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Naily [24]
3 years ago
14

During a fancy fair organized at school, the total amount of money collected was $54 000. The following table represents the amo

unt of money collected from the different items:
Food and drinks: 27 000
Entertainment: 6000
Games: 12 000
Donations: 9000
What percentage of the money collected does each item represent? ​
Mathematics
2 answers:
MA_775_DIABLO [31]3 years ago
5 0

Answer:

Food & drinks = 50%

Entertainment = 11.1%

Games = 22.2%

Donations = 16.7%

Step-by-step explanation:

Basically, divide all of the amounts to $54000.

What I did was divide them, my answers are basically close to the exact percentage.

fredd [130]3 years ago
3 0

Answer: Collected amount÷total amount x 100

Step-by-step explanation:

Food and drinks = \frac{27000}{54000} x 100 = 50%

Entertainment = \frac{6000}{54000} x 100 = 11% (11.1 - .1 recurring)

Games = \frac{12000}{54000} x 100 = 22% (22.2- .2 recurring)

Donations = \frac{9000}{54000} x 100 = 17% (16.6- .6 recurring)

Double checking - 50 + 11 + 22 + 17 = 100%

Hope this helps! :)

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Read 2 more answers
On a recent trip to the convenience​ store, you picked up 4 gallons of​ milk, 5 bottles of​ water, and 8 ​snack-size bags of chi
Otrada [13]

Answer:

One gallon of milk costs $3.64

One  bottle of water costs $2.54

One bag of chips costs $0.79

Step-by-step explanation:

m represent gallons of milk

w represents  bottles of water

c represents snack-size bags of chips

t = total bill

t = m + w + c         ... eq 1

t = $33.60

but

w = 2c

m = 1.90 + w

m = 1.9 + 2c

Slot in the values in eq 1

33.60 = (1.9 + 2c ) + 2c + c

33.60 = 1.9 + 5c

33.60 - 1.9 = 5c

31.7 = 5c

c = $6.34

Slot in the value of c in w and m

w = 2c

w = 2(6.34)

w = $12.68

m = 1.9 + 2c

m = 1.9 + 2(6.34)

m = $14.58

4 gallons of milk cost $14.58

Therefore, one gallon if milk costs = 14.58/4 = $3.64

5 bottles of water cost $12.68

Therefore, one bottle of water costs = 12.68/5 = $2.54

8 bags of chips = $6.34

Therefore, one bag of chips cost = 6.34/8 = $0.79

5 0
3 years ago
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