4. ryan invests a sum of money in a savings account with a fixed annual interest rate of 4.31% compounded 12…
4. ryan invests a sum of money in a savings account with a fixed annual interest rate of 4.31% compounded 12 times per year. after 10 years, the balance reaches $12,855.94. what was the amount of the initial investment? round to the nearest dollar. type a response hints 1. $12,855.94 is a. plug everything else into the formula, use the calculator to find what it is equal to. then divide to get p by itself.
Answer
Explanation:
Step1: Recall compound - interest formula
The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the final amount, $P$ is the principal (initial investment), $r$ is the annual interest rate (in decimal form), $n$ is the number of times compounded per year, and $t$ is the number of years. We are given that $A=$12855.94$, $r = 0.0431$ (since $4.31%=0.0431$), $n = 12$, and $t = 10$.
Step2: Rearrange the formula for $P$
Starting with $A = P(1+\frac{r}{n})^{nt}$, we can solve for $P$ by dividing both sides of the equation by $(1+\frac{r}{n})^{nt}$. So, $P=\frac{A}{(1 +\frac{r}{n})^{nt}}$.
Step3: Substitute the given values
Substitute $A = 12855.94$, $r=0.0431$, $n = 12$, and $t = 10$ into the formula for $P$. First, calculate $(1+\frac{0.0431}{12})^{12\times10}$. $\frac{0.0431}{12}\approx0.00359167$. $1+\frac{0.0431}{12}=1 + 0.00359167=1.00359167$. $(1.00359167)^{120}\approx1.53997$. Then, $P=\frac{12855.94}{1.53997}\approx8349$.
Answer:
$8349$