Mathematics: Post your doubts here!

[i_try9621]

the common difference, r = 2cosθ / √3

-1 < 2cosθ / √3 < 1
multiply with √3, you'll get:

-√3 < 2 cosθ < √3

divide by 2, you 'll get:

-√3/2 < cosθ < √3/2

now taking cosine inverse,

π/6 < θ < 5π/6 (Ans.)

Shouldn't it be only π/6, because of the range given in the question ( 0 < θ < π) ?
Or am I missing something ?

 

[i_try9621]

Component of weight of book is F=mgsin60
normal to plane; R=mgcos60
Use abolute values;
F =0.8*9.8*sqrt(3)/2 = 6.79 N
R = 0.8*9.8*1/2 = 3.92 N.

Ff(Force of friction) = uR = 3.92u, where u is the coefficient of friction, and the applied force of 5N. ie
F = R + 5
6.79 = 3.92u + 5
1.79 = 3.92u
u = 2.19.

Ans. is 0.457 or 0.46 ( 2.d.p)
there is a mistake while dividing.

 

[Anum96]

Ans. is 0.457 or 0.46 ( 2.d.p)
there is a mistake while dividing.

Corrected. Thanks

 

[Rizwan Javed]

Shouldn't it be only π/6, because of the range given in the question ( 0 < θ < π) ?
Or am I missing something ?

The values of theta calculated are already in that range. 5π/6 is smaller than π.

 

[Pixie Twinkles]

Component of weight of book is F=mgsin60
normal to plane; R=mgcos60
Use abolute values;
F =0.8*9.8*sqrt(3)/2 = 6.79 N
R = 0.8*9.8*1/2 = 3.92 N.

Ff(Force of friction) = uR = 3.92u, where u is the coefficient of friction, and the applied force of 5N. ie
F = R + 5
6.79 = 3.92u + 5
1.79 = 3.92u
u = 0.456

Thank you so much!!

 

[Anum96]

Thank you so much!!

You're welcome!!

 

[i_try9621]

Corrected. Thanks

Anytime

 

[i_try9621]

The values of theta calculated are already in that range. 5π/6 is smaller than π.

yeah right, my bad.

 

[The Sarcastic Retard]

6(ii)

 

[The Sarcastic Retard]

Any idea where can I get S2 topical past papers?

 

[The Sarcastic Retard]

9(ii)

 

[i_try9621]

Ques. 7 and Ques. 14 please?

 

[farhan141]

9(ii)

Please tag me when someone solves it. I want to know abt this too.

 

[Anum96]

9(ii)

Please tag me when someone solves it. I want to know abt this too.

 

[Maayee]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdf
can someone help me with q7b
and
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s14_qp_33.pdf ....Q7bi

 

[Eugene99]

the last part (v), how do I solve this?

 

[The Sarcastic Retard]

Thanks.

 

[Mr.Physics]

The weight, X grams, of soup put in a tin by machine A is normally distributed with a mean of 160 g and a standard deviation of 5 g. A tin is selected at random.
(a) Find the probability that this tin contains more than 168 g

Help plz

 

[Eugene99]

The weight, X grams, of soup put in a tin by machine A is normally distributed with a mean of 160 g and a standard deviation of 5 g. A tin is selected at random.
(a) Find the probability that this tin contains more than 168 g

Help plz

z = (X-Mean)/Standard deviation
z value corresponding to X=168 is
z = (168-160)/5 = + 1.6
The area under the standard normal curve right to z = + 1.6 indicates the required probability.
This area lies in the extreme right tail of the normal curve.
The area corresponding to z = 1.6 is 0.4452
The area right to z = 0 is 0.5000
The area beyond z = + 1.6 is or
P(X > 168) = 0.5000 - 0.4452 = 0.0548

 

[Rizwan Javed]

The weight, X grams, of soup put in a tin by machine A is normally distributed with a mean of 160 g and a standard deviation of 5 g. A tin is selected at random.
(a) Find the probability that this tin contains more than 168 g

Help plz

μ = 60 g
σ = 5 g

We're to find the probability that X>168, so

P(X>168)
Standardizing X using Z = X- μ / σ

P (Z > 168 -160 / 5)
= P ( Z > 1.6)
= 1- P(Z<1.6)
= 1 - Ф (1.6)
Using Normal Distribution tables, Ф (1.6) = 0.9452
= 1 - 0.9452
= 0.0548 Ans.