To calculate the balance after the withdrawal, we first need to calculate the total amount in the savings account after 3 years of earning interest.
To calculate the total amount in the savings account after 3 years, we use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment/loan, including interest
P = the principal investment amount (initial deposit)
r = annual interest rate (decimal)
n = number of times that interest is compounded per year
t = number of years the money is invested/borrowed for
Given:
P = $1,763.25
r = 4.5% or 0.045
n = 1 (compounded annually)
t = 3 years
A = $1,763.25(1 + 0.045/1)^(1*3)
A = $1,763.25(1 + 0.045)^3
A = $1,763.25(1.045)^3
A = $1,763.25(1.141630125)
A = $2,010.97365313
After 3 years, the total amount in the savings account is approximately $2,010.97.
Next, we subtract the withdrawal amount of $4,360.00:
$2,010.97 - $4,360.00 = -$2,349.03
Therefore, after the withdrawal, the balance in the savings account is approximately -$2,349.03.
Round any decimals to the nearest penny.
Your fixed expenses are $1,763.25/month. You saved 6 months' worth for an emergency fund in a savings account earning a 4.5% APR over 3 years. After 3 years, you withdrew $4,360.00 because of losing your job. What is your balance after the withdrawal?
3 answers
are you sure
Apologies for the miscalculation.
Let's recalculate:
To calculate the total amount in the savings account after 3 years, we use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment/loan, including interest
P = the principal investment amount (initial deposit)
r = annual interest rate (decimal)
n = number of times that interest is compounded per year
t = number of years the money is invested/borrowed for
Given:
P = $1,763.25
r = 4.5% or 0.045
n = 1 (compounded annually)
t = 3 years
A = $1,763.25(1 + 0.045/1)^(1*3)
A = $1,763.25(1 + 0.045)^3
A = $1,763.25(1.045)^3
A = $1,763.25(1.141630125)
A = $2,009.31369188
After 3 years, the total amount in the savings account is approximately $2,009.31.
Next, we subtract the withdrawal amount of $4,360.00:
$2,009.31 - $4,360.00 = -$2,350.69
Therefore, after the withdrawal, the balance in the savings account is approximately -$2,350.69.
Let's recalculate:
To calculate the total amount in the savings account after 3 years, we use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment/loan, including interest
P = the principal investment amount (initial deposit)
r = annual interest rate (decimal)
n = number of times that interest is compounded per year
t = number of years the money is invested/borrowed for
Given:
P = $1,763.25
r = 4.5% or 0.045
n = 1 (compounded annually)
t = 3 years
A = $1,763.25(1 + 0.045/1)^(1*3)
A = $1,763.25(1 + 0.045)^3
A = $1,763.25(1.045)^3
A = $1,763.25(1.141630125)
A = $2,009.31369188
After 3 years, the total amount in the savings account is approximately $2,009.31.
Next, we subtract the withdrawal amount of $4,360.00:
$2,009.31 - $4,360.00 = -$2,350.69
Therefore, after the withdrawal, the balance in the savings account is approximately -$2,350.69.
1 answer