motet
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Proof: P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n} (Bayes theorem) =P{T>n+m}/P{T>n} =((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m (1-p)^m = {T>m} So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.
divided by what? m will be equals to t divided by v
distance = (rate) x (time), so time = (distance)/(rate).If the rate = 30 m/s & distance = 165 m, then: t = 165 m / (30 m/s) = 5.5 seconds
5m-t/7
B+T+M We need more info otherwise it is just equal to itself. What does the rest of the problem say?