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Mathematics: Post your doubts here!

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Hey,
i wanted some good notes for my paper 1 and paper 6 because i'm done with revision and past papers from 2003-2011
but still i need to revise, so please :)
 
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1) Open the brackets.
(1-4x+4x^2) (1+6ax+15a^2 x^2...) [you don't need anymore since we want the first three terms]
2) The first term we already have 1
Second term- find all the combinations between the brackets that will give us "-x"
(4x*1)+(1*6ax) = -x [divide both sides by 'x']
4+6a= -1
a=1/2
3) The third term- find all the combinations between the brackets that will give us "bx^2"
(4x^2*1)+(-4x*6ax)+(1*15a^2 x^2)= bx^2
4x^2 - 24 ax^2 + 15 a^2 x^2 = bx^2 [divide both sides by x^2)
4 - 24a +15a^2 = b [put the value of 'a' in]
b = -41/2

I'm not sure whether i have explained it in a proper manner, if you still have a doubt i'll try to explain it in another way
dude how bout tht anthr way....m still kinda confused here.... :cautious:
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_61.pdf
Someone please help me solve question no. 6 - the whole question. I'll appreciate an explanation..

Thanks

Probability she is on time = p = 1-0.75-0.05
p= 0.2
q= 1-0.2 = 0.8
n=96

We will be using normal approximation to solve this question...

Using Normal Approximation To Binomial

mean = np = 96*0.2 = 19.2
SD = √npq = √96*0.2*0.8 = √15.36

P(X<20)
Continuity Correction P(X<19.5)
P(z<[19.5-19.2]/√15.36)
P(z<0.07654)
Check the value of z from the normal distribution table.
= 0.531


ii)
She eats a banana if early = 0.7
She doesn't eat if early = 0.3
She eats a banana if on time = 0.4
She doesn't eat if on time = 0.6
She doesn't eat a banana if late = 1

probability she is late and doesn't eat a banana = 0.75*1 = L-B'
probability she is on time and doesn't eat a banana = 0.2*0.6 = OT-B'
probability she is early and doesn't eat a banana = 0.05*0.3 E-B'


(Given She Doesn't Eat A Banana| She Is On Time) (B'|E)

It means that all conditions of (her not eating a banana) will go in the denominator while (her being on time and not eating the banana) will go in the numerator

OT-B'/(L-B' + E-B' + OT-B')

(0.2*0.6)/[(0.75*1) + (0.05*0.3) + (0.2*0.6)]

= 8/59 Answer
 
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PLEASE HELP THIS QUESTION
A manufacture of biscuits produces 3 times as many cream one as chocolate one.Biscuits are chosen randomly and packed into boxes of 10.Find the probability that a box contain equal number of cream and chocolate biscuit.
There are total 10 biscuits, and equal number means, 5 cream and chocolate.
Probability cream is 3/4 [ because ratio is 3:1 ]
and Probability not cream (that is chocolate) is 1/4

now to get 5 cream we use binomial distribution method.

10C5 x (3/4)^5 x (1/4)^5 = 0.0584
 
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can anybody help me with quest 6 ii. Im stuck. Thanks
 

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i need some p4 revision notes plz ..... am really stuck in mechanics especially in work.energy,power thingy :'(
 
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Anyone have some helpful notes for p5 (mech 2)? Especially for the equilibrium topics and maybe elastic spring and strings
 
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Probability she is on time = p = 1-0.75-0.05
p= 0.2
q= 1-0.2 = 0.8
n=96

We will be using normal approximation to solve this question...

Using Normal Approximation To Binomial

mean = np = 96*0.2 = 19.2
SD = √npq = √96*0.2*0.8 = √15.36

P(X<20)
Continuity Correction P(X<19.5)
P(z<[19.5-19.2]/√15.36)
P(z<0.07654)
Check the value of z from the normal distribution table.
= 0.531


ii)
She eats a banana if early = 0.7
She doesn't eat if early = 0.3
She eats a banana if on time = 0.4
She doesn't eat if on time = 0.6
She doesn't eat a banana if late = 1

probability she is late and doesn't eat a banana = 0.75*1 = L-B'
probability she is on time and doesn't eat a banana = 0.2*0.6 = OT-B'
probability she is early and doesn't eat a banana = 0.05*0.3 E-B'


(Given She Doesn't Eat A Banana| She Is On Time) (B'|E)

It means that all conditions of (her not eating a banana) will go in the denominator while (her being on time and not eating the banana) will go in the numerator

OT-B'/(L-B' + E-B' + OT-B')

(0.2*0.6)/[(0.75*1) + (0.05*0.3) + (0.2*0.6)]

= 8/59 Answer
Thank you very much bro, you have really helped me here.

I have one more problem in the same paper. I don't understand how to solve qn no. 7 (b ii) and 7 (c).
Could you please help me understand the solutions?

Hope I'm not bothering you..
 
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It is 90 degree. You can draw a graph and look at the value for which the function is one to one.
Use tan=sin/cos first and then open the square. The combine the fraction and use sin^2=1-cos^2. Use the difference of squares identity and cut the '1-cosx' from the numerator and denominator. And you are done.
 
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second chapter
linear combination of random variables

Three discs are drawn at random from a box containing a number of red and blue discs.
the mean and variance of the number of blue discs drawn are 2 and 0.5 respectively.what are the mean and variance of the red discs drawn?
 
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