A student borrows $10 000 at an interest rate of 4% per annum, compounded monthly. a) If the student pays the loan off in 10 years, how much does the student pay in interest? # of compounding periods/year=12 total # of compounding periods = 12 x 10 = 120 Interest rate/compounding period = 4/12 = .33 Formula = 10,000 (1+1/300)^ ^ 120 or 1/3-in calc 1/300 = 14908 = 4,908 in interest was paid in interest (10,000 b) If the entire loan costs the student $20 000 to pay back, how many years was the loan for? 4908 20 000 10.000(1+1/3)^12N N=208.29 10 10,000 2) = (1+1/3+)^/QN 10g (2) = Log (1+1/3)²N 109 (2) 2210109 (1.3033) - -17.54 years 208.29 12 = 17.35 years.

A student borrows $10 000 at an interest rate of 4% per annum, compounded monthly. a) If the student pays the loan off in 10 years, how much does the student pay in interest? # of compounding periods/year=12 total # of compounding periods = 12 x 10 = 120 Interest rate/compounding period = 4/12 = .33 Formula = 10,000 (1+1/300)^ ^ 120 or 1/3-in calc 1/300 = 14908 = 4,908 in interest was paid in interest (10,000 b) If the entire loan costs the student $20 000 to pay back, how many years was the loan for? 4908 20 000 10.000(1+1/3)^12N N=208.29 10 10,000 2) = (1+1/3+)^/QN 10g (2) = Log (1+1/3)²N 109 (2) 2210109 (1.3033) - -17.54 years 208.29 12 = 17.35 years.

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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