S1/17 WAM Predictions (4 Viewers)

[Esse]

myUNSW down for maintenance I can't even check oml

 

[PrideG]

I think all 4 core commerce subjects have been included in the wam because a mate (first year) got 70.75 and I'm fairly sure it's impossible to get 0.75 if less than 4 subjects have been used in the calculation.

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How is that impossible? It makes zero sense for that to be impossible.

 

[DoYouEvenCheeto]

How is that impossible? It makes zero sense for that to be impossible.

If your wam includes only 1, 2 or 3 subjects, you can't get a wam with a decimal place of 0.75 or 0.25.

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[aoc]

i think acct1501, econ1101 and econ1203 were uploaded according to some friends

 

[leehuan]

How is that impossible? It makes zero sense for that to be impossible.

Um, yes, it is impossible.

Because the A in WAM stands for average. The average of 3 courses has a decimal value of .333 or .667, not .25, .5, .75

 

[RenegadeMx]

myUNSW down for maintenance I can't even check oml

wouldnt update ur WAM over weekend tho

 

[matchalolz]

How is that impossible? It makes zero sense for that to be impossible.

If it's on .0 it could be for any number of subjects, but for all others, if you multiply your WAM by the number of subjects, one of these numbers should be very close to a full number. This should indicate how many subjects are included in that WAM.

 

[matchalolz]

I did Fed Con, Land Law, RCD and a Commerce subject.

2 subjects have been added for me. I'm thinking it's one of the Laws (Fed Con being least likely) and my Commerce subject.

Lol is this infs? I think I flopped that shit :')

 

[2xL]

5th year Law electives?

yep!

 

[Zoinked]

If your wam includes only 1, 2 or 3 subjects, you can't get a wam with a decimal place of 0.75 or 0.25.

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So you're saying no combination of numbers can be divided by three to give a decimal place of .25 or .75? lol.

 

[Esse]

So you're saying no combination of numbers can be divided by three to give a decimal place of .25 or .75? lol.

yeh

 

[DoYouEvenCheeto]

So you're saying no combination of numbers can be divided by three to give a decimal place of .25 or .75? lol.

They have to be whole numbers...

 

[InteGrand]

So you're saying no combination of numbers can be divided by three to give a decimal place of .25 or .75? lol.

The total of n integers will have a remainder of either 0, 1, 2, 3, …, or (n-1) upon division by n. This implies that the possible fractional parts for the average of n whole numbers are just: 0, 1/n, 2/n, 3/n, …, (n-1)/n.

E.g. If 1 subject: only can have decimal of 0.

If 2 subjects: only can have decimal of 0 or 0.5 (i.e. 0 or 1/2).

If 3 subjects: only can have decimal of 0, 0.333..., or 0.666..., (i.e., 0, 1/3, or 2/3).

If 4 subjects: only can have decimal of 0, 0.25, 0.5, or 0.75 (i.e. 0, 1/4, 2/4, or 3/4).

Etc.

Furthermore, if 0 < f < 1 and the smallest positive integer for which f occurs as a possible fractional part in the average is n₀, then f can occur as a fractional part in the average of N subjects (N a positive integer) if and only if N is a (positive) multiple of n₀ (easy exercise).

So for instance, a fractional part of 0.25 in the average means the no. of subjects used in the average must have been some multiple of 4, since 4 is the smallest positive integer for which 0.25 can occur as a fractional part.

(Assumptions: weights in the average are equal and subject marks are whole numbers. If one or more of these assumptions is violated, then other decimal values may be possible.)

 

[DoYouEvenCheeto]

The total of n integers will have a remainder of either 0, 1, …, (n-1) upon division by n. This implies that the possible fractionals parts for the average of n whole numbers are just: 0, 1/n, 2/n, 3/n, …, (n-1)/n.

E.g. If 1 subject: only can have decimal of 0.

If 2 subjects: only can have decimal of 0 or 0.5 (i.e. 0 or 1/2).

If 3 subjects: only can have decimal of 0, 0.333..., or 0.666..., (i.e., 0, 1/3, or 2/3).

If 4 subjects: only can have decimal or 0, 0.25, 0.5, or 0.75 (i.e. 0, 1/4, 2/4, or 3/4).

Etc.

So for instance, a fractional part of 0.25 in the average means the no. of subjects used in the average must have been 4 or a multiple of 4.

(Assumptions: weights in the average are equal and subject marks are whole numbers. If one or more of these assumptions is violated, then other decimal values are possible.)

Legend

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[RivalryofTroll]

Lol is this infs? I think I flopped that shit :')

I think it's been 2 law subjects (based on speaking with my law mates) that've been added so far for me rather than 1 law + INFS1602.

 

[BHLee]

Actually, the WAM reflected on the stream declaration still isn't fully revised right? Like, it still may not include every courses you have taken in this semester correct?
That WAM is sending me some vibe that I might fail ECON1203... -_-

 

[RivalryofTroll]

Actually, the WAM reflected on the stream declaration still isn't fully revised right? Like, it still may not include every courses you have taken in this semester correct?
That WAM is sending me some vibe that I might fail ECON1203... -_-

You can figure this out with some calculating but yes, for most people it is likely that not all courses have been added yet.

 

[leehuan]

Actually, the WAM reflected on the stream declaration still isn't fully revised right? Like, it still may not include every courses you have taken in this semester correct?
That WAM is sending me some vibe that I might fail ECON1203... -_-

As RoT explains it may not be fully revised but if you go back a page I think we have some evidence that ECON1203 has been entered.

 

[arob259]

any ideas about MATH2011, MATH2501 and MATH2801? My WAM has not changed and I find it highly unlikely i failed anything.

 

[Zoinked]

The total of n integers will have a remainder of either 0, 1, …, (n-1) upon division by n. This implies that the possible fractionals parts for the average of n whole numbers are just: 0, 1/n, 2/n, 3/n, …, (n-1)/n.

E.g. If 1 subject: only can have decimal of 0.

If 2 subjects: only can have decimal of 0 or 0.5 (i.e. 0 or 1/2).

If 3 subjects: only can have decimal of 0, 0.333..., or 0.666..., (i.e., 0, 1/3, or 2/3).

If 4 subjects: only can have decimal or 0, 0.25, 0.5, or 0.75 (i.e. 0, 1/4, 2/4, or 3/4).

Etc.

Furthermore, if 0 < f < 1 and the smallest positive integer for which f occurs as a possible fractional part in the average is n₀, then f can occur as a fractional part in the average of N subjects (N a positive integer) if and only if N is a (positive) multiple of n₀ (easy exercise).

So for instance, a fractional part of 0.25 in the average means the no. of subjects used in the average must have been some multiple of 4, since 4 is the smallest positive integer for which 0.25 can occur as a fractional part.

(Assumptions: weights in the average are equal and subject marks are whole numbers. If one or more of these assumptions is violated, then other decimal values may be possible.)

Lol didnt realise that you could only get a whole number