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7800 dollars is placed in an account with an annual interest rate of 6.…

Question

7800 dollars is placed in an account with an annual interest rate of 6.5%. how much will be in the account after 29 years, to the nearest cent?

Explanation:

Step1: Identify the formula for compound interest (assuming compound interest, as simple interest isn't specified and compound is more common for long - term accounts)

The compound interest formula is $A = P(1+\frac{r}{n})^{nt}$, where:

  • $A$ is the amount of money accumulated after $n$ years, including interest.
  • $P$ is the principal amount (the initial amount of money).
  • $r$ is the annual interest rate (in decimal form).
  • $n$ is the number of times that interest is compounded per year. If not specified, we assume $n = 1$ (annual compounding).
  • $t$ is the time the money is invested for in years.

Given:

  • $P=\$7800$
  • $r = 6.5\%=0.065$
  • $n = 1$
  • $t = 29$ years

Step2: Substitute the values into the formula

Substitute $P = 7800$, $r=0.065$, $n = 1$, and $t = 29$ into the formula $A=P(1 +\frac{r}{n})^{nt}$:

$A=7800\times(1+\frac{0.065}{1})^{1\times29}$

First, calculate the value inside the parentheses: $1+\frac{0.065}{1}=1 + 0.065=1.065$

Then, calculate the exponent: $1\times29 = 29$

So we have $A = 7800\times(1.065)^{29}$

Step3: Calculate $(1.065)^{29}$

Using a calculator, $(1.065)^{29}\approx5.316$ (more precise value can be obtained using a calculator: $1.065^{29}\approx e^{29\ln(1.065)}\approx e^{29\times0.0629}\approx e^{1.824}\approx6.20$ (Wait, let's calculate it more accurately. Using a calculator, $1.065^{29}$:

We can calculate step - by - step or use a calculator function. Let's use a calculator: $1.065^{29}\approx6.2008$

Step4: Calculate $A$

$A=7800\times6.2008$

$A = 7800\times6.2008=48366.24$

Answer:

$\$48366.24$