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STALIN [3.7K]
4 years ago
7

FIRST ONE TO ANSWER GETS BRAINLIEST!!!!!!!! PLZ HELP ME!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
babymother [125]4 years ago
6 0

b 21 PLAYERS PER TEAM

o-na [289]4 years ago
6 0

Answer:

basic division the answer is: b 21

Step-by-step explanation:

252 divided by 12 is 21

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Paula’s entire family came to her birthday party. There were 17 people there. Paula’s parents bought enough mini-sandwiches for
djverab [1.8K]

Answer:

102

Step-by-step explanation:

Paula's parents bought 102 mini-sandwiches. When I multiplied 17 by 6, I got 102. To check my answer, I divided 102 by 6 and got 17.

17x6=102

102/6=17

Hope this helps:)

PLEASE MARK ME AS BRAINLIEST

3 0
2 years ago
Point A is located at -4/6 and point B is located at -1/6 what is the distance between points A and B
gladu [14]

Answer:

3/6 units

Step-by-step explanation:

distance is always positive

____-4/6_______________-1/6__0___________

                 |-4/6 - (-1/6)| = 3/6

7 0
3 years ago
Please help me find the surface area and volume
Lady_Fox [76]

Answer:

3) SA= 64 V=28

4) SA= 192 V=144

5) SA=179 V=144

6) SA=60 V=24

3 0
2 years ago
A)<br>(+4) + ( - 3)<br>b) (-2) + (-3)<br><br>c)<br>(+2) + (+4) <br>d) (-6) + (+2) ​
dolphi86 [110]

Answer:

c ) 6

d) 1 0 this are correct answer ok

8 0
3 years ago
You own 17 CDs. You want to randomly arrange 5 of them in a CD rack. What is the probability that the rack ends up in alphabetic
amid [387]

The probability that all five end up in alphabetical order is; 1/120.

<h3>What is the probability that the rack ends up in alphabetical order?</h3>

To evaluate the given probability; first, the number of possible arrangements is;

17P5 = 742,560 possible arrangements.

However, the chance of an alphabetical order arrangement in each case is; 1 out of 5! possible arrangements.

Hence, we have that the number of possible alphabetical arrangement is; (1/120) × 742,560 = 6188.

Hence, the required probability is; 6188/742,560 = 1/120

Read more on probability;

brainly.com/question/251701

#SPJ1

5 0
2 years ago
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