Asked by sandra

I can not solve this problem it's kind of hard for me can anyone please help me out. Thanks

1.If inflation is 6% a year compounded annually, what will it cost in 21 years to buy a house currently valued at $230,000? Round to the nearest cent.

Answered by Marth
Use the equation A = Pe^(rt), where A is the final value, P is the initial value, r is the rate, and t is time.

The time is 21 years, the rate is .06% per year, and the initial value is $230,000.

Therefore, we have
A = 230,000*e^(.06*21)
A = $810,846.94
Answered by Alyssa
I don't know...That doesn't look correct. I got this:

A = 230,000*e^(.06*21)
= 289,800
NOW we add that to the 230,000 to get

A = 519,800???? I think this makes more sense? Any thoughts? ThankS! :)
Answered by Marth

230,000*e^(.06*21) does not equal 289,800. You forgot about the e.
Answered by sandra
Well you see that is what I did and but still don't come up with the answer these are my answer options.

A)$427,867.75
B)$737,641.16
C)$828,813.61
D)$781,899.63
I even did this since there are 12mths in a year I mult. 12 by 21 which I got 252. So what is the answer though????
Answered by Reiny
Just use the compound interest formula,
Amount = princ(1+i)^n

amount = 230000(1.06)^21
= 781899.63
Answered by Reiny
Marth's formula is used for "continuous" rate of interest.
Answered by sandra
So what you did was mult. 230,000 by 1.06 right which I got 368000. But from there what do I do did you mult. 368000 by 21? I'm lost in that step right there.
Answered by Reiny
no,
follow the order of operation, the power has to be done first

do (1.06)^21 on your calculator to get
3.3995636 , then multiply by 230000
Answered by sandra
Yeah I already figured it out thank you. I was wondering how you did it solved the problem I was getting fusterated b/c I tried it so many ways and I just wouldn't come up with an answer, I was way off the mult. choice. Once again thanks.
Answered by femstar
1. Find the amount of time to the nearest day that it would take for a deposit of $1000 to grow to $500,000 at 10% compounded continuously. The equation for continuous compounding is given below:
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