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I'm an IGCSE Maths Teacher - Post your questions here.

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I keep getting PMs about this so I thought I'd send a general message:

I do not have access to Paper 4 until the exam starts. I know as much as all of you about it's content. An even if I did know something, I would not under any circumstances share the information.

So stop trying to cheat and focus on revising for the exam. If you have something to ask me, post it on one of the threads - you do not need to PM me.

(I will answer all questions that I haven't answered yet before tomorrow).
why don't you answer anymore??
 
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why don't you answer anymore??
I'm not permanently at a computer answering questions 24 hours a day. This is not my job.

I think I've done pretty well to answer as many questions as I can over the last week. I will answer every unanswered question on this thread before tomorrow.
 
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ok
This is really lengthy and would take a lot of my time as i am having my boards too
i will try to summarize it
Take the midpoint of each range for example the first onw will be (0.5+0)/2 =0.25 this is for the first one take for all
and then use the formula fx/f
which means (0.25*8)+(o.75*27)+ and so on till the end and then you divide all this by 200 which is the f
this is how you find estimate of the mean
you have made a graph right?? to find median 200/2 =100
on the y axis is the cumulative frequency so 100 of that on x axis will be the answer
that is it! I hope i helped sorry about not explaining the whole thing as i a preparing too
ok thanx im getting correct mean the only wrong thing i ws doing was (0.5-o)/2 but im still not getting the median which does not involve graph only table
 
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what does this mean The equation ()= has one solution when the line = intersects () at one point only. This happens approximately when <1.9 but we can choose any range so to be safe we will write <1.5 . from the attachment
I can't really explain it more than I have done in the revision sheet. What don't you understand about it?
 
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in paper http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_w11_qp_43.pdf
Q.10d i and ii,we have to draw that,which I did,and I have attached the picture of what I drew which i checked more than once here,now for finding least cost of boxes,I know that method of substitu
attachment
ting the points in the unshaded area with the 5x + 2y but do we have to choose and try EACH and every point in the unshaded area?as the question is only for one mark,also,if you take point 6.5,6.5 and subtitute in the formula you would get 45.5 which is less than 47,does that mean we can never take any numbers other than whole numbers while subsituting in these kind of inequality questions?
thank you!!
I can't see this image or the image in the post below it.
 
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Do i have to learn the matrices representing the transformation which maps is there no other way? Btw my teacher did teach me a way but i don't remember
 
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sorry its E(ii) not D(ii)
n(sports club only) = 23
n(music club only) = 13
n(total) = 90.

Either the 1st attends sports club only and the 2nd attends music club only or the 1st attends music club only and the 2nd attends sports club only.

So answer = P(1st person attends sports club only) * P(2nd person attends sports club only) + P(1st person attends music club only)*P(2nd person attends sports club only)

= (23/90 * 13/89) + (13/90 * 23/89) = 598/8010
 
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thank you verry much i was confused because i thought we just have to do (23/90*13/89).. thank u again.
 
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