Mathematics: Post your doubts here!

[Thought blocker]

-_- May I take the glass and pour it on your head?

Permit granted

 

[Talha Irfan]

Can you please plug in the values given in the table for any one country cause I am a little confused...

Thank you so much for the time and effort

The formula to calculate median (50th
percentile or any other percentile) without the
cumulative frequency curve is :
lower class boundary +
((50n/100 - cf (till prev. class)/ f (of that
interval))
where n is the total number of observed items
or maximum cumulative frequency. First
calculate the corresponding class by checking
in which class the n/2th value lies then
perform the relative ascribed function


For country A
median lies where 300/2 = 150th frequency lies which lies in the interval 20 < x < 35 (or x< 35) and it's corresponding frequency is 159 - 68 = 91

applying in formula
20 + (150 - 68) / 91 = 20.9 (different from your answer*)

similar method of median for B

but in this 2 marks question, all this working is not required, we just need to simply state that median frequency of A lies in interval 20<x<35 while median frequency of B lies in interval 50<x<70 hence median for B is greater than median of A

 

[sitooon]

question 9
Hope my doubts make you also benefit
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
part (i) i got till - 1 / {( 1-x) (1+x}^1/2 * 1 / (1+x)^2 , and its right but i cant simplify it
part (ii) didnt know how to solve it

 

[Hassan Ali Abid]

Can anyone tell me how to find the distance b/w two planes?? Its a question from vectors.(A2)

suppose there are two planes ...
Plane 1: a(x)+b( y )+c(z)+D1=0
Plane 2: a1(x)+b2( y )+c3(z)+D2=0

So what will be the procedure to calculate the distance b/w?? Do we need to calculate the distance of each plane from the origin first and then to subtract the distance of each planes to get the actual distance b/w places or there's an other method to do it ???
and can't we use this formula ie . d= [ D1-D2 / under root a1+b1+c1 .under root a2+b2+c2 ] to calculate the distance ???

 

[midha.ch]

question 9
Hope my doubts make you also benefit
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
part (i) i got till - 1 / {( 1-x) (1+x}^1/2 * 1 / (1+x)^2 , and its right but i cant simplify it
part (ii) didnt know how to solve it

first part after getting dy/dx I used (a+b)(a-b) = a^2 - b^2 to get (1- x^2)^0.5

 

[midha.ch]

question 9
Hope my doubts make you also benefit
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
part (i) i got till - 1 / {( 1-x) (1+x}^1/2 * 1 / (1+x)^2 , and its right but i cant simplify it
part (ii) didnt know how to solve it

for part (ii) they asked for max. value of gradient
so I considered the equation for gradient as m and took dm/dx
tip for these kinda sums
when in calculus you are asked to find maximum or minimum value of something, first differntiate it and then equate that to 0
you get one of the variables this way and then use it to find any other information the question indicates

And yes! your doubt helps me a lot
I am quite in a messy situation regarding chem so I don't really pay the required attention to maths but solving your doubts helps me practice the critical sums
so Thank YOU

 

[sitooon]

for part (ii) they asked for max. value of gradient
so I considered the equation for gradient as m and took dm/dx
tip for these kinda sums
when in calculus you are asked to find maximum or minimum value of something, first differntiate it and then equate that to 0
you get one of the variables this way and then use it to find any other information the question indicates

And yes! your doubt helps me a lot
I am quite in a messy situation regarding chem so I don't really pay the required attention to maths but solving your doubts helps me practice the critical sums
so Thank YOU

 

[sitooon]

 

[midha.ch]

View attachment 38159

 

[Faaiz Haque]

2) , 5 i)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_4.pdf

ty

 

[Muskan Achhpilia]

The formula to calculate median (50th
percentile or any other percentile) without the
cumulative frequency curve is :
lower class boundary +
((50n/100 - cf (till prev. class)/ f (of that
interval))
where n is the total number of observed items
or maximum cumulative frequency. First
calculate the corresponding class by checking
in which class the n/2th value lies then
perform the relative ascribed function


For country A
median lies where 300/2 = 150th frequency lies which lies in the interval 20 < x < 35 (or x< 35) and it's corresponding frequency is 159 - 68 = 91

applying in formula
20 + (150 - 68) / 91 = 20.9 (different from your answer*)

similar method of median for B

but in this 2 marks question, all this working is not required, we just need to simply state that median frequency of A lies in interval 20<x<35 while median frequency of B lies in interval 50<x<70 hence median for B is greater than median of A

Thanks!

 

[Muskan Achhpilia]

when line and curve intersects
equation of curve = equation of line
x^2 - 2x = kx - 4
x^2 +(-k-2)x + 4 = 0
a=1, b= -k-2, c=4
given, line intersects curve at 2 distinct points
therefore, b^2 -4ac >0
after solving the inequality, take the 2 critical values of x
write the possible ranges
find the value of individual term in that range
eg- (k+6) is -ve when k<-6
then multiply the signs for each possible range
eg- (k+6)(k-2) = -ve multiplied by -ve = +ve
our initial inequality was greater than 0
thus the range is k<-6 or k>2

Thank you so much!

 

[wohooo]

Anyone can help with question 10iii and 7iii http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

 

[unique111]

Anyone can help with question 10iii and 7iii http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf

For question # 7iii) Given: u = 1 + 2i/1-3i
Taking argument on both sides we have; arg(u)=arg(1 + 2i/1-3i)
or,arg(u)=arg(1+2i)-arg(1-3i)
[From 7i: arg(u)=tan-1(-.5/.5)--->3pi/4] or, 3pi/4=tan-1(2)-tan-1(-3)
Therefore, 3pi/4=tan-1(2)+tan-1(3) shown

 

[Thought blocker]

For question # 7iii) Given: u = 1 + 2i/1-3i
Taking argument on both sides we have; arg(u)=arg(1 + 2i/1-3i)
or,arg(u)=arg(1+2i)-arg(1-3i)
[From 7i: arg(u)=tan-1(-.5/.5)--->3pi/4] or, 3pi/4=tan-1(2)-tan-1(-3)
Therefore, 3pi/4=tan-1(2)+tan-1(3) shown

A2 level ??

 

[unique111]

yes, bro.

 

[Talha Irfan]

Thanks aloot can u help me in ths too http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_63.pdf Q2 ii)

1) Find variance using the assumed mean
formula
2) Use this variance value in the formula
which calculates variance with actual values
3) solve and you will get the required answer

 

[Adiizz]

Hi
Please help me with number 7 vector.
Thank you!

 

[unique111]

Hi
Please help me with number 7 vector.
Thank you!

For 7i) General Point of line: x=s y=1-2s and z=1+s
Place these values in the equation of the plane, you see that everything on both sides of the euation gets cancelled out. Therefore, line l lies in the plane p.
For 7ii) Directions of the new plane will be the same as the line's(since they are parallel) and the other direction will be the normal of plane p (as it is perpendicular). From these two directions you find the common perpendicular which will be the normal of new plane.
It is given that point (2,1,4) lie on the new plane, so with a point and a normal you can find the equation of the new plane. a.n=r.n
where, a=(2,1,4) n=normal and r=(x,y,z).
hope you get it.

 

[sitooon]

Part (ii) of both questions . . .
I know how to draw argand diagram , but cant figure out how to find K