To determine the balance after 10 years when [tex]$900 is deposited into an account with a 4% continuous interest rate, we use the formula for continuous compounding. The formula is given by:
\[ A = P \cdot e^{rt} \]
Where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (\$[/tex]900 in this case).
- [tex]\( r \)[/tex] is the annual interest rate (decimal) (4% or 0.04 in decimal form).
- [tex]\( t \)[/tex] is the time the money is invested or borrowed for, in years (10 years in this case).
- [tex]\( e \)[/tex] is the base of the natural logarithm, approximately equal to 2.71828.
Let's plug in the given values into the formula:
[tex]\[ P = 900 \][/tex]
[tex]\[ r = 0.04 \][/tex]
[tex]\[ t = 10 \][/tex]
[tex]\[ A = 900 \cdot e^{0.04 \cdot 10} \][/tex]
Now, we perform the exponentiation first:
[tex]\[ 0.04 \cdot 10 = 0.4 \][/tex]
Thus, we need to compute [tex]\( e^{0.4} \)[/tex].
[tex]\[ A = 900 \cdot e^{0.4} \][/tex]
Using the value of [tex]\( e^{0.4} \approx 1.49182 \)[/tex]:
[tex]\[ A = 900 \cdot 1.49182 \][/tex]
Finally, performing the multiplication:
[tex]\[ A \approx 1342.64 \][/tex]
Therefore, the balance after 10 years, rounded to the nearest cent, would be:
[tex]\[ \$1342.64 \][/tex]
