Question

Inflation is running 2% per year when you deposit $1000 in an account earning interest of 13% per year compounded annually. In constant dollars, how much money will you have two years from now? (Hint: First calculate the value of your account in two years' time, and then find its present value based on the inflation rate. Round your answer to the nearest cent.)
$ 1

Answer 1

The instructions in your questions tell you how to do it.
Give it a try.

Answer 2

first part?1276.9

Answer 3

final answer?1225.82?

Answer 4

first part:
1000(1.13)^2 = 1276.90

You are right.

Second part:
1276.90(.98)^2 = 1226.33

One step calculation:
1000(.98)^2(1.13)^2 = 12236.33

How did you get 1225.82 ?

Answer 5

I multipied first part times .02 and subtracted that twice

Answer 6

wheres the .98 from?

Answer 7

the answer was wrong

Answer 8

Just like when money increases by 1+i,
when the money decreases you would take 1-i
in this case i = .02
so value(1-.02)^2

when the rate was 13% did you not take 1 + .13 ?

suppose we do it step by step

start with 1000
that loses 2% so amount left = .98(1000) = 980
that gains 13%, so 980(1.13) = 1107.40

that loses 2% , so amount left = 1107.4(.98) = 1085.25
which gains 13% , 1085.25(1.13) = 1226.33

Answer 9

you said you multiplied your answer by 2% and then subtracted it twice ...

let's try that

2% of 1276.90 = 25.54
leaving 1251.36

2% of 1251.36 = 25.03
subtracting that leaves 1226.33

Well, well, what do you say now?

Answer 10

i put that answe ina and it was wrong