Faith invested \$1,100 in an account paying an interest rate of 5.1\% compounded continuously. Assuming no deposits or withdrawals are made, how much money, to the nearest ten dollars, would be in the account after 18 years?

Compounded Continuously:

A=Pe^{rt}

A=Pe

rt

P=1100\hspace{35px}r=0.051\hspace{35px}t=18

P=1100r=0.051t=18

Given values

A=1100e^{0.051(18)}

A=1100e

0.051(18)

2,750

Explanation:

Plug in

A=1100e^{0.918}

A=1100e

0.918

Multiply

A=2754.70450683

A=2754.70450683

Use calculator (with e button)

A\approx 2750

A≈2750

Round to nearest ten dollars

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answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*

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step-by-step explanation: