Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Tanisha is saving money and plans on making quarterly contributions into an account earning an annual interest rate of 
6.7% compounded quarterly. If Tanisha would like to end up with 
$96,000 after 11 years, how much does she need to contribute to the account every quarter, to the nearest dollar? Use the following formula to determine your answer.

A=d(((1+i)^(n)-1)/(i))

A= the future value of the account after 
n periods

d= the amount invested at the end of each period

i= the interest rate per period

n= the number of periods
Answer:

Tanisha is saving money and plans on making quarterly contributions into an account earning an annual interest rate of 6.7% 6.7 \% compounded quarterly. If Tanisha would like to end up with $96,000 \$ 96,000 after 1111 years, how much does she need to contribute to the account every quarter, to the nearest dollar? Use the following formula to determine your answer.\newlineA=d((1+i)n1i) A=d\left(\frac{(1+i)^{n}-1}{i}\right) \newlineA= A= the future value of the account after n n periods\newlined= d= the amount invested at the end of each period\newlinei= i= the interest rate per period\newlinen= n= the number of periods\newlineAnswer:

Full solution

Q. Tanisha is saving money and plans on making quarterly contributions into an account earning an annual interest rate of 6.7% 6.7 \% compounded quarterly. If Tanisha would like to end up with $96,000 \$ 96,000 after 1111 years, how much does she need to contribute to the account every quarter, to the nearest dollar? Use the following formula to determine your answer.\newlineA=d((1+i)n1i) A=d\left(\frac{(1+i)^{n}-1}{i}\right) \newlineA= A= the future value of the account after n n periods\newlined= d= the amount invested at the end of each period\newlinei= i= the interest rate per period\newlinen= n= the number of periods\newlineAnswer:
  1. Identify Given Values: Identify the given values from the problem.\newlineAA (future value of the account) = $96,000\$96,000\newlineii (interest rate per period) = 6.7%6.7\% annual interest rate compounded quarterly, which means i=6.7%4=0.0674i = \frac{6.7\%}{4} = \frac{0.067}{4} per quarter\newlinenn (number of periods) = 1111 years 4* 4 quarters/year = 4444 quarters\newlineNow, we can use these values to find dd (the amount invested at the end of each period).
  2. Convert Interest Rate: Convert the annual interest rate to a quarterly interest rate. \newlinei=0.0674=0.01675i = \frac{0.067}{4} = 0.01675\newlineThis is the interest rate per quarter.
  3. Calculate Number of Periods: Calculate the number of periods nn.n=11 years×4 quarters/year=44 quartersn = 11 \text{ years} \times 4 \text{ quarters/year} = 44 \text{ quarters}
  4. Calculate (1+i)n(1+i)^n: Use the formula A=d((1+i)n1i)A = d\left(\frac{(1+i)^{n}-1}{i}\right) to find dd. First, calculate (1+i)n(1+i)^n. (1+i)n=(1+0.01675)44(1+i)^n = (1 + 0.01675)^{44}
  5. Calculate (1+i)n(1+i)^n: Calculate (1+i)n(1+i)^n using a calculator.\newline(1+0.01675)442.03009(1 + 0.01675)^{44} \approx 2.03009
  6. Calculate ((1+i)n1)((1+i)^n - 1): Calculate ((1+i)n1)((1+i)^n - 1).\newline((1+i)n1)2.030091=1.03009((1+i)^n - 1) \approx 2.03009 - 1 = 1.03009
  7. Calculate Denominator: Calculate the denominator of the formula, which is ii. i=0.01675i = 0.01675
  8. Calculate Right Side: Calculate the entire right side of the formula, which is ((1+i)n1)/i\left(\left(1+i\right)^{n}-1\right)/i. ((1+i)n1)/i1.030090.0167561.5493\left(\left(1+i\right)^{n}-1\right)/i \approx \frac{1.03009}{0.01675} \approx 61.5493
  9. Solve for d: Solve for d using the formula A=d((1+i)n1i)A = d\left(\frac{(1+i)^{n}-1}{i}\right).
    $96,000=d×61.5493\$96,000 = d \times 61.5493
    Now, divide both sides by 61.549361.5493 to find dd.
    d$96,00061.5493d \approx \frac{\$96,000}{61.5493}
  10. Calculate d: Calculate d using a calculator.\newlined$(96,000)/61.5493$(1559.68)d \approx \$(96,000) / 61.5493 \approx \$(1559.68)\newlineSince we need to round to the nearest dollar, d$(1560)d \approx \$(1560).

More problems from Compound interest