If \( \$ 700 \) are deposited into an account with a \( 10.5 \% \) interest rate, compounded monthly, what is the balance after 16 years? \[ \mathrm{F}=\$[?] \] \( \mathrm{F}=\mathrm{P}\left(1+\frac{r}{n}\right)^{\mathrm{nt}} \) Round to the nearest cent.

To find the future balance \( F \) of an account with a principal \( P \) of \$700, an interest rate \( r \) of 10.5% (or 0.105), compounded monthly (\( n = 12 \)), over \( t \) of 16 years, we can use the formula: \[ F = P\left(1+\frac{r}{n}\right)^{nt} \] Plug in the values: \[ F = 700\left(1+\frac{0.105}{12}\right)^{12 \times 16} \] Calculate \( \frac{0.105}{12} \): \[ \frac{0.105}{12} \approx 0.00875 \] Now substitute and calculate \( nt = 12 \times 16 = 192 \): \[ F = 700\left(1 + 0.00875\right)^{192} \] Calculating \( 1 + 0.00875 = 1.00875 \): \[ F = 700(1.00875)^{192} \] Now compute \( (1.00875)^{192} \): \[ (1.00875)^{192} \approx 5.043267772 \] Thus: \[ F \approx 700 \times 5.043267772 \approx 3530.287443 \] Rounding to the nearest cent: \[ F \approx \$3530.29 \] So, the balance after 16 years will be: \[ \mathrm{F}=\$3530.29 \]