
Mark M. answered 01/28/15
Tutor
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Mathematics Teacher - NCLB Highly Qualified
Continuously compounded interest
A = Pert, where P = principal, r = rate, t = time, A = amount
1000000 = Pe0.0595(45)
1000000 = Pe2.6775
ln 1000000 = ln Pe2.6775
ln 1000000 = ln P + ln e2.6775
13.8 ≈ ln P + 2.6775 ln e
13.8 ≈ ln P + 2.6775
11.12 ≈ ln P
67507.90 ≈ P
For option one, $67,507.90 needs to be invested.
Option B
1000000 = p(1 + r/n)nt, where p = principal, r = rate, n = number of calculations per year, t = years
1000000 = p(1 + 0.059)45
1000000 = p 1.05945
1000000 ≈ 13.19p
75815 ≈ p
Option B needs to invest $75,815.

Mark M.
A = p(1 + r/n)t , where p = principal, r = rate, n = number of calculations a year, t = number of years
1000000 = p(1 + 0.059)45
1000000 = p(1.0595)45
1000000 ≈ p(13.475)
74211.50 ≈ p
$74,211.50 must be invested in option two.
Report
01/28/15
Carolina R.
01/28/15