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Mariana [72]
2 years ago
8

William has 24 cans of fruit and 60 cans of vegetables that he will be putting into bags for a food drive. He wants each bag to

have the same number of cans of each type of food. He uses all the cans. Use the drop-down menus to complete the statements below about the number of bags of food William can make. CLEAR CHECK The greatest number of bags William can make is . Each of these bags will have cans of fruit and cans of vegetables. If he made fewer bags, . He could use bags, but this is not the greatest number he could use.
Mathematics
1 answer:
nataly862011 [7]2 years ago
7 0

Answer:

See explanation

Step-by-step explanation:

William has 24 cans of fruit and 60 cans of vegetables that he will be putting into bags for a food drive.

Factor number 24 and 60:

24=2\cdot 12=2\cdot 2\cdot 6=2\cdot 2\cdot 2\cdot 3=2^3\cdot 3\\ \\60=2\cdot 30=2\cdot 2\cdot 15=2\cdot 2\cdot 3\cdot 5=2^2\cdot 3\cdot 5

Find the greatest common factor

GCF(24,60)=2^2\cdot 3=12

Hence, the greatest number of bags William can make is 12.

Each of these bags will have 24\div 12=2 cans of fruit and 60\div 12=5 cans of vegetables.

If he made fewer bags, 6 bags, each of these bags will have 24\div 6=4 cans of fruit and 60\div 6=10 cans of vegetables.

If he made fewer bags, 4 bags, each of these bags will have 24\div 4=6 cans of fruit and 60\div 4=15 cans of vegetables.

If he made fewer bags, 3 bags, each of these bags will have 24\div 3=8 cans of fruit and 60\div 3=20 cans of vegetables.

If he made fewer bags, 2 bags, each of these bags will have 24\div 2=12 cans of fruit and 60\div 2=30 cans of vegetables.

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n = \frac{11!}{(11-11)!} = 11! = 39916800\\  

2) If Blue balls are distinguishable, but the red balls are identical

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n = \frac{11!}{4!}  = \frac{39916800}{24} = 1663200

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Answer:

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2. Thirty times the sum of eight and fifty-two

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      52+8 (The product of thirty and eight more than fifty-two.)

       

2) 52+8 (The sum of eight and fifty-two)

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3 years ago
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