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Mega A Level Maths Thread MK II

Reply 5980

username937760

16

Original post by justinawe

Just googled "telescoping series", and realised that's exactly the same thing I did :colonhash:

Telescoping series is (I think) the popular name in America. I've always just called it the method of differences.

Reply 5981

justinawe

19

Original post by DJMayes

Telescoping series is (I think) the popular name in America. I've always just called it the method of differences.

Ah. iirc, they call the trapezium rule the "trapezoidal rule" or something funky like that :colone:

Reply 5982

L'Evil Fish

OP

18

S1 shall be obliterated :h:

Although, some of these probability ones throw me off a little :s-smilie:

Why is:

P(A|AnB) = P(A)/P(AnB)?????

I thought it should be:

P(A n AnB)/P(AnB)

Reply 5983

ryanb97

15

Has anyone got access to the edexcel emporium ... like all the resources on there??

Ryan

Reply 5984

username937760

16

Original post by L'Evil Fish

S1 shall be obliterated :h:

Although, some of these probability ones throw me off a little :s-smilie:

Why is:

P(A|AnB) = P(A)/P(AnB)?????

I thought it should be:

P(A n AnB)/P(AnB)

What you've written there is "The probability that event A happens given that event A and B happen" which is quite clearly 1.

(edited 10 years ago)

Reply 5985

Robbie242

Original post by L'Evil Fish

S1 shall be obliterated :h:

Although, some of these probability ones throw me off a little :s-smilie:

Why is:

P(A|AnB) = P(A)/P(AnB)?????

I thought it should be:

P(A n AnB)/P(AnB)

And what is the probability A intersects AnB, why its the probability of A

Reply 5986

L'Evil Fish

OP

18

Original post by DJMayes

What you've written there is "The probability that event A happens given that event A and B happens" which is quite clearly 1.

Okay, bad example... I copied it wrong. But why wasn't it the probability of both happening over the second half happening? :s-smilie:

Reply 5987

username937760

16

Original post by L'Evil Fish

Okay, bad example... I copied it wrong. But why wasn't it the probability of both happening over the second half happening? :s-smilie:

It is. The rule is:

P(A|B) = \dfrac{P(A \cap B)}{P(B)}

There is a second formula for conditional probability (Derived from this one) but it is not one you need to know.

Reply 5988

MathsNerd1

17

Now that I'm at home could someone tell me which page the question was on as I can't remember myself :-/

I'm going to fight for my place now and not give in just because of one certain outcome, Warwick will be begging for me to stay after my 4 years! :biggrin:

Posted from TSR Mobile

Reply 5989

justinawe

19

Original post by MathsNerd1

Now that I'm at home could someone tell me which page the question was on as I can't remember myself :-/

I'm going to fight for my place now and not give in just because of one certain outcome, Warwick will be begging for me to stay after my 4 years! :biggrin:

Posted from TSR Mobile

Here:

Original post by Felix Felicis

Here's a question :colone:

Simplify (with justification :rolleyes:

)

\displaystyle \sum_{r=1}^{n} \frac{r}{(r+1)!}

Reply 5990

Zaphod77

1

Original post by MathsNerd1

Now that I'm at home could someone tell me which page the question was on as I can't remember myself :-/

I'm going to fight for my place now and not give in just because of one certain outcome, Warwick will be begging for me to stay after my 4 years! :biggrin:

Posted from TSR Mobile

Good, stay positive! :smile:

Reply 5991

purplemind

6

I know this might be a stupid question but I can't get my head around it now. :tongue:

If I get something like (2x+3)/(x-1)
How the hell am I supposed to know how to express it in the form
a+b/(x-1)
(a, b are constants)? :biggrin:

Reply 5992

MathsNerd1

17

Original post by justinawe

Here:

Okay thanks, let me just do my first C3 paper and then I'll try this out, nothing's going to keep me from achieving what I know I deserve! :biggrin:

Reply 5993

Felix Felicis

13

Original post by purplemind

I know this might be a stupid question but I can't get my head around it now. :tongue:

If I get something like (2x+3)/(x-1)
How the hell am I supposed to know how to express it in the form
a+b/(x-1)
(a, b are constants)? :biggrin:

\displaystyle \frac{2x+3}{x-1} = \frac{2x - 2 +2 + 3}{x-1} = \frac{2x-2}{x-1} + \frac{5}{x-1}

Reply 5994

MathsNerd1

17

Original post by Zaphod77

Good, stay positive! :smile:

I'm eager to hear your response about this morning.

Reply 5995

hassassin04

17

Original post by MAyman12

If the final answer is wrong, you must lose some A1 marks along the way.
I had the same problem with M1 this May. :frown:

I feel you mate.

Hmm...if that's the case then what is Error carried forward? It seems that you would get method marks anyway regardless of whether you got answers wrong in the previous part or elsewhere.

Reply 5996

Zaphod77

1

Original post by MathsNerd1

I'm eager to hear your response about this morning.

Yeah, I really want to talk about it! I'm tempted to stay up to midnight to talk, but I'm not very good at keeping awake :tongue:

I'm going to try though, or set an alarm for midnight, I'd be interested to find out how everyone else found it - only 3 people sat the exam in my school!