Using the chart ‘Amount of Stat Compound Interest from previous reading section ‘More Compound Interest,

Using the chart "Amount of Stat Compound Interest from previous reading section "More Compound Interest, find the total amount and amount of interest paid in the following compound interest problems.
$3.000 at 8% for 5 years
Compounding Total Amount Interest Amount
annually
$
$
semiannually
Quarterly
$

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This Post Has 5 Comments

  1. [tex]PV = C* \frac{1- (1+r/n)^{-nt} }{r/n}[/tex]

    Where, PV = Present vale, C = Quarterly payments =$52,000, r = Annual interest rate = 8% = 0.08, n = Number of  payments per year = 4, t= time of payment in years = 5

    Substituting,
    [tex]PV = 52000* \frac{1- (1+0.08/4)^{-4*5} }{0.08/4} = $85,027.53[/tex]

    The closest answer is $85,027.44

  2. 🙂
    Use the formula of the present value of annuity ordinary
    Pv=pmt [(1-(1+r/k)^(-kn))÷(r/k)]
    Pv=5,200×((1−(1+0.08÷4)^(−4
    ×5))÷(0.08÷4))=85,027.45answer

    Hope it helps

  3. If P = Periodic payments done quarterly (that is, four time in a year) = $5,200, n = number of years (5), R = Annual interest rate = 8% = 0.08, and A = Amount to be set aside today, then

    A = P [1-(1+R/4)^4n]/(R/4) = 5200 [1-(1+0.08/4)^4*5]/(0.08/4) = $85,027.45

    Therefore, Valdez construction should set aside $85,027.44 now to satisfy the capital requirement.

  4. Step-by-step explanation:

    the formula used for any interest other then continuously is

    A=P(1+r/n)^nt

    so... r=rate in decimal form ONLY n=how often interest is paid t=time passed

    A=3000(1+.08/1)^(1)(5) = annually

    A=3000(1+.08/2)^(2)(5)

    A=3000(1+.08/4)^(4)(5)

    use you calculator to get the answer

    it is good practice to use your calculator now because soon you will be using it for population questions and those need to be entered very carefully.

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