Se escriben, cada uno en un papel, los dígitos desde el 1 al 9. Si se eligen al azar 2 papeles ¿Cual es la probabilidad de obten er como diferencia entre los dígitos el número 3?
1 answer:
This can be solved by a contingency table to enumerate all possibilities. 1 2 3 4 5 6 7 8 9 1 - 1 2 3 4 5 6 7 82 1 - 1 2 3 4 5 6 73 2 1 - 1 2 3 4 5 64 3 2 1 - 1 2 3 4 55 4 3 2 1 - 1 2 3 46 5 4 3 2 1 - 1 2 37 6 5 4 3 2 1 - 1 28 7 6 5 4 3 2 1 - 19 8 7 6 5 4 3 2 1 - The diagonal entries are void because we cannot pick two tags with the same number. We count 72 outcomes (excluding diagonal) and 12 of them are threes. ThereforeProbability that the difference is 3 = 12/72 = 1/6
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Midpoint formula =
M = ( X1 + X2) divided by 2 , (Y1 + Y2) divided by 2 = m=(x ,y)
So then, we plug in our numbers .
M= (3+ -9)/2 , (7+3)/2
M= (-3, 5)
Answer:
C) 16
Step-by-step explanation: Please brainliest if possible!
Answer:
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Step-by-step explanation:
4% duh because ne year is 12 months and you get 2% every six months
X>6 or x€[6,+Infiniti] that's what I got