Mega A Level Maths Thread MK II

I don't have one

I always have to derive my inverse trig from first principles as a result...

Hi can someone help me with S2 hypothesis testing please.






I done the part to the get the first 3 marks but im not seeing where to get the 2 and the 10.

I would assume P(X<=A) < 0.025 but I dont know to go from there.

Like for two marks if they ask:

Differentiate sin^-1(7x) I do it from scratch

Yeah :/ it's never that high though, most I've done is... About 6 of them

Okays thank youu I'll try it then post it

You want to scan the tables for where P(X<=Critical value1) is as close to 0.05 as possible. Similarly, scan the table for where P(X>=Critical value2) is as close to 0.05 as possible. As the tables have P(X<=x), you want to look for when 1-P(X<=Critical value2 - 1) ~ 0.05. Your critcal regions will be X<=critical value 1 or X>= critical value 2

Haha
Here's a nice logs question.
Show that the solution for x in the equation.
$6^x = 5 \times 9^x$
Can be written as $x=\dfrac{\log_3(5)}{\log_3(2)-1}$

Okey dokey, I may think up another one if you want

Oh I see, thanks so much!

How come it its x<=2 not x<=1 since 0.0500 is below

Hi guys i'm sitting Ocr stats 1 exam and am stuck on these perms and combs questions - can anyone please help me , I'd really appreciate it , thanks perms and combs.ppt

Can't get further than this:

$6^x = 5 \times 9^x[br][br]xlog6=log(5 x 9^x)[br][br]x=\dfrac{\log5\times 9^x}{\log6}[br]$

How do you change the base from 10 to 3? :s



Haha thank you :P i'm not too good at it though!

Oh, you don't get one? You're supposed to, for WJEC.

http://www.wjec.co.uk/uploads/publications/10908.pdf

Well that's useful

Why not just start by taking $\log_3$?

Your school is supposed to supply you with one for every maths exam

Do they not do this?

How do you do that we've never changed it lool

so $6^x=xlog6=\log_{3}2$ How did you figure it out :s is it because: $log6= 2(log3)= \log_{3}2$?

They do have a yellow booklet there... I've just never thought to use it

They do have a yellow booklet there... I've just never thought to use it

you must be joking!

Here we were, labouring under the assumption that you were a clever, sensible chap

I've done okay then

If I forget things, I derive them...

Proof of series and that

No . . .
It's fairly straight forward to derive:
Take:
$y=\log_ax \Rightarrow a^y = x [br]\Rightarrow y\log_b(a) = \log_b(x)[br]\Rightarrow y = \dfrac{\log_b(x)}{\log_b(a)}$
As we said that $\ y=\log_ax$
$\Rightarrow \log_ax = \dfrac{\log_b(x)}{\log_b(a)}$

Also $\log(6) \not= 2\log(3)$
Instead: $\log(6) = \log(3) + \log(2)$