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If it's the average speed of [a,b] where a is less than b, both are on the x axis (time), and both are in the domain of x then the equation is(f(b)-f(a))/(b-a)where f(b) is the distance when time=b and f(a) is the distance when time=a(?)
All the time
12 = -4/2 * -2 + b 12 = -2 * -2 + b 12 = 4 + b 8 = b
The speed upstream is B - C where B is the speed of the badge in still water and C is speed of the current The speed downstream is B + C. Velocity = Distance/Time : therefore Time = Distance/Velocity. Time for upstream journey = 6/(B - C) Time for downstream journey = 6/(B + C) BUT Total time for journey = 2 = 6/(B - C) + 6/(B + C) = 12B/(B2 - C2) Therefore 2B2 - 2C2 = 12B : However, B = 8kph so substituting gives, 128 - 2C2 = 96 : 2C2 = 32 : C2 = 16 : C = 4 The speed of the current is 4kph.
Let A= getting an ace the first timeand B= getting an ace the secondWe are looking to for the probaliity of getting A and B that is P(A and B)We know P(A and B) = P(A) . P(B|A)= (4/52) . (3/51) = 1/122 = .00452NOTE that P(B|A) is the conditional probability of getting an ace the second time given that you got an ace the first time.