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Mathematics: Post your doubts here!

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xhizors

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Ashique

Ashique

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Soldier313 said:
Aoa wr wb
Can someone please tell me in qn 1 of this paper, do i need to draw separate bar charts for males and females?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_6.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_ms_6.pdf
Click to expand...

Yup. You do. The ms says you could draw a pie or a bar chart. Pie charts have been removed from our syllabus. Anyhow, what you need to do is- on the SAME graph, draw two bars. You should shade one and leave the other one unshaded. The shaded one could represent males or vice versa. You get the point right?
 
Soldier313

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Ashique said:
Yup. You do. The ms says you could draw a pie or a bar chart. Pie charts have been removed from our syllabus. Anyhow, what you need to do is- on the SAME graph, draw two bars. You should shade one and leave the other one unshaded. The shaded one could represent males or vice versa. You get the point right?
Click to expand...
ohh right, that's cool (y)
Thanx a lot!
 
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kiara15

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The points A and B have position vectors, relative to the origin O, given by
−−→OA = i+2j+3k and −−→OB = 2i+j+3k.
The line l has vector equation
r = (1−2t)i+ (5+t)j+ (2−t)k.
(i) Show that l does not intersect the line passing through A and B. [4]
(ii) The point P lies on l and is such that angle PAB is equal to 60◦
. Given that the position vector
of P is (1 − 2t)i + (5 + t)j + (2 − t)k, show that 3t
2
+ 7t + 2 = 0. Hence find the only possible
position vector of P.
anyone help???
 
Esme

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xhizors said:
Thanks lol its not messy but neither the Q10 of w12 31 :)
but thanks again
Click to expand...

Here :D Sorry for the delay
And.. ummm hope you can understand what I've written lol
 

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kiara15

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XPFMember said:
Assalamoalaikum!!

For permutations and combinations watch this whole series of the tutorial, and you're good to go! ;) http://www.examsolutions.co.uk/maths-re ... rial-1.php


I see people finding permutations and combinations hard, but honestly i enjoy it a lot..

And one more thing, (you may understand this better may be after you've gone through the topic, but anyways)
to make out when you have to use permutation (nPr of the calculator) and when to use combinations (nCr) , checck if the different arrangements of the selections will make any difference or not...if it does then you've to use nPr, and if it doesn't then u must use nCr.

Eg. lets say I want to choose two numbers from a total of three..how many possible selections are there?

now see i may choose 1 2 ; 1 3 ; 2 3
now the no. i've chosen can be written as 2 1 ; 3 1 ; 3 2

and obviously, this is giving me different numbers, so here arrangement does matter..so go for nPr => 3P2 = 6

but say if i got to chose 2 people from a total of 3 (A, B and C)

so possibilities are A B ; A C ; B/ C
and if i make this as B A ; C A ; C B
it doesn't make any difference as in the end there are the same people whether A first or B first....so that means arrangement does not matter...we'll use nCr => 3C2 = 3

P.S. These were pretty simple examples, I used them to explain the concept...you'll find this part of my explanation helpful when later on you tackle some more harder situations...this part really makes this thing easy...it did for me ;)
Click to expand...
[plz explain permutations too by using quest from p.papers?
 
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