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Tems11 [23]
2 years ago
5

Each week, Mrs.Mack buys exactly 3 5/8 pounds of cheese for her mac-and-cheese recipe. She chooses from these kinds of cheese. M

rs.Mack buys as many packages as she needs— up to three packages of any kind. What packages of cheese (and how many of each kind) does Mrs.Mack buy? 7/8pounds of american cheese. 0.5 pounds of cheddar cheese. 0.2 pounds of swiss cheese. HELP PLEASE
Mathematics
1 answer:
ryzh [129]2 years ago
7 0

9514 1404 393

Answer:

  • 3 packages of American (7/8 lb)
  • 2 packages of cheddar (1/2 lb)

Step-by-step explanation:

The amount of cheese Mrs Mack needs is an odd number of eighths of a pound. So, she must buy an odd number of the 7/8 pound packages.

If she were to buy any 1/5 pound packages, she would have to buy a multiple of 5 of them so as to have an integral number of eighths. The only suitable multiple is 0.

If Mrs Mack buys 1 of the 7/8 pound packages, she cannot make up the difference with half-pound packages. She must buy 3 of the 7/8 pound packages, for a total of 2 5/8 pounds of American cheese. The remaining pound can be bought as 2 of the 1/2 pound packages.

To buy 3 5/8 pounds, Mrs Mack must purchase ...

  3 packages of American cheese (7/8 lb each)

  2 packages of Cheddar cheese (1/2 lb each)

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Answer:

see explanation.

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8 0
2 years ago
Can someone help me? I figured out part B however, I am struggling with part A and I would be so happy if any of you helped me.
UkoKoshka [18]
<h3>Answer:  262,785</h3>

Note: you may need to delete the comma

============================================================

Explanation:

The info "5 months" is never used to compute the mean. We could easily replace it with "6 months" or "7 months" or any stretch of time, and still get the same answer. So we'll ignore this value.

What we'll do is add up the 21 items given to us, and then divide by 21.

Because there are so many values, and it's easy to get lost, I'm going to add up across the rows

  • Row One: 256,229+253,657+218,747+246,163+235,626+288,694 = 1,499,116
  • Row Two: 316,265+196,721+319,620+285,077+215,152+253,291 = 1,586,126
  • Row Three: 315,011+199,901+265,443+291,806+303,556+215,359 = 159,1076
  • Row Four: 258,554+293,658+289,935 = 842,147

Those subtotals then add up to this

1,499,116+1,586,126+1,591,076+842,147 = 5,518,465

This is the same as adding up all 21 values.

Finally, we divide that sum over 21 as there are 21 values in this list

(5,518,465)/21 = 262,784.047619048

That value then rounds up to 262,785

If your teacher just wanted things to the nearest whole number (without rounding up), then the answer would be 262,784

Side note: using a spreadsheet program would probably be the most efficient/fastest method for this type of problem.

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never [62]
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3 years ago
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If you divide 70 by 10, you get the value of the second seven, 7.

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