Author Topic: help needed..its urgent!!!!histograms  (Read 2528 times)

Offline vids

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help needed..its urgent!!!!histograms
« on: May 18, 2009, 06:52:07 am »
maths p4
nov 03 q 8(b)

Offline vids

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Re: help needed..its urgent!!!!histograms
« Reply #1 on: May 18, 2009, 06:54:09 am »
histograms???? ???

Offline MissDE

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Re: help needed..its urgent!!!!histograms
« Reply #2 on: May 18, 2009, 07:11:42 am »
I really would like to help but am not sure ::)

I think we have to first make the table:
less than 10
less than 15
less than 20
...............................

sry for not helping the way you need :-\

Offline astarmathsandphysics

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Re: help needed..its urgent!!!!histograms
« Reply #3 on: May 18, 2009, 07:35:36 am »
Find frquency densities by dividing frequencies by the lengths in the intervals

Length         Frequency density
0-10 12/10=1.2
10-15 32/5=6.4
15-20 28/5=5.6
20-25 24/5=4.8
25-40 24/15=1.6

Now draw bars, Length on the x axis and frequency density on the y axis. Make sure you get even scales.

Offline Ghost Of Highbury

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Re: help needed..its urgent!!!!histograms
« Reply #4 on: May 18, 2009, 07:55:26 am »
RANGE :0<m<=10            10<M<=15           15<M<=20               20<M<=25           25<M<=40
WIDTH :   10                          5                      5                                5                      15
FREQU.:  12                         32                     28                              24                     25

THE SCALE 2cm - 5 kg
                1cm - 2.5kg

the standard width taken is 2.5
now...
---------------------RULE------------------------------------------------------------------
if class width = n * standars width, then the height of the rectangle = frequency/n
---------------------------------------------------------------------------------------------
for the 1st.) 10 = n * 2.5 ...therefore n = 4 => height = 12/4 = 3
          2nd.) 5  = n * 2.5....therefore n = 2 => height = 32/3 = 16
          3rd.) 5  = n * 2.5....therefore n = 2 => height = 28/2 = 14
          4th.) 5  = n * 2.5....therefore n = 2 => height = 24/2 =  12
          5th.) 15  = n * 2.5....therefore n = 6 =>height = 24/6 =  4

v have been taught this method
is this correct.
plz chk the markscheme
i don't have it

divine intervention!

Offline astarmathsandphysics

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Re: help needed..its urgent!!!!histograms
« Reply #5 on: May 18, 2009, 07:57:06 am »
No the height is the freoency dividing by the length of the interval so for the first one the height is 12/10=1.2

Offline astarmathsandphysics

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Re: help needed..its urgent!!!!histograms
« Reply #6 on: May 18, 2009, 07:58:21 am »
The height is called the frequency density for histograms.

Offline Ghost Of Highbury

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Re: help needed..its urgent!!!!histograms
« Reply #7 on: May 18, 2009, 08:01:59 am »
ohh..
thanks a lot  :)
By the way...wat does the markscheme say..
can u plz upload it
i don't have it
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Offline sweetsh

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Re: help needed..its urgent!!!!histograms
« Reply #8 on: May 18, 2009, 08:02:30 am »

Offline Ghost Of Highbury

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Re: help needed..its urgent!!!!histograms
« Reply #9 on: May 18, 2009, 08:05:35 am »
ohh..yaa..i'll try in that...
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Offline astarmathsandphysics

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Re: help needed..its urgent!!!!histograms
« Reply #10 on: May 18, 2009, 08:06:15 am »
The ms is there but is in 1 file for all the papers.

Offline astarmathsandphysics

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Re: help needed..its urgent!!!!histograms
« Reply #11 on: May 18, 2009, 08:07:26 am »
Hey sweetsh your avater is really good!

Offline sweetsh

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Re: help needed..its urgent!!!!histograms
« Reply #12 on: May 18, 2009, 08:08:59 am »
Haha! Nice idea right?

Offline Ghost Of Highbury

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Re: help needed..its urgent!!!!histograms
« Reply #13 on: May 18, 2009, 08:09:25 am »
i got the markscheme ..
it says..

1 mark for - horizontal scale 2cm = 5cm
1 mark for - k =1
5 marks for the heights - 3k, 16k, 14k, 12k, 4k

??
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Offline astarmathsandphysics

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Re: help needed..its urgent!!!!histograms
« Reply #14 on: May 18, 2009, 08:10:00 am »
Where did you get it? I might get one.