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Svet_ta [14]
3 years ago
5

PLEASE HELP!!!!! In a card game, Alfred picks a card from a deck, notes the color, and then puts it back into the deck. Later, V

icky picks a card from
the same deck
In this experiment, the event that Alfred picks a card from the deck is independent of the event that Vicky picks a card from the
same deck. What can be changed in this experiment so that these events become dependent?
A. It is not possible to make the second event dependent of the first event.
B.
Alfred should not put the card back into the deck after picking a card.
C.
Vicky should pick a card from another deck.
D.
The deck should be split into two piles, and Alfred should pick from one pile, while Vicky picks from the other.
Mathematics
1 answer:
leva [86]3 years ago
3 0
I think it’s d but I’m not sure
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5. 
First plug the value of x into the equation
 2(-2y-12)+y=9
then solve for y
2(-2y)+2(-12)+y=9 \\ -4y+(-24)+y=9 \\ -4y-24+y=9 \\ -3y-24=9 \\ -3y=33 \\ \frac{-3y}{-3}= \frac{33}{-3} \\ y = -11

6. Do the same steps 
5(-y-1)+y=-13 \\ 5(-y)+5(-1)+y=13 \\ -5y+(-5)+y=13 \\ -5y-5+y=13 \\ -4y-5=13 \\ -4y=18 \\ \frac{-4y}{-4}= \frac{18}{-4} \\ y=-4.5 

8 0
3 years ago
It is common to leave a 10% tip when the service was just mediocre. At one such occasion, a patron left a 10% tip on a $23.45 bi
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You can win concert tickets from a radio station if you are the first person to call when the song of the day is played, or if y
Anit [1.1K]

The song of the day is announced at a random time between 7:00 and 7:30, so you have a 30-minutes span of time where the song could be announced. If you start listening to the radio 20 minutes in, you're at 2/3 of the time. So, there is a 2/3 chance that the song has already been chosen, and a 1/3 chance that it will be announced in the remaining 10 minutes.

Similarly, the trivia question is asked at a random time between 7:15 and 7:45 so there is again a 30-minutes span of time. But this time, we're only 5 minutes late, so there is a 5/30=1/6 chance that the question was already asked, and a 25/30=5/6 chance that the question will be asked in the remaining 25 minutes.

So, there are four scenarios:

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\dfrac{1}{3}\cdot\dfrac{1}{6} =\dfrac{1}{18}

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Your result is thus

\dfrac{10}{18}+\dfrac{1}{18} = \dfrac{11}{18}

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