# Suppose you deposit \$1000 in an account paying 3% annual intrest, compounded continuously. Use A=Pe^rt to find the balance after 10 years

F \$20,085.54
G \$1300
H \$1349.86
J \$1068.65

## You are given the formula AND all the data to sub into that formula.

Do you have a scientific calculator??
Took me about 9 seconds to finish the calculations.
Give it a try.

## amount = 1000 e^(.03(10))

= 1000 e^.3
= 1000 (1.349858..)

= \$1349.86

How does your teacher expect you to do these calculations?

## To find the balance after 10 years when an account pays 3% annual interest compounded continuously, we can use the formula A = Pe^rt, where:

A = the ending balance
P = the initial principal (amount deposited)
e = the mathematical constant approximately equal to 2.71828
r = the annual interest rate (in decimal form)
t = the time period (in years)

In this case, we have:
P = \$1000
r = 3% = 0.03 (converted to decimal form)
t = 10 years

Substituting these values into the formula, we get:
A = 1000 * e^(0.03 * 10)

Simplifying the exponent, we have:
A = 1000 * e^0.3

Calculating e^0.3, we find:
A ≈ 1000 * 1.3498588

Multiplying the values, we get:
A ≈ 1349.86

Therefore, the balance after 10 years is approximately \$1349.86.

Hence, the correct answer is option H: \$1349.86.