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skelet666 [1.2K]
3 years ago
15

Raymond wants to make a box that has a volume of 360 cubic inches. He wants the height to be 10 inches and the other two dimensi

ons to be whole numbers of inches. How many different-sized boxes can he make?

Mathematics
1 answer:
agasfer [191]3 years ago
7 0
The volume of the box is l*w*h, or l*w*10, and also the volume is 360. so, 10lw=360, meaning l*w=36, so we are looking for whole number factors of 36. 
length: 1    2  3  4 6 9 12 18 36width:36  18 12 9 6 4   3   2   1
there are a total of 9 possible ways to make the base of the box so your answer would be  9

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Step-by-step explanation:

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Answer:

The number of students in the key club is 490

Step-by-step explanation:

The given parameters are;

The number of student in the school = 1000

The total number of student in the debate club = 310

The total number of student in the student council = 650

The total number of student who are in debate and student council = 170

The total number of student who are in both debate and the key club = 150

The total number of student who are in both student council and the key club = 180

The number of students who are in all three clubs = 50

Therefore, we have;

Let A represent the number of students in the debate club

Let B represent the number of students in the student council

Let C represent the number of students in the key club

We have;

n(A∪B∪C) = n(A) + n(B) + n(C) -  n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C)

Where;

n(A∪B∪C) = 1000

n(A) = 310

n(B) = 650

n(A∩B) = 170

n(B∩C) = 180

n(C∩A) = 150

n(A∩B∩C) = 50

Therefore;

n(C) = n(A∪B∪C) - (n(A) + n(B) -  n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C))

Substituting the values gives;

n(C) = 1000 - (310 + 650 -  170 - 180 - 150 + 50) = 490

The number of students in the key club, n(C) = 490.

5 0
3 years ago
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