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Hoochie [10]
3 years ago
5

Victoria searched her house for quarters. She started with 17 quarters and found 6 more quarters. She then goes to the arcade. I

f a game costs 2 quarters to play, how many games can she play and how many quarters are left over? Victoria can play times and there will be quarters left over.
Mathematics
2 answers:
Viefleur [7K]3 years ago
7 0

Answer:

11 games

Step-by-step explanation:

17+6=23

23/2=11.5

Victoria can play 11 games. You'd usually round up, but in this case you can't play half a game so she'll play 11 games and have one quarter leftover.

xxTIMURxx [149]3 years ago
5 0

Answer:

Victoria can play 11 games and have 1 quarter left over.

Step-by-step explanation:

17 given quarters +6 found quarters = 23 total quarters.

Since 23 is an odd number and we want to divide by 2, we have to go down to the last even number: 22. (This gives us the left over quarter)

22 quarters/2quarters per game = 11 games

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Micheal buys two bags or chips and three boxes of pretzels for $5.13 . He then buys another bag of chips and 2 boxes of pretzels
trapecia [35]

Answer:

So a bag of chip costs $2.39

Step-by-step explanation:

Let x = cost of a bag of chip

Let y = cost of a box of pretzel

Micheal buys two bags of chips and three boxes of pretzels for $5.13

This means

2x + 3y = 5.13 - - - - - - - - - -1

He then buys another bag of chips and 2 boxes of pretzels for $3.09

This means

x + 2y = 3.09 - - - - - - - - - - - -2

Solving equation 1 and equation 2 simultaneously and using the elimination method,

Multiply equation 1 by 1 and equation 2 by 2

2x + 3y = 5.13

2x + 6y = 6.18

Subtracting,

-3y = -1.05

y = -1.05/-3= $0.35

Put y= 0.35 in equation, x + 2y = 3.09

x + 2(0.35)= 3.09

x + 0.7= 3.09

x = 3.09-0.7 = $ 2.39

So a bag of chip costs $2.39

5 0
3 years ago
Suppose that E and F are two events and that N(E and F)=450 and N(E)=700. What is P(F|E)​?
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Answer:i really dont know

Step-by-step explanation:

6 0
3 years ago
Am i correct pls help
a_sh-v [17]

Answer:

yea ur correct

Step-by-step explanation:

use PhotoMath or something like that to check next time, it'll save a lot of time

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