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November 27, 2014

November 27, 2014

Posted by **mike** on Monday, March 30, 2009 at 1:16am.

- math -
**Damon**, Monday, March 30, 2009 at 3:35amDo you mean

A(t) = Ao e^-(.0063 t) ????

If so then

1 = 2 e^-(.0063 t)

.5 = e^-(.0063 t)

ln .5 = -(.0063 t)

-.69315 = -(.0063 t)

t = 110.02 days which is about 110 days and one hour

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