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Mathematics: Post your doubts here!

asd

asd

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itallion stallion said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf question 10 part1

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_61.pdf q6 part II.


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_33.pdf Q7 part iii.

Thanks a lot!
Click to expand...
Well, For the stats question:
They say there's 600 feathers, with 63 less than 6 cm, 155 more than 12 cm.
so for P(X<6)= 63/600=0.105 and for P(X>12)= 155/600=0.258
(by checking from the normal distribution table you'd see theres no 0.105 or 0.258! so you gotta subtract them from 1)
1-0.105=0.895 and 1-0.258=0.742
You get 2 equations; (i) (6-u)/S.D = -1.253* & (ii) (12-u)/S.D = 0.65
Solve simultaneously and they go down to:
6 + 1.253S.D = 12 - 0.65.D
=S.D = 3.15
and you can calculate the mean (u) now.

*Note that I put the negative sign with 1.253 in equation (i) because if you have a case where there's a probability for something less than a certain value e.g: P(X<a), and its probability is less than 0.5 e.g: P(X<a) = 0.2, you put a -ve sign. While there's no negative sign in equation (ii) because the given probability is for something greater than a certain value e.g: P(X>a) and its probability is less than 0.5 e.g: P(X>a)=0.6. I can't explain the reason online without diagrams, but there's a general rule, if there's a less than "<" sign, and the probability is less than 0.5, you put -ve sign. Also if there's a greater than ">" sign with the probability greater than "0.5" you put a -ve sign. And if there's a greater than ">" sign, and the probability is less than 0.5 you put a +ve sign. Also, when theres a less than "<" sign with probability greater than 0.5, you put a +ve sign.

IGNORE THIS IF YOU ALREADY KNOW ABOUT THIS. IT MAYBE REALLY CONFUSING.
 
itallion stallion

itallion stallion

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asd said:
Well, For the stats question:
They say there's 600 feathers, with 63 less than 6 cm, 155 more than 12 cm.
so for P(X<6)= 63/600=0.105 and for P(X>12)= 155/600=0.258
(by checking from the normal distribution table you'd see theres no 0.105 or 0.258! so you gotta subtract them from 1)
1-0.105=0.895 and 1-0.258=0.742
You get 2 equations; (i) (6-u)/S.D = -1.253* & (ii) (12-u)/S.D = 0.65
Solve simultaneously and they go down to:
6 + 1.253S.D = 12 - 0.65.D
=S.D = 3.15
and you can calculate the mean (u) now.

*Note that I put the negative sign with 1.253 in equation (i) because if you have a case where there's a probability for something less than a certain value e.g: P(X<a), and its probability is less than 0.5 e.g: P(X<a) = 0.2, you put a -ve sign. While there's no negative sign in equation (ii) because the given probability is for something greater than a certain value e.g: P(X>a) and its probability is less than 0.5 e.g: P(X>a)=0.6. I can't explain the reason online without diagrams, but there's a general rule, if there's a less than "<" sign, and the probability is less than 0.5, you put -ve sign. Also if there's a greater than ">" sign with the probability greater than "0.5" you put a -ve sign. And if there's a greater than ">" sign, and the probability is less than 0.5 you put a +ve sign. Also, when theres a less than "<" sign with probability greater than 0.5, you put a +ve sign.

IGNORE THIS IF YOU ALREADY KNOW ABOUT THIS. IT MAYBE REALLY CONFUSING.
Click to expand...
Thank u so much for the detailed answer.awesome.
Can u plz solve the other too?
 
SitiPutri

SitiPutri

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asd

asd

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itallion stallion said:
Thank u so much for the detailed answer.awesome.
Can u plz solve the other too?
Click to expand...
q10 part 1 for oct/nov
(Lim from 0 to 1) Int(u^(n+2) + u^n)

converting the limits, tan(pi/4)= 1 tan(0)=0
converting dx into du, u=tanx => du/dx= sec^2(x)= 1+tan^2(x)
dx= du/1+u^2

Now, lim(0-1) int[u^n(u^2 + 1)]/ (u^2 + 1)
(u^2 +1) cancels out. so youre left with int(u^n)du

= u^n+1/n+1
put in the limits and you get 1/n+1
 
Rutzaba

Rutzaba

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an extremely idiotic atempt
Heres the deal

If you have an equation

Ax²-bx +c =0

Frst make the coefficient of x²=1

We ignore the c and see only the a and b part

To remove the a and make it one we take common a that wud bring us to

A (x²- (b/A)x) +c

Then we divide the b wali term by 2 ( so that we can justify the 2ab part when we expand (a -b)^2)

A(x² - (b/2A)x) +c

Add the (b/2) square and subtract b/2 square

[A(x²-(B/2A)- (B ²/4A) +(B ²/4A)]+C

Next we take out the +(b ²/4A) out of the bracket

Remember in the process it will be multiplied by A

b²/4A wen multiplied by A will give B ²/4

thus

A[(x²-(B/2A) +(B ²/4A)] -B ²/4 +c= 0

Can u see the statement inside the []?

Does it resemble a ² - 2ab +c ?

So A[x- (B/2A)] ²] +c-B ²/4
robinhoodmustafa
 
Y

yuv2404

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asd said:
Gradient = dy/dx = PN/TN= tanx = 1/2(PN)(TN)

For part (i): Rearrange PN/TN = tanx to give TN= PN/tanx
Substitute this into Area of the triangle (1/2 PN*TN) which gives 1/2* (PN)*(PN/tanx)
Note that PN= y.
So it goes down to 1/2. y^2. cotx

Now for part (ii): Separate the variables and integrate.
int(2/y^2) dy = int(cotx)dx
= -2/y = int(cosx/sinx) dx
= -2/y = ln(sinx) + c

Plug in (Pi/6, 2) to evaluate c. c= -0.31 Idk why they have it as 0.3 in the ms.
then rearrange to express y interms of x.
Click to expand...
asd said:
Gradient = dy/dx = PN/TN= tanx = 1/2(PN)(TN)

For part (i): Rearrange PN/TN = tanx to give TN= PN/tanx
Substitute this into Area of the triangle (1/2 PN*TN) which gives 1/2* (PN)*(PN/tanx)
Note that PN= y.
So it goes down to 1/2. y^2. cotx

Now for part (ii): Separate the variables and integrate.
int(2/y^2) dy = int(cotx)dx
= -2/y = int(cosx/sinx) dx
= -2/y = ln(sinx) + c

Plug in (Pi/6, 2) to evaluate c. c= -0.31 Idk why they have it as 0.3 in the ms.
then rearrange to express y interms of x.
Click to expand...
thank you very much :)
 
F

Fahm Deen

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May/June 2010 /11 no.1 .i and ii
Can anyone please give the solution with explanation and provide me with formulas like this related with sin and cos also.
 
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