# 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.

## No, I do not, unfortunately. Does the 0.03 go into the r?

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

= 1000 e^.3

= 1000 (1.349858..)

= $1349.86

How does your teacher expect you to do these calculations?

## With a calculator, I just can't afford one. Thank you very much I appreciate the help!

## 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.