ordinary annuity formula

[tpowell]

I need help with problem can't seem to get the correct answer.



Use a calculator to evaluate an ordinary annuity formulaA = m

1 + r n nt − 1

r n

for m, r, and t (respectively). Assume monthly payments. (Round your answer to the nearest cent.) $150; 4%; 40 yr
A = $1

 

[Deleted member 4993]

I need help with problem can't seem to get the correct answer.



Use a calculator to evaluate an ordinary annuity formulaA = m

1 + r n nt − 1

r n

for m, r, and t (respectively). Assume monthly payments. (Round your answer to the nearest cent.) $150; 4%; 40 yr
A = $1

Your post is undecipherable....

In addition:

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[Deleted member 4993]

And try google...

That's too many tries - more than a zillion - any short-cuts??

 

[soroban]

Hello, tpowell!

Did you ever PREVIEW your work?
Did you see what it looked like?

I need help with problem can't seem to get the correct answer.

Use a calculator to evaluate an ordinary annuity formula A = m

1 + r n nt − 1

r n

for m, r, and t (respectively). Assume monthly payments.
(Round your answer to the nearest cent.)
$150; 4%; 40 yr
A = $1 .??

Can't read anything you wrote . . .

I will assume you want the Annuity Formula: .\(\displaystyle A \;=\;D\,\dfrac{(1+i)^n-1}{i}\)

. . \(\displaystyle \text{where: }\:\begin{Bmatrix}A &=& \text{final amount} \\ D &=& \text{periodic deposit} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}\)


We are given: .\(\displaystyle \begin{Bmatrix}D &=& \$150 \\ i &=& \frac{4\%}{12} &=& \frac{1}{300} \\ n &=& 40\cdot12 &=& 480 \end{Bmatrix}\)


Therefore: .\(\displaystyle A \;=\;150\,\dfrac{(1+\frac{1}{300})^{480}-1}{\frac{1}{300}} \;\approx\;\$177,\!294.20\)