Common factor of 10 and 20
1 answer:
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We can find the number of extra clubs in the deck from the given probability, since the probability of drawing a diamond and a club is
but this would imply that
, which suggests we're taking 4 clubs out of the original deck and contradicts the problem statement that there are "too many clubs".
Perhaps the question is providing the probability of drawing a diamond, THEN a club, or vice versa. In that case, we have
and solving gives
, which makes more sense. Then the number of extra spades in the deck must be 2.
but this would imply that
Perhaps the question is providing the probability of drawing a diamond, THEN a club, or vice versa. In that case, we have
and solving gives
Answer:
x = -75
Step-by-step explanation: