Question

Melanie invested $5,500 in an account paying an interest rate of 6.9% compounded continuously. Assuming no deposits or withdrawals are made, how much money, to the nearest dollar, would be in the account after 11 years?

Answer 1

Answer:

$A=\ 11,748.76$

Step-by-step explanation:

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

we have  

$t=11\ years\\ P=\ 5,500\\ r=6.9\%=6.9/100=0.069$  

substitute in the formula above

$A=5,500(e)^{0.069*11}$  

$A=5,500(e)^{0.759}$  

$A=\ 11,748.76$

Answer 2

the other user didn't round to the nearest dollar..

answer is actually 11749.

Compounded Continuously:

A=Pe^{rt}

A=Pe  

rt  

P=5500\ {35px}r=0.069\ {35px}t=11

P=5500r=0.069t=11

Given values

A=5500e^{0.069(11)}

A=5500e  

0.069(11)

 

Plug in

A=5500e^{0.759}

A=5500e  

0.759

 

Multiply

A=11748.7645718

A=11748.7645718

Use calculator (with e button)  

A≈11748.76

Round to nearest dollar

A=11749