Mathematics: Post your doubts here!

[Alice123]

7i) P(white box A) = 1/6
The clip taken from box A is placed in box B BEFORE a clip is taken from box B. So as the total number of clips in box B is 9 initially, it comes 10 later.
P( red clip from Box B) = 7/10
P (W,R) = 1/6 * 7/10 = 7/60

ii) Possible probabilities (W,R) (R,R)
The first combo was calculated in part i, for the second combo as a Red clip is taken from box A and placed in box B the number of red clips in Box B is now 8, and the total no of clip is not 10.
P (R,R) = 5/6 * 8/10 = 40/60
total probability =7/60 +40/60 = 47/60

iii) Find the conditional probability. P(R,R) = 40/60
probability that clip from box B red = 47/60
therefore P (R|R) = (40/60)/(47/60) = 40/47

iv) Since a clip is taken from each box, the probability of taking a red clip can be 0 , 1, or 2
for O P(W,W) = 1/6 * 3/10 = 3/60
For 1-
P (W,R) = 1/6 * 7/10 = 7/60
P (R, W) = 5/6* 2/10 = 10/60
Total = 17/60

For 2-
P(R,R) = 40/60

U're fast!!!! THANKS ALOT!!!

 

[Alice123]

7i) P(white box A) = 1/6
The clip taken from box A is placed in box B BEFORE a clip is taken from box B. So as the total number of clips in box B is 9 initially, it comes 10 later.
P( red clip from Box B) = 7/10
P (W,R) = 1/6 * 7/10 = 7/60

ii) Possible probabilities (W,R) (R,R)
The first combo was calculated in part i, for the second combo as a Red clip is taken from box A and placed in box B the number of red clips in Box B is now 8, and the total no of clip is not 10.
P (R,R) = 5/6 * 8/10 = 40/60
total probability =7/60 +40/60 = 47/60

iii) Find the conditional probability. P(R,R) = 40/60
probability that clip from box B red = 47/60
therefore P (R|R) = (40/60)/(47/60) = 40/47

iv) Since a clip is taken from each box, the probability of taking a red clip can be 0 , 1, or 2
for O P(W,W) = 1/6 * 3/10 = 3/60
For 1-
P (W,R) = 1/6 * 7/10 = 7/60
P (R, W) = 5/6* 2/10 = 10/60
Total = 17/60
For 2-
P(R,R) = 40/60

Do u mind explaining the second part?? If a Red clip is taken from box A and placed in box B and the number of red clips in Box B is 8, then how is the total no of clips not 10???

 

[littlecloud11]

Do u mind explaining the second part?? If a Red clip is taken from box A and placed in box B and the number of red clips in Box B is 8, then how is the total no of clips not 10???

Sorry, that was a typo, it is 10. I used 10 in calculations as you can see.
I meant to write that the total no. of clips is 'now' 10 not 'not' 10.

 

[Alice123]

Sorry, that was a typo, it is 10. I used 10 in calculations as you can see.
I meant to write that the total no. of clips is 'now' 10 not 'not' 10.

yea i understood that later... thanks for takin the pain to explain

 

[littlecloud11]

yea i understood that later... thanks for takin the pain to explain

No problem

 

[Esme]

One more thing, the perpendicular bisector and the line y= π/4 x should be dotted lines and not solid ones. The question says < not < and equal to.

You're right Thank you!

 

[suhaib05]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf

Q10 (iii). I have no idea how they worked out the answer and both the mark scheme as well as the er are not helpful :/

 

[PhyZac]

One more thing, the perpendicular bisector and the line y= π/4 x should be dotted lines and not solid ones. The question says < not < and equal to.

So should it be all dotted lines?
like the vertical line equaling 1 is dotted?
and both green lines dotted also?
I really sorry for those question.
Jazaki Allah khiaran for pointing that out..! May Allah s.w.t have mercy on your family and bless you with highest grades in both worlds Aaameen!!

