SOLUTION: Example: P=P0[1+r]^t
what is:
in 2009 is p1=60,943,912[1+0.9934]^(1)
in 2010 is p2=60943,912[1+0.9934]^2
I need the same answer for 2011 and 2012 please explain and show
Algebra.Com
Question 157702: Example: P=P0[1+r]^t
what is:
in 2009 is p1=60,943,912[1+0.9934]^(1)
in 2010 is p2=60943,912[1+0.9934]^2
I need the same answer for 2011 and 2012 please explain and show me how to figure this our.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'll do the first two to get you started
p1=60,943,912[1+0.9934]^(1) = 60,943,912[1.9934]^(1) = 60,943,912[1.9934] = 121,485,594.1808
p2=60,943,912[1+0.9934]^(2) = 60,943,912[1.9934]^(2) = 60,943,912[3.97364356]= 242,169,383.44
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