Mathematics: Post your doubts here!

[PhyZac]

questions abt statistics
thanks a lot！！

See.
What we have to find is Probability between 7.8 and 11

Now, since the variable is modelled normally, no need for continuity correction ( adding or subtracting 0.5) ( u only need to do tht for binomial approximation.)

So, since 7.8 is the mean , so the probability will be 0.5

and 11 u do it normal way 11-mean/standard deviation. get probability.
Subtract the 0.5 from answer and u get it.

 

[tompapaya]

See.
What we have to find is Probability between 7.8 and 11

Now, since the variable is modelled normally, no need for continuity correction ( adding or subtracting 0.5) ( u only need to do tht for binomial approximation.)

So, since 7.8 is the mean , so the probability will be 0.5

and 11 u do it normal way 11-mean/standard deviation. get probability.
Subtract the 0.5 from answer and u get it.

Thank you very much for your answer! I'm sorry I still have some doubts.
I am very confused why the mean is probability is 0.5？
another question is
Isn't it as long as it is modelled normally, I should not adding or subtracting 0.5？
Thanks!

 

[PhyZac]

Thank you very much for your answer! I'm sorry I still have some doubts.
I am very confused why the mean is probability is 0.5？
another question is
Isn't it as long as it is modelled normally, I should not adding or subtracting 0.5？
Thanks!

See the mean is right in the middle of the graph, and so it is 0.5,
you can try it,
7.8 - 7.8 / 2.8
= 0/ 2.8
= 0
now by seeing the table u will find 0 as 0.5

When the variable is modelled normally DO NOT add or subtract 0.5

When you make a binomial approximation you HAVE to add or subtract.

In that question it is clearly mentioned that the variable is normally distributed.

Anymore doubts, feel free.

 

[tompapaya]

See the mean is right in the middle of the graph, and so it is 0.5,
you can try it,
7.8 - 7.8 / 2.8
= 0/ 2.8
= 0
now by seeing the table u will find 0 as 0.5

When the variable is modelled normally DO NOT add or subtract 0.5

When you make a binomial approximation you HAVE to add or subtract.

In that question it is clearly mentioned that the variable is normally distributed.

Anymore doubts, feel free.

perfect interpretation！I got it！I would be very grateful！

 

[GorgeousEyes]

4.i) In this one we use the formula F=ma instead of F, i will use T, since T(tension) is the force working here.
So for Particle A , (there is no friction, only weight and tension) 2-T=0.2a
For particle B , (there is friction and tension) T - uR= 0.3a << in this equation i used Friction=uR(u is the coefficient of friction)
Now we have 2 equations 2-T=0.2a
T-uR=0.3a
Cancel out the Ts. And since its not moving(limiting equilibrium) acceleration = O
We end up with 2-u3=o , u= 2/3 (3 is the normal contact force=R)

4.ii)Since there is upward force acting upon it, we need to find the new normal contact force. So, R = downward force - upward = 3 - 1.8 =1.2
Now also find out the friction =uR = 2/3 x 1.2 = 0.8
We make the equation for new force X.(its going to right away from pulley) X = Frictional Force + Tension
We don't have the tension we find it out by the help of particle A.(tension is same on the both sides)
For particle A, only weight is working. so Weight - Tension = 0 (cause there is no acceleration)
2 - T = 0 , T= 2
Now we can find out X, X= Frictional Force + Tension
= 0.8 + 2 = 2.8
Hope it helped

Thank youu Raiyaan

 

[GorgeousEyes]

variant 42

No.4
i) By resolving .. Fcos=12cos30 in x-axis direction(1)
Fsin+12sin30=10 in y-axis direction.(2)
Make Fsin the subject in the second equation .. Fsin=1o-12sin30=4
Do you know that Sin /cos = tan , right ? So to get rid of sin and cos in the equation ..we will divide equation 2 over equation 1 .. F will be cancelled from the equation and sin/cos will be tan ---> Tan=4/12cos30 ... Shift tan you willl find the angle = 21.1 .. and by putting this value in the fist or the second equation your F will be 11.13
ii) 12 is not here anymore . right ? so you have the force f to resolve :
So in x-axis direction : Fcos = 11.1cos21.1 (1)
in y-axis direction : Fsin-10 ---> 11.1sin21.1-10(2)
R = 1square +2square under square root .
and direction : tan (2)/(1)
Hope u understand incha'allah

 

[GorgeousEyes]

No.4
i) By resolving .. Fcos=12cos30 in x-axis direction(1)
Fsin+12sin30=10 in y-axis direction.(2)
Make Fsin the subject in the second equation .. Fsin=1o-12sin30=4
Do you know that Sin /cos = tan , right ? So to get rid of sin and cos in the equation ..we will divide equation 2 over equation 1 .. F will be cancelled from the equation and sin/cos will be tan ---> Tan=4/12cos30 ... Shift tan you willl find the angle = 21.1 .. and by putting this value in the fist or the second equation your F will be 11.13
ii) 12 is not here anymore . right ? so you have the force f to resolve :
So in x-axis direction : Fcos = 11.1cos21.1 (1)
in y-axis direction : Fsin-10 ---> 11.1sin21.1-10(2)
R = 1square +2square under square root .
and direction : tan (2)/(1)
Hope u understand incha'allah

sorry I don't know how to solve no.5 unfortunately =/

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_61.pdf

Question 2: Explain it as much as you can and solve it completley .. since I don't understand it at all :S

Dug PhyZac and anyone else who has statistics

EDIT: Ah Got It.. It was tricky .. but opening the bracks did the trick...

