question\n3300 dollars is placed in an account with an annual interest rate of 6.75%. to the nearest tenth…

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 (in decimal form), and $t$ is the number of years. Here, $P = 3300$, $r=0.0675$, and $A = 10700$. So we have the equation $10700=3300(1 + 0.0675)^t$.

Step2: Simplify the equation

First, divide both sides of the equation by 3300: $\frac{10700}{3300}=(1.0675)^t$, which simplifies to $\frac{107}{33}=(1.0675)^t$.

Step3: Take the natural logarithm of both sides

$\ln(\frac{107}{33})=\ln(1.0675^t)$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(\frac{107}{33}) = t\ln(1.0675)$.

Step4: Solve for $t$

$t=\frac{\ln(\frac{107}{33})}{\ln(1.0675)}$. Calculate $\ln(\frac{107}{33})\approx\ln(3.2424)\approx1.176$ and $\ln(1.0675)\approx0.0653$. Then $t=\frac{1.176}{0.0653}\approx18.0$.

Answer:

$18.0$ years