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Median: 12+13/2 = 12.5Mode: 11 (because it has the highest frequency)
Median: 12+13/2 = 12.5
Mean = (11*35 + 12*28 + 13*22 + 14*18 + 15*14 + 16*9)/(35 + 28 + 22 + 18 + 14 + 9) = 12.8
shouldnt it be 13+14/2
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Median: 12+13/2 = 12.5Mode: 11 (because it has the highest frequency)
Median: 12+13/2 = 12.5
Mean = (11*35 + 12*28 + 13*22 + 14*18 + 15*14 + 16*9)/(35 + 28 + 22 + 18 + 14 + 9) = 12.8
Thank you so much I actually get it now and it feels good!
I don't understand logs cuz it's not in the igcse mathematics extended syllabus! But thank you for trying to help!![]()
Nopes. The two numbers you get are 63 (which comes under 12) and 64(which comes under 13)Median: 12+13/2 = 12.5
shouldnt it be 13+14/2
thanx very muchNopes. The two numbers you get are 63 (which comes under 12) and 64(which comes under 13)
Anything for my followersthanx very much![]()
Well i had a questions which the point was at (-1 , 1) and moved to (-1,2) it was shear y axis invarient but the scale factor was -1 why :S ?negative shear simply moves to the left side instead of right
Transformations:http://www.gcsemathstutor.com/transformations.php
http://www.s-cool.co.uk/gcse/maths/transformations/revise-it/transformations
just take the points in the corners of the unshaded area (the regions he wants ) then make a table put the x's and y's of the corners and multiply it by the amount given for y and x and find the highest or lowest costPlease i want your help. When you are given a bunch of inequalities and you draw them and you shade the unwanted regions. How do you find for example the least possible cost (if the inequalities represent number of boxes) of total boxes? this was a question in october november 2011 paper 43 question 10 please check it out and inform me how
Thanks in Advaaance!!!
Ladies and gentlemen!!
I thought about doing that but then i figured why would the question be for only 1 mark and with so little space? but if thats the way then thankk you so much!! Wish you the best!just take the points in the corners of the unshaded area (the regions he wants ) then make a table put the x's and y's of the corners and multiply it by the amount given for y and x and find the highest or lowest cost![]()
http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s10_qp_42.pdf
Help ! Question 10, (c ,i)
I know I am sending this too late.However, i found this in the examiner's reportThank you!
Good luck to you too and I'll pray you get an A* as well!![]()
Very few candidates scored this single mark. h(x) maps a value in one set onto a specific value in
a second set; h−1(x) reverses the process and maps the second value back onto the first. So if
h−1(x) = 2, the “2” has come from h(2) = 9. The function h−1(x) need not be considered.
i know reading it, makes it look so easy. but i am wondering whether is it always the case that theOk, so I was freaking myself out when all I had to do was ignore the inverse... -.-
But thank you so much! This was actually REALLY helpful!
Good luck!![]()
Thank you I really appreciate you work!!@slayer
very easy, draw a line (that covers all the the graph) between a point in the triangle and it's corresponding point in the image, then do it again with another point in the triangle and it's corresponding point in the image, the two lines you drew intersects at a point, this point is the center
in your question:
draw a line between (3,1), (-3,-5) (extend to cover all the graph)
draw a line between ( 6,3), (6,1) (extend to cover all the graph)
center: (6,4)
maybe not the best way, but with a little of common sense, it always work (atleast with positive enlargements)
EDIT: if someone have a better way, share please, i do have a faster way but it's error prone (and so i used the line drawing way)
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