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 Stephanie Oswalt
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#64377
Hello! We recently received the following question from a student. An instructor will respond below!
The drill is on page 381 and the answer is on pages 388-389 in the LGB. It’s the drill, Chapter 6, Grouping/linear about:

A television executive must schedule 6 programs and the times are from 8 p.m. in the evening up until 11 p.m., but the times go from 8 p.m. - 10 p.m. hence. I am respectfully asking why the answer to the question is E., when both S and U can clearly either one could be the correct answer? Is it because S occupies the spot of 9 p.m., while the remaining spot is 10 pm for U?

Thank you
 Brook Miscoski
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#64427
Hi,

I think stepping through the rules can help resolve your question. Because you've reached the conclusion that S and U will be shown, I think you are simply missing one rule. The first rule states that if S and U are both used, then S is before U, which explains why E is the answer.

If you didn't realize that S and U are both used, you need to make sure you understood the second and third rules properly.

2nd rule. This is a conditional rule, so you need to write it and its contrapositive:
R :arrow: T
-T :arrow: -R

Because the question tells us that T is unscheduled, we know that both T and R are unscheduled.

3rd rule. P or Q is at 8 PM, and the other one is unscheduled.

As you've indicated, there are only three slots, so that leads us to conclude that all of the unscheduled slots are taken. Thus, S and U both have to be scheduled, and you have to apply the first rule.
 leslie7
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#83829
Hi pg. 396 ed. 2020 Q1

S + U - > (S-U)

The book states that "if one of the two is not scheduled, then there are no restrictions on the scheduling of the other program" . When I read and try to interpret that into a logic diagram I think, /S or /U -> open ended? not sure what this is supposed to look like, it appeared to me to look like a mistaken negation of the sufficient condition?

On that note, to verify ... is the negation of S + U -> (U-S)

/ U-S -> /S or /U but in practice what does this diagram mean?

if U does not come before S then we might not have S or we might not have U or we might not have both?

(I realize this might just be a refresher question from previous chapters just not sure where to relocate it)


Someone please clarify, I dont understand the negation of this
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 Dave Killoran
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#83842
Hi Leslie,

Some thoughts below!
leslie7 wrote: Fri Feb 05, 2021 9:03 am Hi pg. 396 ed. 2020 Q1

S + U - > (S-U)

The book states that "if one of the two is not scheduled, then there are no restrictions on the scheduling of the other program" . When I read and try to interpret that into a logic diagram I think, /S or /U -> open ended? not sure what this is supposed to look like, it appeared to me to look like a mistaken negation of the sufficient condition?
The book comment relates to the fact that the sufficient condition uses both S and U. The trigger there is having both, so if one or both is missing, all bets are off and you are free to do whatever.

Kind of like, "If you are tall and have red hair, then you must be in the second group." Well, if you are missing one of those traits or both, then you could be in group two, or anywhere else; you are, at that point, free to do what you want (or, open-ended in your movement).




leslie7 wrote: Fri Feb 05, 2021 9:03 am On that note, to verify ... is the negation of S + U -> (U-S)

/ U-S -> /S or /U but in practice what does this diagram mean?

if U does not come before S then we might not have S or we might not have U or we might not have both?

(I realize this might just be a refresher question from previous chapters just not sure where to relocate it)


Someone please clarify, I dont understand the negation of this
The technical contrapositive is:

  • ..... ..... ..... U
    S :longline: U :arrow: or
    ..... ..... ..... S
In other words, if it is not S :longline: U, then one or both of U and S are out. So, if you saw U in the first spot, you'd know it couldn't be S :longline: U, and then S would have to be out. Otherwise, whenever you have both, it's always U :longline: S. Note, you can't have U :longline: S because that's both of them, and once you have both it has to be S :longline: U.

Thanks!
 leslie7
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#83863
This is so clear, thank you so much Dave !

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