With 4 jacks in the deck of 52, there is a 4/52 = 1/13 probability of drawing 1 jack.
With 13 clubs in the deck, there is a 13/52 = 1/4 probability of drawing 1 card of clubs.
1 of the cards in the deck is both a jack and of suit of clubs, which has a 1/52 probability of being drawn.
P(club OR jack) = P(club) + P(jack) - P(club AND jack) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13
So the answer is B.
Answer:
The zeros are x=0,3,-2
There is a multiplicity of 1 for all of them.
Step-by-step explanation:
The control group and the experimental groups are not comparable. They are selected from different districts.
The answer is 10080 you can use a calculator for this
