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Mathematics: Post your doubts here!

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The underlined phrase means the weight of 94% of the letters have a standard deviation of 12g. So let's say the mean is 20g. 94% of the letters have a weight of something between 18g and 42g, i.e. 12g off from the mean value.

As for the second question..

(i) Let the probability of the spinner landing on the blue side be 'x'. Since the probability of the spinner landing on the red and green sides are three and four times P(blue) respectively, their probabilities are '3x' and '4x'. Probabilities always add up to 1, so..

x + 3x + 4x = 1
8x = 1
x = 1/8

Hence the probability of the spinner landing on the blue side is 1/8.

(ii) You want a red AND a green AND a blue, so you multiply their probabilities:

1/8 * 4/8 * 3/8 = 12/512.

Also, since there are 3! different ways for this to happen (you could have RBG, BGR, GRB, etc.), you multiply the result by 3!

12/512 * 3! = 9/64.

(iii) Use normal approximation for this.

n = 136
p (success of landing on blue) = 1/8
q (failure of landing on blue) = 7/8

mean = 136 x 1/8 = 17
standard deviation = √(17 * 7/8) = √14.875

Also note that you will use continunity correction, so your value becomes 19.5, not 20. Using formula,

z = (19.5 - 17) / √14.875

Once you figure out the value of z, use the table to figure out the probability.

I hope this helped. :)
Thanks a lot, it really helped. :)
But I still don't get w11/p61/Q5.. The marking scheme has taken the probability of <32 to be equal to 0.97 in solution to part (i). How's that so?
 
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Need some help again
June11/p62 Q4 (iii)s11.p62.JPG
N10/p63/Q1
How's the mean 60 kg and variance 90kg^2 as stated in the marking scheme?n10.p63.JPG
n10/p61/Q6n10.p61.JPG
 
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can somebody plz help in november 2011 q6 variant 41? i dont know how to show that the speed at c is 29.9
 
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can somebody plz help in november 2011 q6 variant 41? i dont know how to show that the speed at c is 29.9
AOA,
as we know....
wrkdne by drivngfrce - wrkdne by resistive frce=gain in kinetic energy-loss in potential energy
so,
dis. BC=45/sin1
=2578.44
now putng values in above eq....
wrkdne by drv ffrce-resistive force X dis.bc =1/2m(v^2-u^2)-mgh
1660000-(2578.44x697.24)=1/2x1200x(v^2-15^2)-1200x45x10
-137791.5=600(v^2-225)-540000
v^2-225=402208.5/600
v^2=670.34+225
V=(895.34)^1/2
V=29.9 m/s ..................proved (y):cool:
hope u will get it:LOL:
 
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9709_s10_qp_1

Q5 part ii

help me plzz

9709_w10_qp_41
Q3
i need a detailed solution to this plz..


q52) you have obtained 2-5cos^2x ................ U have the values of x ranging frm 0 to π/ (0,90,180)

substitute the values in the eq. u obtained above u will get,
2-5cos^2(0) = -3
2-5cos^2(90)= 2
2-5cos^2(180)=-3

now frm these the largest is -3 and smallest is 2 Ans.

 
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q52) you have obtained 2-5cos^2x ................ U have the values of x ranging frm 0 to π/ (0,90,180)

substitute the values in the eq. u obtained above u will get,
2-5cos^2(0) = -3
2-5cos^2(90)= 2
2-5cos^2(180)=-3

now frm these the largest is -3 and smallest is 2 Ans.
why f(90)? btw the answer is wrong
 
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Aoa, posting here after almost a year :)
I missed a class during school, and since I'm giving composite Math, it cost me a topic. I can't seem to understand the questions in M1 regarding tension and weight, with 3 masses, such as Question 3 of this http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_qp_41.pdf.
ANY help at all would be hugely appreciated. Also, please tell me if this topic is covered in the course book ( the blackish one, that's CIE approved), as I don't have it. A link to understanding this topic online, or any help would be invaluable.
Expecting a prompt response.
Regards.
 
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Screenshot_2012-04-22-10-02-16-1.pngThis is from oct/nov 2011 p11. Can someone explain the iii) part.
From the graph we can see that there are four roots for part ii) That is correct.
The ans for prt iii) is 20 (4 x 5) Please eplain why ??
 
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View attachment 7063This is from oct/nov 2011 p11. Can someone explain the iii) part.
From the graph we can see that there are four roots for part ii) That is correct.
The ans for prt iii) is 20 (4 x 5) Please eplain why ??
in 2pi there are 4 solutions
in 1pi there are 2 solutions
for the domain 10pi to 20pi
there are 10 x 2 solutions

and how do u draw a graph for cos^2 and sin^2 functions?
 
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Thanks a lot.
I dont knw as well. I had the samee doubt. I realised that i told you to drw the graph. Sorry i dint read dat it was cos^2.
 
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Someone plz draw the graph of F(x)= 3-2tan(1/2x) for 0<x<pi
if possible also explain how to draw tan graphs.
Grateful for any help
 
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