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Mathematics: Post your doubts here!

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I have faced a problem in Statistics 1. I hope I can find help here. "Two discs are drawn without replacement from a box containing 3 red and 4 white discs. If X is the random variable "The number of white discs drawn", construct a probability distribution table and then find E(x), E(x^2) and Var(x)." Thanks :)
 
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I have faced a problem in Statistics 1. I hope I can find help here. "Two discs are drawn without replacement from a box containing 3 red and 4 white discs. If X is the random variable "The number of white discs drawn", construct a probability distribution table and then find E(x), E(x^2) and Var(x)." Thanks :)
this means u can draw :
- o discs and here the probabilitiy will equal to 3/7 x 2/6= 1/ 7
- 1 disc and here the probablitiy equals (3/7 x 4/6)+(4/7 x 3/6) =4/7
- 2 discs and the probability equals (4/7 x 3/6)=2/7
you put these values in a table
to find E(X) , you multiply the probability by the value of X= (o x 1/7)+(1 x 4/7)+ (2 x 2/7)
E(X^2) : u square each X value before multiplying it by it's probability as above
Var(X)= E(X^2)- (E(X))^2
hope this helps
 
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how do you do part (ii)?

{3^y}^3=27^y, {3^y}^2=9^y .....So if u compare the f(x) and equation in (ii), its the same... if u substitute the value of x as 3^y in f(x), u will get the same equation shown in (ii)... So, do this way..

Given equation:12*27^y+25*9^y-4*3^y-12=0
or, 12* {3^y}^3 + 25* {3^y}^2 - 4* 3^y - 12=0
comparing this equation with f(x), we get x=3^y

Now see, the value of x is equal to -2... -2 is the only value which makes f(x) zero...so same is the case there..its given the equation=0...so x=3^y=-2 or, 3^y=-2..use log and find the answer correct to 3sf.
 
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How to do this question especially part ii. Please help me.


Well, u have found out w and w^2...so..plot it in a graph..i assume, its not difficult for u to plot in a graph..then join those two points. Now, the line u can see is a diameter of a circle. so u can put a ruler to find out a length..and divide it into two half...the pt which divides the line into two equal halves is the centre of circle...So lets assume that pt is 3,4 which u can easily find out from the graph. Now, for radius, use distance formula to calculate the length of w to w^2 and half of it is the radius u are gonna use...We measured the length above using ruler just to find out the centre of circle... that length can't be used as a radius of diameter of a circle...So, u got a radius using distance formula and centre from a graph. Now, If Centre if (3,4) and radius is 5, u can simply write it as |z-(3+4i)|=5...got it?
 
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Help!!! I don't understand the working for paper 7 may june 2003 Q7 part ii....
why they need to divide surd n???

My teacher once told me about it, but i think i forget...I think this is what he told me. The experiment is done within a population and then sample is taken..so u have to divide it that way....If sample is taken and find out mean and all..u dont need to do that way....actually, i think this is..but m not sure....!! Btw, u dont need to worry about it. In s2, everytime u solve questions related to Normal distribution, u need to divide...
 
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{3^y}^3=27^y, {3^y}^2=9^y .....So if u compare the f(x) and equation in (ii), its the same... if u substitute the value of x as 3^y in f(x), u will get the same equation shown in (ii)... So, do this way..

Given equation:12*27^y+25*9^y-4*3^y-12=0
or, 12* {3^y}^3 + 25* {3^y}^2 - 4* 3^y - 12=0
comparing this equation with f(x), we get x=3^y

Now see, the value of x is equal to -2... -2 is the only value which makes f(x) zero...so same is the case there..its given the equation=0...so x=3^y=-2 or, 3^y=-2..use log and find the answer correct to 3sf.

Wrong solution to this question, we'll use 3^y=2/3.
 
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Wrong solution to this question, we'll use 3^y=2/3.

Thanks for correction..I didn't realise log of negative values don't exist...Well...the only u need to keep in mind is that, whichever value that makes f(x) zero can be the value of x...!! -2 was given in question..so i didnot think anything else..!! my apologies..m sorry
 

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im stuck at paper 1 may june 2005 question 2
Paper: http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s05_qp_1.pdf
MS: http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s05_ms_1.pdf
could you please explain it to me
i dont understand in the marksheet how '-12(x2 - 4x)to the power -2'
Assalamoalaikum!

This one is a differentiation question! Have you done the differentiation chapter? Differentiate the eqn given, and u'll get this solution...then put x = 3 to find the gradient...
 
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Assalamoalaikum!

This one is a differentiation question! Have you done the differentiation chapter? Differentiate the eqn given, and u'll get this solution...then put x = 3 to find the gradient...

wsalam,
yes i have done the diffrenciation chapter but i cant understand how the chain rule has been used here?
how does -12(x2-4x) come as in the marksheet?
 
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There is no chain rule/product rule in AS-level Maths, just your basic differentiation and bracket differentiation. For that particular question you bring the entire denominator in the numerator and write it's power as -1 and solve it using bracket differentiation like so:
 

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For May/June 2011 paper 12.. question number 4

for 4 (i) i dont face any problems.
ive found the equation of the tangent

y = -3x -4

For 4 (ii)
I couldn't solve it.. check my answer for (i) and check the question (first pic)..

AHMED NASSER
 

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For May/June 2011 paper 12.. question number 4

for 4 (i) i dont face any problems.
ive found the equation of the tangent

y = -3x -4

For 4 (ii)
I couldn't solve it.. check my answer for (i) and check the question (first pic)..

AHMED NASSER
AoA!
The equation should be y + 3x = 8. (You've written -6 as the product of -3 and -2 in your working. It should be 6.)
The angle can be found by the equation tan A = -3. The angle A is thus 108.4*. This is because the tangent of the angle that a line makes with the x-axis is equal to its gradient.
 
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i need help in p3
chp combination n permutation ..
m unable to solve th buk exercise n mixing both of them ..
tell me wht cn i do ???
 
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Can someone please hel me in this question?
Question 4 part ii

I don't get how to find the answer
 

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Can someone please hel me in this question?
Question 4 part ii

I don't get how to find the answer
the ans to part a should be multiplied with (1+kx). pick the terms of x^2 from the answer and equate it to zero. u will get the value of k
 
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Can someone please hel me in this question?
Question 4 part ii

I don't get how to find the answer

In the first part, you'll obtain the answer '64 - 192x + 240x^2'. For the second part, you'll multiply '64 - 192x + 240x^2' with '1 + kx' and equate the coefficients of 'x^2' to 0. Like this:

(1 + kx) (64 - 192x + 240x^2)
64 - 192x + 240x^2 + 64kx - 192kx^2 + 240kx^3

Take the coefficients of 'x^2' and equate them to 0:

240 - 192k = 0
240 = 192k
1.25 = k
 
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