14.6 as a mixed number
2 answers:
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Answer:
A. 3 possible combinations
B. 8 4-ounce's bags and 3 3-ounce's bags
C. 2 4-ounce's bags and 11 3-ounce's bags
D. 8 4-ounce's bags and 3 3-ounce's bags
E. All solutions offer the same revenue.
Step-by-step explanation:
You have been tasked with filling 4 ounce and 3 ounce bags from a 41 ounce container of candy. Let x be the number of 4 ounce bags and y be the number of 3 ounce bags. Then
A. Find all integer solutions:
- When x=0, then 3y=41 - impossible, because 41 is not divisible by 3.
- When x=1, then 3y=37 - impossible, because 37 is not divisible by 3.
- When x=2, then 3y=33, y=11 - possible.
- When x=3, then 3y=29 - impossible, because 29 is not divisible by 3.
- When x=4, then 3y=25 - impossible, because 25 is not divisible by 3.
- When x=5, then 3y=21, y=7 - possible.
- When x=6, then 3y=17 - impossible, because 17 is not divisible by 3.
- When x=7, then 3y=13 - impossible, because 13 is not divisible by 3.
- When x=8, then 3y=9, y=3 - possible.
- When x=9, then 3y=5 - impossible, because 5 is not divisible by 3.
- When x=10, then 3y=1 - impossible, because 1 is not divisible by 3.
You get 3 possible combinations.
B. 1. 2 + 11 = 13,
2. 5 + 7 = 12,
3. 8 + 3 = 11.
The minimal number of bags is 11.
C. 1. 2·7+11·5=69 cents
2. 5·7+7·5=70 cents
3. 8·7+3·5=71 cents
The cheapest is 1st solution.
D. 1. 2·6+11·5=67 cents
2. 5·6+7·5=65 cents
3. 8·6+3·5=63 cents
The cheapest is 3rd solution.
E. 1. 2·2+11·1.50=$20.50
2. 5·2+7·1.50=$20.50
3. 8·2+3·1.50=$20.50
All solutions offer the same revenue.