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Sagot :
Let's go through each part of the problem step-by-step:
a. Finding the equation that gives the amount of money in the account after [tex]\( t \)[/tex] years:
The formula for the amount of money [tex]\( A \)[/tex] in an account after [tex]\( t \)[/tex] years, with an initial deposit [tex]\( P \)[/tex], annual interest rate [tex]\( r \)[/tex], and [tex]\( n \)[/tex] compounding periods per year, is given by:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Here:
- [tex]\( P = 1800 \)[/tex] (the initial deposit)
- [tex]\( r = 0.05 \)[/tex] (the annual interest rate of 5%)
- [tex]\( n = 4 \)[/tex] (compounded quarterly)
Substituting these values into the formula, we get:
[tex]\[ A(t) = 1800 \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
So, the equation that gives the amount of money in the account after [tex]\( t \)[/tex] years is:
[tex]\[ A(t) = 1800 \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
b. Finding the amount of money in the account after 8 years:
We use the equation obtained in part (a) and substitute [tex]\( t = 8 \)[/tex]:
[tex]\[ A(8) = 1800 \left(1 + \frac{0.05}{4}\right)^{4 \cdot 8} \][/tex]
Calculating this expression, we obtain:
[tex]\[ A(8) = 2678.63 \][/tex]
Therefore, after 8 years, there will be \[tex]$2678.63 in the account. c. Finding the number of years it will take for the account to contain \$[/tex]3600:
We need to determine [tex]\( t \)[/tex] such that:
[tex]\[ 3600 = 1800 \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
We can rewrite the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ 3600 / 1800 = \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
[tex]\[ 2 = \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
Next, we take the logarithm of both sides of the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ \log(2) = 4t \cdot \log\left(1 + \frac{0.05}{4}\right) \][/tex]
[tex]\[ t = \frac{\log(2)}{4 \cdot \log\left(1 + \frac{0.05}{4}\right)} \][/tex]
Solving this, we obtain:
[tex]\[ t \approx 13.95 \][/tex]
Therefore, it will take approximately 13.95 years for there to be \[tex]$3600 in the account. d. Finding the amount of money in the account after 8 years with continuous compounding: The formula for continuous compounding is given by: \[ A = Pe^{rt} \] Here: - \( P = 1800 \) - \( r = 0.05 \) - \( t = 8 \) Substituting these values into the formula, we get: \[ A = 1800 e^{0.05 \cdot 8} \] Calculating this expression, we obtain: \[ A \approx 2685.28 \] Therefore, with continuous compounding, there would be \$[/tex]2685.28 in the account after 8 years.
a. Finding the equation that gives the amount of money in the account after [tex]\( t \)[/tex] years:
The formula for the amount of money [tex]\( A \)[/tex] in an account after [tex]\( t \)[/tex] years, with an initial deposit [tex]\( P \)[/tex], annual interest rate [tex]\( r \)[/tex], and [tex]\( n \)[/tex] compounding periods per year, is given by:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Here:
- [tex]\( P = 1800 \)[/tex] (the initial deposit)
- [tex]\( r = 0.05 \)[/tex] (the annual interest rate of 5%)
- [tex]\( n = 4 \)[/tex] (compounded quarterly)
Substituting these values into the formula, we get:
[tex]\[ A(t) = 1800 \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
So, the equation that gives the amount of money in the account after [tex]\( t \)[/tex] years is:
[tex]\[ A(t) = 1800 \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
b. Finding the amount of money in the account after 8 years:
We use the equation obtained in part (a) and substitute [tex]\( t = 8 \)[/tex]:
[tex]\[ A(8) = 1800 \left(1 + \frac{0.05}{4}\right)^{4 \cdot 8} \][/tex]
Calculating this expression, we obtain:
[tex]\[ A(8) = 2678.63 \][/tex]
Therefore, after 8 years, there will be \[tex]$2678.63 in the account. c. Finding the number of years it will take for the account to contain \$[/tex]3600:
We need to determine [tex]\( t \)[/tex] such that:
[tex]\[ 3600 = 1800 \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
We can rewrite the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ 3600 / 1800 = \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
[tex]\[ 2 = \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
Next, we take the logarithm of both sides of the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ \log(2) = 4t \cdot \log\left(1 + \frac{0.05}{4}\right) \][/tex]
[tex]\[ t = \frac{\log(2)}{4 \cdot \log\left(1 + \frac{0.05}{4}\right)} \][/tex]
Solving this, we obtain:
[tex]\[ t \approx 13.95 \][/tex]
Therefore, it will take approximately 13.95 years for there to be \[tex]$3600 in the account. d. Finding the amount of money in the account after 8 years with continuous compounding: The formula for continuous compounding is given by: \[ A = Pe^{rt} \] Here: - \( P = 1800 \) - \( r = 0.05 \) - \( t = 8 \) Substituting these values into the formula, we get: \[ A = 1800 e^{0.05 \cdot 8} \] Calculating this expression, we obtain: \[ A \approx 2685.28 \] Therefore, with continuous compounding, there would be \$[/tex]2685.28 in the account after 8 years.
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