Question 616868
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Hi, 
P(t)=1-e^-0.016t. 
within how many months the probability of failure will be 99.99% OR .9999
 .9999 = 1-e^-0.016t. 
 e^-0.016t = .0001
   -.016t = ln(.0001) 
        t =  ln(.0001)/-.016
        t = 574.65 months