Solutions to Extra Problems 3 a

Solutions to Extra Problems 3 a - ACTSC 231 Solutions Extra...

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Unformatted text preview: ACTSC 231 Solutions Extra Problems 3 a 1. (a) 300a6j = 300 1 (b) 300a6j = 300 1 (c) 300a6j = 300 1 ln 1 + i(m) m m v6 v6 = 300 1 = 300 1 e :04(6) :04 (1:04) ln 1:04 1 6 = 1600: 291 = 1603: 888 3 Note 4(6) = ln (1 + i) Note = v6 = 300 (1+ :04 ) 4 4 ln(1+ :04 ) 4 = 1601: 207 1 (d) 300a6j = 300 1 (e) a 1 v6 = 300 1 (1 :04)6 ln(1 :04) 1 2 = 1596: 508 5 Note = =e 1 2 [ln(1+:01t) ln (1 ln(1)] d) = (t) = e 1 2 eln(1+:01t) = (1 + :01t) Rt :005 ds 0 1+:01s =e 1 2 ln (1+:01s)jt ds 0 300a6j = = = 300 300 300 Z 6 a 6 1 (t) dt 1 2 1 2 (1 + :01t) 2 :01 0 1 2 = 300 (1:06) 2 1 :01 = 1773: 780 8 Z 0 (1 + :01t) dt 6 0 2. a 1 (t) = e 0 s ds = e 0 ::005=(1+:0025s)ds = e : 2 ln(1+:0025s)j0 = e 2 (1 + :0025t) R 10 R 10 2 (400 + t) (1 + :0025t) dt = 400 0 (1 + :0025t) (1 + :0025t) 0R 10 10 1 400 400 0 (1 + :0025t) dt = :0025 ln (1 + :0025t) 0 = 160000 ln (1:025) = 3950:82 3. Let A denote the P.V. Rt Rt t 2 ln(1+:0025) = 2 dt = 1 A= = = 5v + 5v 2 + 5 (1:02) v 3 + 5 (1:02) v 4 + 5 (1:02) v 5 + 5 (1:02) v 6 + 5 (1:02) v 7 + 5 (1:02) v 8 + 5 v + v 2 + 5 (1:02) v 2 v + v 2 + 5 (1:02) v 4 v + v 2 + 5 (1:02) v 6 v + v 2 + 0 1 !2 !3 1:02 1:02 1:02 A 5 v + v 2 @1 + + + 2+ 2 2 (1:05) (1:05) (1:05) • 5 v + v 2 a1jj , where vj = 1 2 3 2 2 3 3 = = = = 5 (1:05) 5 (1:05) + (1:05) + (1:05) 2 1 2 1 1:02 (1:05) = )j= 2= 1+j 1:02 (1:05) 1 :1 vj 1 1 02 1 (1::05)2 2 1 = 8:0882353% 124: 242 42 2 ...
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Solutions to Extra Problems 3 a - ACTSC 231 Solutions Extra...

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