 

[asexamskillme111]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_43.pdf

Q5 part (ii)
So I've found the value of R and I can't get myself to understand where theta is exactly. Can someone help by drawing a diagram or otherwise? Thank you.

 

[GorgeousEyes]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_43.pdf

Q5 part (ii)
So I've found the value of R and I can't get myself to understand where theta is exactly. Can someone help by drawing a diagram or otherwise? Thank you.

You don't need to know where theta is . theta is the direction and there's a simple rule u can calculate it by .. Tan(y-axis/x-axis)
y-axis is the resolving in y-axis direction , and x-axis is the resolving in the x-axis direction .
so u already know x-axis direction=85 and y-axis direction = 42 .. so 42/85 shift tan = 26.3 .. that's it
Hope u get it.

 

[parthrocks]

c u can do it like this :
separate the 4 and the 10s :-

4^4 x (10^-1 x 10^-2 x 10^-3 x 10^1)
= 256 x 10^-1-2-3+1
= 256 x 10^-5
= 0.00256
=0.002560

don't be confused because of the last zero because it is a trick. it does not actually matter how many zeros we put at the end the answer is still 0.002560

get it??

Great work...Hats off!

 

[Tkp]

sorry I don't know how to solve no.5 unfortunately =/

the particles p and q has a initial speed of 12 and 7 and the final velocity would be zero as they are telling Each particle comes to
rest on returning to the ground.so u know v=0,u =12 and a=g.now get the time for p and same as q.now put the set values like t>0.7 or <1.2

2nd 1 the particles are at the same height.so u need to find the time at first for the same height.apply the formula s=ut+.5at^2 that means 3(ut+.5t^2)=8(ut+.5t^2)

now u get time 0.8s.so now apply the formula v=u-gt and u will get the velocity of the two particles

 

[Fatima18]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_S08_qp_01.pdf
Guys, how to do 8ii?

 

[syed1995]

the particles p and q has a initial speed of 12 and 7 and the final velocity would be zero as they are telling Each particle comes to
rest on returning to the ground.so u know v=0,u =12 and a=g.now get the time for p and same as q.now put the set values like t>0.7 or <1.2

2nd 1 the particles are at the same height.so u need to find the time at first for the same height.apply the formula s=ut+.5at^2 that means 3(ut+.5t^2)=8(ut+.5t^2)

now u get time 0.8s.so now apply the formula v=u-gt and u will get the velocity of the two particles

You have maths as well?

 

[Raiyan3]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_S08_qp_01.pdf
Guys, how to do 8ii?

Just substitute one of the value of K found in 8.i) in the equation of fg(x)=x. Which was i believe 36/2-x +2k = x. you will end up with a quadratic equation and solve it then!

 

[syed1995]

Put values of k in f[g(x)]=x

then solve f[g(x)]-x=0

Put 5 or -7 in (x^2 + 2kx - 2x + 36 - 4k) and equate to 0.


x^2 + 8x + 16 =0 or x^2 - 16x + 64 =0

Roots come as -4 when k=5 and 8 when k=-7

 

[Fatima18]

Put values of k in f[g(x)]=x

then solve f[g(x)]-x=0

Put 5 or -7 in (x^2 + 2kx - 2x + 36 - 4k) and equate to 0.


x^2 + 8x + 16 =0 or x^2 - 16x + 64 =0

Roots come as -4 when k=5 and 8 when k=-7

Thank you!
Can you please tell me how to do number 8i) in oct paper 2008...I dont know how to find da gradient through differentiation?

 

[syed1995]

Thank you!
Can you please tell me how to do number 8i) in oct paper 2008...I dont know how to find da gradient through differentiation?

Just differentiate the equation and put the value of x in it to get gradient.

y=5-8x^-1
dy/dx=8/x^2

since co-ordinate is (2,1) and x =2

put x = 2 in dy/dx

gradient = 8/2^2
gradient = 2

normal gradient will be found with m1 x m2 = -1

so normal gradient will be -1/2

 

[Fatima18]

Just differentiate the equation and put the value of x in it to get gradient.

y=5-8x^-1
dy/dx=8/x^2

since co-ordinate is (2,1) and x =2

put x = 2 in dy/dx

gradient = 8/2^2
gradient = 2

normal gradient will be found with m1 x m2 = -1

so normal gradient will be -1/2

So why is it that dx over dy is= 8 over x squared?

 

[syed1995]

So why is it that dx over dy is= 8 over x squared?

Basic Differentiation.

-8/x can be written as -8(x)^-1

Differentiating that will be .. -8(x)^-1-1 * -1

which becomes 8(x)^-2 which can also be written as 8/x^2

Like if y= 2x^n then dy/dx = 2x^n-1 * n = 2n(x^n-1)

working with powers while writing on a forum is difficult.