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You have borrowed $28,000 at an interest rate of 12% compounded annually. Equal payments will be made over a four-year period, with each payment made at the end of the corresponding year. What is the amount of the annual payment foe the second year?

Respuesta :

Answer:

[tex] A = 28000 [\frac{0.12 (1.12)^4}{(1.12)^4 -1}][/tex]

[tex] A = 28000 [\frac{0.12*1.574}{1.574-1}][/tex]

[tex] A=28000*0.3292 = 9218.564[/tex]

So then the annual pay would be $ 9218.564 for this case

Explanation:

For this question we can use the Equivalent annual value (A) given by the following expression:

[tex] A = PV [\frac{i (1+i)^t}{(1+i)^t -1}][/tex]

Where [tex] PV = 28000[/tex] represent the pesent value

[tex] i = 0.12[/tex] since the rate is yearly

[tex] t = 4[/tex] since we have 4 years to pay

So then we have everything to replace and we got:

[tex] A = 28000 [\frac{0.12 (1.12)^4}{(1.12)^4 -1}][/tex]

[tex] A = 28000 [\frac{0.12*1.574}{1.574-1}][/tex]

[tex] A=28000*0.3292 = 9218.564[/tex]

So then the annual pay would be $ 9218.564 for this case

And this amount would be paid each year in order to pay all the money after 4 years.

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