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Alenkasestr [34]
3 years ago
8

Please help!! Picture included!

Mathematics
1 answer:
elena-s [515]3 years ago
8 0

Answer: the answer is c

Step-by-step explanation:brainlist \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]

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b) How many ways can you deal cards (from a deck of 52) to 4 people when each player gets 7 cards. 2 hidden and 5 visible. Assum
Mariana [72]

Answer:

{52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

Step-by-step explanation:

The dealing of the cards can be seen in the following way:

We first have to choose which 7 cards are going to be dealt to the 1st player. So we have to pick 7 cards out of the 52 available cards. This can be done in {52 \choose 7} ways. Now that we have chosen which 7 cards are going to be dealt to the 1st player, we have to choose which 2 of them are going to be the hidden ones. So we have to pick 2 cards out of the 7 cards to be the hidden ones. This can be done in {7 \choose 2} ways. At this point we now know which cards are being dealt to the 1st player, and which ones are hidden for him. Then we have to choose which 7 cards to deal to the 2nd player, out of the remaining ones. So we have to pick 7 out of 52-7=45. (since 7 have been already dealt to the 1st player). This can be done in {45 \choose 7}. Then again, we have to pick which 2 are going to be the hidden cards for this 2nd player. So we have to pick 2 out of the 7. This can be done in {7 \choose 2} ways. Then we continue with the 3rd player. We have to choose 7 cards out of the remaining ones, which at this point are 52-7-7=38. This can be done in {38 \choose 7}. And again, we have to choose which ones are the hidden ones, which can be chosen in {7 \choose 2} ways. Finally, for the last player, we choose 7 out of the remaining cards, which are 52-7-7-7=31. This can be done in {31 \choose 7} ways. And choosing which ones are the hidden ones for this player can be done in {7 \choose 2} ways. At the end, we should multiply all our available choices on each step, to get the total choices or total ways to deal the cards to our 4 players (since dealing the cards is a process of several steps with many choices on each step).

{52 \choose 7}{7 \choose 2}{45 \choose 7}{7 \choose 2}{38 \choose 7}{7 \choose 2}{31 \choose 7}{7 \choose 2}={52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

7 0
3 years ago
12c - 4 = 14c - 10<br><br> c = ?<br><br> !!!
Sergio [31]
The answer to this question is c=3

3 0
3 years ago
Plz help i will mark brainlist
iragen [17]
Y=x+4
3x+Y=-8

3x+x+4=-8
X=-3. Substitute the value of x


Y=-3+4
Y=1

(X,Y)= (-3,1)

So the answer is C
6 0
2 years ago
Read 2 more answers
Three consecutive integers have a sum of –21. Which equation can be used to find the value of the three numbers?
Ksju [112]

Answer:

The third one, x+(x+1)+(x+2)=-21 because x, x+1 and x+2 are three consecutive numbers.

6 0
3 years ago
Read 2 more answers
The height that a properly inflated basketball bounces decreases exponentially by 27.8% with each subsequent bounce after the in
n200080 [17]

Answer:

H (b) = 0.722^b(6)

Step-by-step explanation:

The first time the basketball bounces it reaches a height of (1-0.278)(6) = (0.722) (6).

The second time the basketball bounces it reaches a height of (0.722) (0.722) (6) = [(0.722) ^ 2] (6). Therefore, after b rebounds the height that the basketball will reach is given by:

H (b) = 0.722^b(6)

3 0
3 years ago
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