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Mathematics: Post your doubts here!

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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
 
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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.

 
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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 :D
 
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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 )
 
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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! :)
 
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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
 
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OnPaste.20160227-210011.png
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:
OnPaste.20160227-210556.png
OnPaste.20160227-211301.png
 

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OnPaste.20160227-211240.png

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?
 
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