Mathematics: Post your doubts here!

[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.

 

[Eugene99]

View attachment 59497
the last part (v), how do I solve this?

Rizwan Javed could you answer this as well? the last part

 

[Mr.Physics]

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

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.

Tysm guys

 

[Rizwan Javed]

Rizwan Javed could you answer this as well? the last part

Sorry for getting late.
First calculate those two expressions given using the standard deviation and mean you calculated in the previous part. Then from the graph find the cumulative frequencies at these two times you have calculated. Then the number of people who entered the store will be:

(cumulative frequency found from the graph at time m +0.5s ) - (lower cumulative frequecy from the graph at time m - 0.5s )

 

[Eugene99]

Sorry for getting late.
First calculate those two expressions given using the standard deviation and mean you calculated in the previous part. Then from the graph find the cumulative frequencies at these two times you have calculated. Then the number of people who entered the store will be:

(cumulative frequency found from the graph at time m +0.5s ) - (lower cumulative frequecy from the graph at time m - 0.5s )

oh thanks, I think I get that!

 

[The Sarcastic Retard]

7(ii) temme ma mistake :

 

[Konstantino Nikolas]

View attachment 59503
7(ii) temme ma mistake :

again?
Show your soln ... let's try to decode the mistake you made.

 

[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

 

[Sariya Khan]

Is motion under gravity included in 2016's syllabus for M1?

 

[Rizwan Javed]

Is motion under gravity included in 2016's syllabus for M1?

I think it is.

 

[Wkhan860]

View attachment 59503
7(ii) temme ma mistake :

Hope this helps

 

[Anum96]

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

 

[iqbal]

Using the substitution u=√y^3-1 and then using the substitution u=tanQ evaluate∫1/y√y^3-1 dy, giving your answer in terms of y only

 

[Maayee]

View attachment 59509

Thank u soooo much

 

[Konstantino Nikolas]

The second part please. I seem to just not get it right! :/
Sp = Sq + 5 ... and then while using s = ut + 0.5at^2 for both sides, we need to use (t - 2) for LHS since Q is projected 2s later ... but doing it that way leads me to a wrong answer.

Here's the ms:

 

[Konstantino Nikolas]

In this problem, the weight is not acting on the same plane as the other three forces? Is that why they aren't considering weight while resolving vertically?

 

[Rizwan Javed]

View attachment 59514
The second part please. I seem to just not get it right! :/
Sp = Sq + 5 ... and then while using s = ut + 0.5at^2 for both sides, we need to use (t - 2) for LHS since Q is projected 2s later ... but doing it that way leads me to a wrong answer.

Here's the ms:
View attachment 59515
View attachment 59517

Use time for P to be t+2 and for Q to be t only. You'll get the right answer

 

[Copy Cat]

Can someone please explain the following: