Respuesta :
Answer:
This isn't an answer but it might help
for exponential decay use a(1-r)^t
exponential growth uses a(1+r)^t
compound interest uses p(1+r/n)^(t x n)
and for interest compounded continuously use Pe^(rt)
a= initial amount, r= rate(%), t= time, P= principle (practically the same as a), and n= number of times compounded
for the different # of times compounded it goes like this: yearly n=1, daily n=365, monthly n=12, weekly n=52, quarterly n=4, semi-annually n=2, and bi-monthly (this is the tricky one) can be either n=6 or n= 24.
also as long as you have a calculator that can do semi-complex things like a Ti 30xs multiview or a ti 36x pro, I have both of those and they can do it as long as you input the formula, hope all this advice helps because I saw you post many of the same types of question. good luck.
Step-by-step explanation:
Answer: she would need $123775.5 in her account when she retires.
Step-by-step explanation:
We would apply the formula for determining future value involving constant deposits at constant intervals. It is expressed as
S = R[{(1 + r)^n - 1)}/r][1 + r]
Where
S represents the future value of the investment.
R represents the regular payments made(could be weekly, monthly)
r = represents interest rate/number of payment intervals
n represents the total number of payments made.
From the information given,
Since she would be taking $1500 four times in a year, then
R = 1500
r = 0.04/4 = 0.01
n = 3 × 20 = 60 times in 20 years
Therefore,
S = 1500[{(1 + 0.01)^60 - 1)}/0.01][1 + 0.01]
S = 1500[{(1.01)^60 - 1)}/0.01][1.01]
S = 1500[{(1.817 - 1)}/0.01][1.01]
S = 1500[0.817/0.01][1.01]
S = 1500[81.7][1.01]
S = 1500 × 82.517
S = 123775.5