# 9.1 Graphical representation of data

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## Graphical representation of data

Once the data has been collected, it must be organised in a manner that allows for the information to be extracted most efficiently. One method of organisation is to display the data in the form of graphs. Functions and graphs have been studied in  Functions and Graphs , and similar techniques will be used here. However, instead of drawing graphs from equations as was done in Functions and graphs , bar graphs, histograms and pie charts will be drawn directly from the data.

## Bar and compound bar graphs

A bar chart is used to present data where each observation falls into a specific category and where the categories, this is often for qualitative data. The frequencies (or percentages) are listed along the $y$ -axis and the categories are listed along the $x$ -axis. The heights of the bars correspond to the frequencies. The bars are of equal width and should not touch neighbouring bars.

A compound bar chart (also called component bar chart) is a variant: here the bars are cut into various components depending on what is being shown. If percentages are used for various components of a compound bar, then the total bar height must be 100%. The compound bar chart is a little more complex but if this method is used sensibly, a lot of information can be quickly shown in an attractive fashion.

Examples of a bar and a compound bar graph, for Data Set 1 , are shown in [link] . According to the frequency table for Data Set 1, the coin landed heads-up 44 times and tails-up 56 times.

## Histograms and frequency polygons

It is often useful to look at the frequency with which certain values fall in pre-set groups or classes of specified sizes. The choice of the groups should be such that they help highlight features in the data. If these grouped values are plotted in a manner similar to a bar graph, then the resulting graph is known as a histogram. Examples of histograms are shown in [link] for Data Set 2, with group sizes of 1 and 2.

 Groups $0 1 $1 2 $2 3 $3 4 $4 5 $5 6 Frequency 30 32 35 34 37 32
 Groups $0 2 $2 4 $4 6 Frequency 62 69 69

The same data used to plot a histogram are used to plot a frequency polygon, except the pair of data values are plotted as a point and the points are joined with straight lines. The frequency polygons for the histograms in [link] are shown in [link] .

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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Sherica
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Sherica
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Tamia
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Uday
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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Kristine 2*2*2=8
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is it 3×y ?
J, combine like terms 7x-4y
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what's the easiest and fastest way to the synthesize AgNP?
China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Stotaw
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Azam
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Prasenjit
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Azam
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Damian
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Azam
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Uday
I'm interested in Nanotube
Uday
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Prasenjit
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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