£4000 is invested at 1.5% compound interest.
Show that the value of the investment after 2 years is £4120.90.
Thanks to anyone who can help.
4000*1.015^2 = 4120.90
I have literally just had to do the same ques no...thanks Steve!
To calculate the value of the investment after 2 years with compound interest, we can use the formula:
A = P(1 + r/n)^(nt)
Where:
A = the final amount
P = the principal amount (initial investment)
r = the annual interest rate (expressed as a decimal)
n = the number of times that interest is compounded per year
t = the number of years
In this case, the principal amount (P) is £4000, the interest rate (r) is 1.5% (or 0.015 as a decimal), the compounding frequency (n) is not specified, so let's assume it is compounded annually (n = 1), and we want to calculate the value after 2 years (t = 2).
So, substituting the given values into the formula:
A = £4000(1 + 0.015/1)^(1*2)
= £4000(1.015)^2
= £4000(1.030225)
= £4120.90 (rounded to two decimal places)
Therefore, the value of the investment after 2 years is £4120.90.
Interesting theory Steve, but is that really the answer? I have a feeling instead of 4000*1.015^2 = 4120.90, it will be 4000 /1.015^2=3882.64699, then you times this by 2, and then divide by 1.8821, to get the answer of
4125.86684023.
@steve, how?? also what does ^ mean?? explainnnn
£4000 is invested at 1.5 compound intetest.
Show that the value of the investment after 2 years is £4120.90