506 MrG Compounded Interest!

3085 days ago by MATH4R2013

#1) P=$1000, r=10%, t=1yr find simple interest I=(1000)*(.10)*(1) I 
       
#2) P=$1000, r=10%, t=1yr find interest compounded twice in one year I1=(1000)*(.10)*(1/2) show(I1) I2=(1050)*(.10)*(1/2) show(I2) show(I1+I2) 
       
#3) P=$1000, r=10%, t=1yr find interest compounded four times in one year I1=(1000)*(.10)*(1/4) show(I1) I2=(1000+I1)*(.10)*(1/4) show(I2) I3=(1000+I1+I2)*(.10)*(1/4) show(I3) I4=(1000+I1+I2+I3)*(.10)*(1/4) show(I4) show(I1+I2+I3+I4) show((I1+I2+I3+I4).n(digits=5)) 
       
show((1000*(1+.1/1)^1).n(digits=6)) show((1000*(1+.1/2)^2).n(digits=6)) show((1000*(1+.1/3)^3).n(digits=6)) show((1000*(1+.1/4)^4).n(digits=6)) 
       
#11) p336 var('P') solve(100==P*(1+.06/12)^24,P)[0].rhs().n(digits=4) 
       
#9) p336 show((100*e^(.1*2.25)).n(digits=5)) 
       
#29a) p336 show((ln(2)/12/ln(1+.08/12)).n(digits=4)) 
       
#29b) p336 show((ln(2)/.08).n(digits=4)) 
       
#21) p336 show((1+.0525/4)^4-1) show(100*((1+.0525/4)^4-1)) show((100*((1+.0525/4)^4-1)).n(digits=3))