Recent content by shfreeman

Why not just simplify your problem (a-b)/2+a into (3a-b)/2 and then use the rules for the error of 3a+b. You don't need to worry about the independency of error a to error a. This way I get (3 Δa + Δb)/2 = 1.4 Or using the other rule [ √(Δa2 + Δb2) ] ( √(9*Δa2 + Δb2) ) / 2...