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Mathematics: Post your doubts here!

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A particular solution of the differential equation

3 y^2 dy/dx=4 ( y^3 + 1 )cos^2(x)

is such that y = 2 when x = 0. The diagram shows a sketch of the graph of this solution for 0 ≤ x ≤ 20;
the graph has stationary points at A and B. Find the y-coordinates of A and B, giving each coordinate
correct to 1 decimal place.

october november 2013 paper 33
there you go :)
 

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Hi . .
I am curremtly doing M2 , and started recently doing pastpapers
i tried to ask my teacher to help me but he is not able to solve them !
maybe because this is the first year of teaching a level so they find it difficult
my request if any one does M2 , plz PM me , so i will be able to tell you my doubts
 
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may june 09 paper 62 question 1...how do we find the standard deviation from box and whisker diagram??

i didn't give the paper a look, but I suppose that the box and whisker plot would be symmetrical (means the box is equally divided by the line of the inter quartile value). This occurrence gives rise to the possibility of applying the normal distribution. Apply it and find the standard deviation accordingly by taking areas such as 0. 75 and 0.25 into consideration
 
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i didn't give the paper a look, but I suppose that the box and whisker plot would be symmetrical (means the box is equally divided by the line of the inter quartile value). This occurrence gives rise to the possibility of applying the normal distribution. Apply it and find the standard deviation accordingly by taking areas such as 0. 75 and 0.25 into consideration
Yes its the upper quartile minus mean for z score but if its possible can u post a pic or sumthing of how u solve it further? Why is the area 0.75 or 0.25
Btw thanks
 
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Any help , ?
for second question tell me the procedure , its a different question and assume my dy/dx = 2x +3
 

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Maths doubt of stem and leaf diagram, please help

Answer is also given in the picture in the smaller font

Thanks a tonne!
 

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The following doubt is of finding median without a cumulative frequency graph, please help!

Answers are given in red font

Thank you very much
 

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The following doubt is of finding median without a cumulative frequency graph, please help!

Answers are given in red font

Thank you very much

The formula to calculate median (50th percentile or any other percentile) without the cumulative frequency curve is :

lower class boundary +
((50n/100 - cf (till prev. class)/ f (of that interval))

where n is the total number of observed items or maximum cumulative frequency. First calculate the corresponding class by checking in which class the n/2th value lies then perform the relative ascribed function
 
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Any help , ?
for second question tell me the procedure , its a different question and assume my dy/dx = 2x +3

4(1/2 r^2 [(pie-x)- sin(pie-x)]) = 1/2 pie r^2
since sin 180 - x = sin x
2 r ^2 [(pie-x) - sin x] = 1/2pie r^2

2pie - 2x -2sinx = 1/2 pie
2pie - 1/2pie - 2sinx = 2x
3/2 pie - 2 sinx = 2x
x = 3/4 pie - sinx

For the second question, please send in the whole question and i'll try to solve it
 
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Maths doubt of stem and leaf diagram, please help

Answer is also given in the picture in the smaller font

Thanks a tonne!
Median = Mid-term
we find the nth term where median lies using total number of terms
we have 71 insects for X (odd number)
so Median = {(71+1)/2}th term
= 36th term
start adding up the numbers on the left given in bracket
that's your frquency for the certain range
you"ll have 28th-44th term on the left side of 82 on the stem
5||82 is your median i.e. 0.835

Interquartile range = lower quartie - upper quartile
lower quartile = 25th percentile = {(25/100)x(71+2)}th term
= 18th term
= 0.814
upper quartile = 75th percentile = {(75/100)x(71+2)}th term
= 54th term
= 0.833
thus, IQ = 0.833-0.814 = 0.019
 
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