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Mathematics: Post your doubts here!

O

osama ahmed ibrahim

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s1.JPG 1.JPG
help in part ii
i didnt take log function so how they expect me to use it
also i cant understand the answer help me i always leave this q blank
i want clear explanation and an alternative to log function if possible in part II
 
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Rizwan Javed

Rizwan Javed

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Wkhan860 said:
http://papers.gceguide.com/A Levels/Mathematics (9709)/9709_s13_qp_63.pdf

Question 7(iii) please.

Rizwan Javed
Click to expand...
Consider poodles are represented by 'O'
and spaniels and retrievers both are represent by *
the way they can be arranged such that no poodle is together is:
O * O * O * O * O

Now arrange the spaniels and retrievers in their places represent by *. This will be done in 4! ways.

Now from the 5 possible places of poodles, you are to select 3 since there are 3 poodles. So select these places 5C3. Now 3 poodles can be arranged in their selected places in 3! ways.

So the total no. of possible arrangement are:

4! * 5C3 * 3! = 1440 Ans.
 
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Wkhan860

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Rizwan Javed said:
Consider poodles are represented by 'O'
and spaniels and retrievers both are represent by *
the way they can be arranged such that no poodle is together is:
O * O * O * O * O

Now arrange the spaniels and retrievers in their places represent by *. This will be done in 4! ways.

Now from the 5 possible places of poodles, you are to select 3 since there are 3 poodles. So select these places 5C3. Now 3 poodles can be arranged in their selected places in 3! ways.

So the total no. of possible arrangement are:

4! * 5C3 * 3! = 1440 Ans.
Click to expand...
Thanks!

Same paper qtn 4 part (ii)
The MS says that we have to use continuity correction here but theres no mention of the word approximation in the question. Why is that?
 
Rizwan Javed

Rizwan Javed

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Wkhan860 said:
Thanks!

Same paper qtn 4 part (ii)
The MS says that we have to use continuity correction here but theres no mention of the word approximation in the question. Why is that?
Click to expand...
Though the question doesnot state this, but since you are using NORMAL APPROXIMATION to solve a question which involves a binomial distrbution you'll have to do continuity correction.
 
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Wkhan860

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Rizwan Javed said:
Though the question doesnot state this, but since you are using NORMAL APPROXIMATION to solve a question which involves a binomial distrbution you'll have to do continuity correction.
Click to expand...
But we do not use continuity correction in ever normal distribution question. So why so in this case?
 
Rizwan Javed

Rizwan Javed

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Wkhan860 said:
But we do not use continuity correction in ever normal distribution question. So why so in this case?
Click to expand...
We don't use continuity correction only when it is stated in the question that a certain variable follows a normal distribution. Otherwise, if YOU are using normal distribution to make an approximation for any other distribution like binomial , you have to use continuity correction.
 
Rizwan Javed

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osama ahmed ibrahim said:
also i cant understand what q wants in part ii
View attachment 60261
View attachment 60260
Click to expand...
In part (ii) you are required to prove the inequality.

0<θ<90 means that it lies in the first quadrant. In the first quadrant, sinθ is always positive i.e. sinθ>0. Similarly, tanθ is also greater than 0 in the first quadrant. tanθ>0

If both are greater than 0, then from the identity given above, you can predict this that
tan^2θ * sin^2θ will be also be greater than 0.
So if the RHS ( tan^2θ * sin^2θ ) of the identity is greater than zero, the LHS would also be greater than 0. So, just solve the inequality now:

tan^2θ - sin^2θ > 0
take sin^2θ on the RHS of the inequality
tan^2θ > sin^2θ
take square root on both sides, and you'll end up with:
tanθ > sinθ
 
i_try9621

i_try9621

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Bilal.khan said:
Hey guys.
Can someone please help me with this question
View attachment 60208

I got the answer for the first part
View attachment 60209

But the problem is i don't get why did they not calculate mg sinθ as resistive force in the second part although they used it in the first part of the question. The engine would also do work against this component of weight so why dont we use it in our calculation.
View attachment 60210
please help....
Click to expand...
I think that the component of weight parallel to the slope for a distance of 500 metres is the gain in Potential Energy . Because if you notice, ''mgsinθ'' times ''500'' is 375000 J. And the gain in Potential energy is "mg" times "500sinθ" is also 375000 J. Although I may not be sure...
 
techgeek

techgeek

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qwertypoiu said:
One of the integration questions...
View attachment 60254
Click to expand...
congrats.gif
congrats.gif
 
areeba240

areeba240

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upload_2016-4-13_9-29-21.png
how to solve the last two parts i.e. (b) and (c) of this qtn...plzzzzz helppppp
thanks in advance:)
 
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