Alg_Complete

# 10000 500 e0195 t 2007 paul dawkins 311

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Unformatted text preview: interest is compounded monthly. [Solution] (c) interest is compounded continuously. [Solution] Solution Before getting into each part let’s identify the quantities that we will need in all the parts and won’t change. P = 100, 000 r= 7.5 = 0.075 100 t= 54 = 4.5 12 Remember that interest rates must be decimals for these computations and t must be in years! Now, let’s work the problems. (a) Interest is compounded quarterly. In this part the interest is compounded quarterly and that means it is compounded 4 times a year. After 54 months we then have, æ 0.075 ö A = 100000 ç1 + ÷ 4ø è ( 4)( 4.5) = 100000 (1.01875 ) 18 = 100000 (1.39706686207 ) = 139706.686207 = \$139, 706.69 Notice the amount of decimal places used here. We didn’t do any rounding until the very last © 2007 Paul Dawkins 308 http://tutorial.math.lamar.edu/terms.aspx College Algebra step. It is important to not do too much rounding in intermediate steps with these problems. [Return to Problems] (b) Interest is compounded monthly. Here...
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## This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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