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Mumz [18]
3 years ago
9

How much more cheese is in one serving of Nancy’s dish than in Carlos dish?

Mathematics
1 answer:
kirill115 [55]3 years ago
3 0
I’m sorry I don’t know what you are asking
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Which is the area of a rectangle with a length of 21 feet and a width of 15 feet
muminat

Answer:

315 ft^2

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei
MAXImum [283]

Answer:

There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.

Step-by-step explanation:

Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

 The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is {11 \choose 3} = 165 . As a result, we have 165 ways to distribute the blackboards.

If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is {7 \choose 3} = 35. Thus, there are only 35 ways to distribute the blackboards in this case.

4 0
3 years ago
Frank cut a 52 inch string into 3 pieces. The first piece is 20 inches long. The second piece is at least 12 inches long, but no
Eddi Din [679]
We need to figure out how much string would be left, after taking away the first two pieces.
We know that the first piece is 20 inches long, so we can say that there is 52-20 inches left, or 32 inches.

The second piece is between 12 and 18 inches, meaning that there would be between 32-12 and 32-18 inches left for the third piece, or 20 and 14 inches. This means that the third piece would be at least 14 inches long, but no more than 20, since we don’t have more string than that (20+12+20=52, and 20+14+18=52)

So we can say that x is greater or equal to 14, but less than or equal to 20, or:

14<=x<=20 (“<=“ is written like a normal “<“ sign with a line _ right under it)
4 0
3 years ago
Select a counter-example that makes the conclusion false.
Softa [21]
A counter-example could be

There are more than 3 marbles in the bag, and the other ones are red.

Hope this helps!
4 0
3 years ago
8 + 4a + 6.2 - 9a = ​
Strike441 [17]

Answer: 14.2 - 5a

Work: I just did "CLT" of "Combine Like Terms". I did 8 + 6.2 First because 8 is first, then I did 4a - 9a which is -5a, and I combined them, and got the answer. I hope this helps, and stay safe! :)

7 0
2 years ago
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