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?

Step-by-step explanation:

we know that

The formula to calculate continuously compounded interest is equal to

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  

substitute in the formula above

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

∞ this is your sign

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}