 

[littlecloud11]

So should it be all dotted lines?
like the vertical line equaling 1 is dotted?
and both green lines dotted also?
I really sorry for those question.
Jazaki Allah khiaran for pointing that out..! May Allah s.w.t have mercy on your family and bless you with highest grades in both worlds Aaameen!!

Yup. The should.
Seriously, it okay. I don't mind the questions.

 

[AlishaK]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_61.pdf
Please help me asap. PhyZac Q3 ii) Why do we take for short interval X<a , and for long interval X>a...Why can't we do vice versa.
And also, Q5 a (ii)
Jazak Allah khair!
Cheers!

 

[Esme]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdf

Q10 (iii). I have no idea how they worked out the answer and both the mark scheme as well as the er are not helpful :/

Equation of a line is y=mx+c
Here the y-intercept is zero as it is passing through the origin so c=0
Now take the equation of the curve y=x^2 e^-x
let x be as it is and m will be the equation of gradient that you found in the previous step
i.e m= e^-x(2x-x^2)
So you have
y=mx
x^2 e^-x = e^-x(2x-x^2) x
x^2 = x(2x-x^2)
x= 2x -x^2
x^2 = x
x=1

 

[studen12345]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf
Q4 How to obtain In Terms Of Y
Q10 Second Part
Q3 How To Take Dx/dt Of The Paremateric Eq
Also Plzzz Explain The Question Having Substitution Method Involve Thanku Plz Smbdy Explain Posting Secnd Tym

 

[PhyZac]

Yup. The should.
Seriously, it okay. I don't mind the questions.

Jazaki Allah khiaran!!! Thanks alot ! Alot!

May Allah S.w.T reward you for all you help, and shower his blessing upon you and your family, and bless all of you with a happiness and grant you all Jannatul Firdous Aaameeeen!

In Sha Allah you get all A*'s ! Aameeen

 

[ravaneous]

Why do we equate the denominator of the derivative to zero for value of y

 

[Esme]

View attachment 26119
Why do we equate the denominator of the derivative to zero for value of y

Are you asking about the second part ?

 

[Esme]

View attachment 26119
Why do we equate the denominator of the derivative to zero for value of y

That's because gradient of any line parallel to the y-axis is 1/0
So when you say dy/dx = 1/0
y/(x (3y^3-1) ) = 1/0
Cross multiply and automatically y becomes 0.

 

[Alice123]

(20+12<x<20-12)
now you have to imagine the graph in your head, and you have to come into a conclusion that it is symmetrical.

Now imagine the z of 20+12 that is 32 is x
So the z of 20-12 that is 8 will be a negative number -x

Both of them are x but one is negative, because they are the same distance from the mean

Let the probability of x is p
so the probability of -x is 1-p

we know that difference of the two probability is 0.94

so p - (1-p) = 0.94
p - 1 + p = 0.94
2p = 0.95 + 1 = 1.94
p = 0.97

Now is p is 0.97 z is from the table, 1.882 (or 1.881)

now continue normally

1.882 = 32-20 / sd
sd = 12/1.882
=6.38
This was done by Phyzac before
littlecloud11

 

[littlecloud11]

Thank you both so much!! Much appreciated
PhyZac Alice123

 

[Esme]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_ms_31.pdf

Q7iii) I still don't know how to do these type of questions How to pin point the exact point ? I know we have to make a tangent from the origin to the lowest point on the circle, but which point to take ?
Anika Raisa or PhyZac... Please

 

[PhyZac]

7iii) View attachment 21999
As the question states that the argument of z is least, z must represent the lowest possible point in the shaded region. Draw a tangent from the center to the lowest point on the circle (this doesn't have to be accurate) as shown by the blue line. This line represents |z|. The angle between the radius and the tangent is always 90 so a right-angled triangle is formed. The opposite side, with is equal to the radius is 1 unit. The magnitude of complex u is the hypotenuse and the adjacent represents |z|. you can easily find |u|, √(2^2+2^) = √8.
So, |z| =√(√8)^2 -1^2 = √8-1 =√7

Esme
There you go, littlecloud11 have answered this before.