David W. answered • 01/19/18

Experienced Prof

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Given that n(ξ)= 200, n(G)= 120, and n(C)= 90

n(G) = 120 = x + y [120 students who joined uniformed groups]

n(ξ)= 200 = w + x + y + z [all students in Secondary One] [This is the "universe"]

The math:

n(ξ)= w + (x + y) + (z + y) - y = 200 [rearrange; include y that cancels]

w + n(G) + n(C) - y = 200 [ah, we know those]

w + 120 + 90 - y = 200

y - w = 10

So, if w=0 [no uninvolved Secondary One students],

**y=10**[This

**the minimum (smallest possible) value for y**]. Note (to answer your question): y cannot equal 0 because that means that w = -10.

We also have:

so, if x=0 then y = 120 thus y <= 120

and

y + z = 90

so, if z=0 then y = 90 thus y <= 90

Use y <=90 (because they

**both**must be true)

y-10 <= 80

w <= 80 (because w-y=-10 means w=y-10)

**The largest (maximum value) for w is 80.**

So, it could be that w=80, x=30, y = 90, and z=0.

Other values of x, y, or z make w smaller.

David W.

01/19/18

David W.

**Let's try that again.**

01/19/18

Amy S.

01/19/18

Amy S.

01/19/18

David W.

01/19/18

Amy S.

_{I forgot to add, but there is a part c) which is to find the value of x and z if w=30. Could you help me with this part too? Thanks a lot!}01/19/18