NEED HELP ASAP PLEASE!!
A savings account starts with $600 and pays 5% interest per year, compounded four times per year.
a) A function that models the amount in dollars in the bank account after m years is A(m)=____________?
b) How much will be in the savings account after 3 years? $______?
c) The amount in the account will exceed $1400 after ________ years.
amount = 600(1 + .05/4)^(4m)
b) A = 600(1.0125)^12 = ...
c) set A = 1400
1400 = 600(1.0125)^n, remember n is quarter years.
2.33333 = (1.0125)^n
do you know how to do logs? If so, ..
n = log 2.33333/log 1.0125 = 68.2 quarter years or about 17 years.
notice 17 years ---> n= 68
amount = 600(1.0125)^68 = $1396.41
so you will need about 68 and 1/4 years.
a) To find the function that models the amount in dollars in the bank account after m years, we can use the compound interest formula:
A = P(1 + r/n)^(nt)
Where:
A is the amount in the account after m years,
P is the initial amount in the account,
r is the annual interest rate (expressed as a decimal),
n is the number of times interest is compounded per year,
t is the number of years.
In this case, the initial amount (P) is $600, the annual interest rate (r) is 5% (or 0.05 as a decimal), and it is compounded four times per year (n = 4).
Therefore, the function that models the amount in dollars in the bank account after m years is:
A(m) = 600(1 + 0.05/4)^(4m)
b) To find the amount in the savings account after 3 years, we can substitute m = 3 into the function A(m):
A(3) = 600(1 + 0.05/4)^(4*3)
Now, we can calculate the value:
A(3) = 600(1 + 0.0125)^(12)
= 600(1.0125)^(12)
≈ $663.77
Therefore, the amount in the savings account after 3 years will be approximately $663.77.
c) To find the number of years it takes for the amount in the account to exceed $1400, we can set up the inequality:
A(m) > 1400
Substitute the function A(m) into the inequality:
600(1 + 0.05/4)^(4m) > 1400
Now, we need to solve for m. It's easier to solve the inequality taking the logarithm of both sides:
log(600(1 + 0.05/4)^(4m)) > log(1400)
Simplifying further:
log(600) + 4m*log(1 + 0.05/4) > log(1400)
From here, isolate m to find the number of years it takes for the amount to exceed $1400.