Who's looking forward to math? :) (3 Viewers)

[juicystar07]

I don't know if I'm a massive outlier or something, but I am so scared for maths tomorrow.
I love studying maths, and it is far the easiest and fun-nest thing to prepare for (doing the question and knowing if you're exactly right or wrong) but I just get so intimidated.
Ugh, why can't I be like you guys?

 

[crizk]

Actually looking forward to my 2U and 3U exam!!

2U - majority of it will be easy!
3U - I know the majority will be challenging.

When you know what difficulty to expect in exams, there's less stress because you kind of know how you'll go

 

[LoveHateSchool]

I don't know if I'm a massive outlier or something, but I am so scared for maths tomorrow.
I love studying maths, and it is far the easiest and fun-nest thing to prepare for (doing the question and knowing if you're exactly right or wrong) but I just get so intimidated.
Ugh, why can't I be like you guys?

This is me haha, I think we have to be more confident but it;'s hard on a board full of maths gods. Some of you BOSers I'm surprised are even sitting Mathematics, I thought you would have done 4U for sure.

I just want the work to pay off, I have almost a full drawer of done maths past papers for the year!

 

[mickk794]

good luck to all doing maths this week.
Myself, i have basically shot myself in the foot. I havent done 1 bit of revision since trials and also for tomorrow, because i convinced myself that it was not needed since it was my 12th unit and worst subject i do. Hopefully nobody does what i did, because i'm feeling crap right now.

 

[someth1ng]

2U should be hard but 3U should be easy/average - obviously, compared to respective past papers.

 

[Rawf]

Are we allowed to use 3u methods for 2u? Or will we get no marks?
Like for probability I'm using nCk

 

[onedaysnotice]

good luck tmo everyone

 

[Carrotsticks]

The questions will be written such that knowledge of nCk will not help very much.

 

[Rawf]

The questions will be written such that knowledge of nCk will not help very much.

I saw a question for probability asking for 4 cards being drawn, all of different suits.
rather than having to do a crap load of fractions, can I just do:
(13C1)^4 / 52! ?
The final answer is the same.