Evaluate the expression 0.03^3
1 answer:
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Given:
Desmond deposits $ 50 monthly.
Yearly he deposits = $50×12 = $ 600
Rate of interest compounded monthly = 4.7%
To find the amount he will receive after 10 years and the rate of change the value of his account after 10 years.
Formula
where,
A be the final amount
P be the principal
r be the rate of interest
t be the time and
n be the number of times the interest is compounded.
Now,
Taking,
P = 600, r = 4.7, n = 12, t = 10 we get,
or,
or,
Now,
At starting he has $ 600
At the end of 10 years he will be having $ 959.1
So,
The amount of change in his account = $ (959.1-600) = $ 359.1
Therefore the rate of change =
= 59.85%
Hence,
a) His account will contain $ 959.1 after 10 years.
b) The rate of change in his account is 59.85% after 10 years.