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Flauer [41]
3 years ago
5

For a school fundraiser students are selling snack bags and candy bars to raise money on Wednesday the students or 23 snack bags

and 36 candy bars that raised $114.75 on Thursday the seventh so 37 snack bags and 36 candy bars that raised $146.25 what is the selling for each candy bar
Mathematics
1 answer:
meriva3 years ago
3 0

Answer:

$1.75

Step-by-step explanation:

The selling for each candy bar may be determined by  a set of linear equations. This pair of linear equations may be solved simultaneously by using the elimination method. This will involve ensuring that the coefficient of one of the unknown variables is the same in both equations.

It may be solved by substitution in that one of the variable is made the subject of the equation and the result is substituted into the second equation .

Let the cost of a snack bag be s and that of a candy bar be c, then if on Wednesday the students or 23 snack bags and 36 candy bars that raised $114.75 on Thursday the seventh so 37 snack bags and 36 candy bars that raised $146.25

23s + 36c = 114.75

37s + 36c = 146.25

14s = 31.5

s = $2.25

23(2.25) + 36c = 114.75

36c = 114.75 - 51.75

36c = 63

c = 63/36

= $1.75

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laiz [17]

1.

Let the sandwich be represented by = x

Let the drink be represented by = y

As given, Joe charges $10.00 for two sandwiches and one drink, so equation becomes 2x+y=10    ....(i)

He charges $6.50 for one sandwich and one drink, so equation becomes x+y=6.50  ....(ii)

Subtracting (ii) from (i) we get

x=3.50

And as x+y=6.50

y=6.50-3.50=3.00

Hence, cost of a sandwich is $3.50 and a drink is $3.00

2.

the question is incomplete. coordinates are missing

3.

Let x represent the number of performances

cost = 8000+300x

revenue = 700x

To break even, the cost must be equal to the revenue.

8000+300x=700x

400x=8000

so x=20

Hence, 20 performances are needed to break even.

4.

Let us take x for 20% material, y for 50% material

\frac{20x+50y}{200}=30

20x+50y=6000    ......... (1)

Accounting for volumes to use, x+y=200 or x=200-y ................ (2)

Putting the value of x in (1)

20(200-y)+50y=6000

4000-20y+50y=6000

4000+30y=6000

30y=2000

y=66.67

x=200-y = 200-66.67 =133.33

Hence, 133.33 ml of 20% acetic acid and 66.67% of 50% acetic acid is needed.





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