assgn1-soln__updated_

assgn1-soln__updated_ - re-t/ Afro - &@) = /4'eD l...

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re-t- 'A/ Afro -- &@) - //4'eD - v\'-- / - T.@ = llo l'ozzVzJ lA- l+1 | t'zoo fr(ES-Joo ( 20G):3 ( l)trt :) 1 L /oo ffir=/Lo I to a&) =/t --\ . .=/ a6)=-f ii-:(? , *l=1,,t ly;pw , ]b 'o -r:6'o . b.di tL, aqw Lt!r+e,4 v44ae- 6,t f /l4e V _&ALlpo ">,Aled r\f ^\ t 9l/, vt\-{)_ Jtt-, l\/(bt, 77t-" -l l/boa '':--'- -,rg/ I
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Q3 . Label time t = 0 for March 1, 1998. At time 2 (i.e., March 1, 2000), Robin’s balance is $6 , 300(1 + 5%) 2 = $6 , 945 . 75. After the withdrawal of $2,000 on March 1, 2000, the balance becomes $[6,945.75-2,000]=$4,945.75. Let t 1 be the time length needed for $4,945,75 to grow to $5,000 at an annual interest rate of 3.5%. Then, we have the equation 4 . 945 . 75(1 + 3 . 5%) t 1 = 5 , 000 . Solving the above equation, we obtain t 1 = 0 . 317117. This implies that the balance after 2.317117 years (since t=0), is $5,000. Therefore, on March 1, 2005 (i.e., t=7), right before the deposit of $1,000, the balance will be $(5 , 000)(1 + 5%) 7 - 2 . 317117 = $6 , 283 . 43 . On March 1, 2010 (t=12), the balance is therefore $(6 , 283 . 43 + 1 , 000)(1 + 5%) 12 - 7 = $9 , 295 . 71.
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g+ +& 4z (|ry, 4*,u , - tlo [rq- l+f t+t ttt tt- t"6 D'Ua-?"1 13 tvw 't1 rro "d- '/t*-- '-Q
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G +,il +!* -ufue, +t ^Lrue ;-;;, i" ,o*ido'r : i r*v=io6+ grtz>6t,otI+{te)(,01)t+(z+) (/,olf +', r ' _ . r '',? + Q+e)(/,,1f+(B+)(,01 ;*tlttoS (/,o?)?,,,, (4 ,l/ rt\-i - " I rl --l .t | -tt /lntiup\iy b"!A siier +tk aLwe_ efw(c'r- w;Lft lof t"e /na tvu I I
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assgn1-soln__updated_ - re-t/ Afro - &@) = /4'eD l...

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