find the annual interest rate (to the nearest tenth).\nthe annual interest rate, $r$ is \nquestion help: ebook

suppose a deposit of $2,000 in a savings account that paid an annual interest rate $r$ (compounded yearly) is worth $2,222 after 2 years. using the formula $a = p(1 + r)^t$, we have\n$2,222 = 2,000(1 + r)^2$\nsolve for $r$ to find the annual interest rate (to the nearest tenth).\nthe annual interest rate, $r$ is \nquestion help: ebook

Answer

Explanation:

Step1: Isolate $(1 + r)^2$

Divide both sides of the equation $2222 = 2000(1 + r)^2$ by 2000. We get $\frac{2222}{2000}=(1 + r)^2$. Simplifying $\frac{2222}{2000}$ gives $\frac{1111}{1000}=(1 + r)^2$.

Step2: Take square - root of both sides

$1 + r=\pm\sqrt{\frac{1111}{1000}}$. Since $r$ represents an interest rate and cannot be negative in this context, we consider only the positive square - root. $\sqrt{\frac{1111}{1000}}\approx\sqrt{1.111}\approx1.054$. So $1 + r\approx1.054$.

Step3: Solve for $r$

Subtract 1 from both sides: $r\approx1.054 - 1=0.054$.

Answer:

$5.4$