let us get CAP firstworst case: A all have 3 MCs, all C have 4 MCs, suppose you have 1 B which has 3 MCs;
cap = (2*5*3 + 3*2*4 + 1*3.5*3)/(6+12+3)
= (30 + 24 + 10.5)/21
= 64.5/21>3(saft mode)
so, don't worry, at least your cap >3, you won't be fired.
but you should do better next time...
All the best
是这样的,
2 A all have 3 AUs, 1 C has 1 AU, 1 C has 4 AUs, 1 C has 3 AUs,
2 B have 1 AUs, 1 B has 3 AUs.
I am in NTU,
所以,是不是
(2*5*3+1*2*1+1*2*4+1*2*3+2*3.5*1+1*3.5*3)/(6+1+4+3+3+2)
=63.5/19
=3.35
我从来没有考这么差过, 其实我原来的主课连B都很少拿, 这一次...
我都不知道该怎么办了. sigh
2 B have 1 AUs, 1 B has 3 AUs.
I am in NTU,
所以,是不是
(2*5*3+1*2*1+1*2*4+1*2*3+2*3.5*1+1*3.5*3)/(6+1+4+3+3+2)
=63.5/19
=3.35
我从来没有考这么差过, 其实我原来的主课连B都很少拿, 这一次...
我都不知道该怎么办了. sigh
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