Chapter 3 Graphics Mathematics

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Chapter 3 Graphics Mathematics

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Title: Chapter 3 Graphics Mathematics

1
Chapter 3Graphics Mathematics

Computer graphics needs mathematics to enable the
depiction of points, lines and other objects on
the computer screen and through the printer or
plotter.

A point is a position on a plane or in space.
Graphics involves first,
the location of points
next, the surfaces as bounded by lines
finally the representation of solids as bounded
by surfaces.
The point itself is just the location and it has
no size.

A line is a set of points. It is theoretically
made up of an infinite number of points because
it is a finite length divided by zero- the
theoretical length of the point.
But in a computer, the smallest dot that can be
displayed on the screen is a pixel, usually
rectangular in shape. This has finite dimensions,
however small. Hence, a computer line is made up
of a finite number of dots.

On this basis, computer graphics involves various
branches of mathematics in its development,
implementation, and interpretation.
The range of the mathematics spans simple algebra
and geometry to very sophisticated differential
and integral calculus.

5
Coordinate System

All graphics, whether manual or computer,
involves a reference point called the origin, and
linear or angular distances measured along one or
more reference directions called the coordinates,
a base line along which distances are measured
and from which angles are measured is called an
axis. The origin and axes together are called
Coordinate Systems.
The system of coordinates that will be used by
computer to display images on the screen or to
plot them on paper will be different from the
standard Cartesian (x, y, z) or other World
Coordinates in which we locate and describe real
world objects or our own pictorial
representations.

The following coordinate systems are commonly
used to depict images in two dimensions (2D) on a
plane, as on paper or on a screen, or in three
dimensions (3D) in space, as a solid prototype
(real-life object) or its model.
Two-Dimensional Cartesian (x, y)
Polar (r, ?)
Three-Dimensional Cartesian (x, y, z)
Cylindrical (r, ?, z)
Spherical (r, ?, ?)

7
Basic Mathematical Concepts

To get a computer to draw a circle would be a
stupendous task-because computers cannot see or
think in the sense that humans can.
In general, everything we draw with a computer
has to be represented by means of dots (pixels)
totaling tens of thousands, even millions. Their
individual locations are entered, or scanned from
an existing picture, or calculated by formulas
from a few key data, and are connected by special
line-drawing algorithms.

With improvements in hardware technology and
software engineering, more and more of this
computational labour is being built-in as part of
the computer hardware and software.
For instance, it is quite easy these days to let
the computer draw the circle and the straight
line on the monitor, and then digitize (determine
in terms of numbers) the coordinates of the
intersection point or points and output their
values.

However, even to use the sophisticated computer
graphic capabilities available, the user of
graphics software should have a sense of form and
shape should know the physical configuration and
be able to draw at least a freehand sketch of the
object to be drawn, which implies a good
understanding of its geometrical relationships.

Thus, in the application and programming of
geometrical principles and techniques for
computer graphics, knowledge and even some
experience with geometry (locations) and topology
(their spatial relationships) are essential. In
addition, a basic understanding of the
mathematics underlying computer graphics will be
a definite plus.
A lot of analytical geometry and trigonometry is
used in computer graphics.

11
Cubic Splines

Equation for Cubic splines
Splines are smooth curves passing through a given
set of points. They are essential and convenient
substitutes to the French Curves (plastic
templates used to draw curves) which was so
essential to manual drafting.
Cubic splines are popular because
they are fairly simple to compute
provide continuity of the curve
its slope
its curvature at a point (i.e. 2nd derivative)

Any curve may be built up of a series of cubic
segments.
The general equation for the most basic (natural)
2-dimensional cubic spline is
y a0 a1 . x a2 . x2 a3 . x3
We need 4 boundary conditions to evaluate the 4
unknown constants a0, a1, a2, and a3.

These may be 4 points through which the curve
must pass
three points and the slope at any one of them
or,
as is common, the locations of the two end points
and the slopes at these end points.
Occasionally, the curvature at a point may also
be specified.
The slope equation is
y dy/dx a1 2a2 . x 3a3 . x2
The curvature equation is
y d2y/dx2 2a2 . x 6a3 . x
For single segment we will have 4 boundary
conditions and the solution may be quite
straightforward.

14
Example

Determine the equation for a cubic spline segment
passing through 2 end points A(0,5) and B(10,20),
having slope of 2 and curvature of 0.5 at A.

15
Solution

At A,
with x 0, y 5 ? y 2, and y -0.5, we
have
5 a0, from the cubic spline equation
2 a1, from the 1st derivative
-0.5 2 a2, or a2 -0.25
At B,
with x 10, and y 20, we have
20 5 2(10) 0.25(102) a3(103) or, a3
0.02
Thus the desired cubic is y 5 2x 0.25x2
0.02x3, and with the equation now available, as
many points as needed for graphic representation
may be computed.

16
Parametric Form

The parametric form of the cubic spline will be
as follows
x a0 a1 . u a2 . u2 a3 . u3 (a)
y b0 b1 . u b2 .u2 b3 . u3 (b)
where, the parameter u will vary between
specified min. and max. values to cover the
desired range of the curve.

17
Concatenation of Multiple Cubic Spline Segments

To concatenate (to attach smoothly) a number of
cubic spline segments making up curve passing
through a series of n points, we need to insert
(n1) cubic spline segments, requiring the
evaluation of 4(n 1) coefficients.
The value of y at the 2 end points each of the
(n1) segments yield 2(n 1) equations.
We may specify the slopes y at the first and the
last points, giving 2 more equations.
This still leaves 4(n 1) 2(n 1) 2, or
2(n 2) boundary conditions to be specified,
which may conveniently be obtained as the
matching of slope y and curvature y at the (n
2) internal points, blending the successive
segments smoothly with one another.

The computations get pretty tedious for more than
a few points, and many computerized numerical
analysis techniques such as parameterization,
normalization, tri-diagonalization, and other
matrix manipulations are invoked for faster and
more efficient computation.
There are thus many formulations of the
technique, but the basic principles are the same.

19
Example

Let us consider a first-order cubic spline made
up from 2 segments AB and BC, through the 3
points A(0,10), B(10,20), and C(20,30), with
slopes of 1 at A and 1 at C.
meaning that the curve and slope are continuous

Let the segments AB and BC respectively have the
equations
y a0 a1 . x a2 . x2 a3 . x3 (1)
and y b0 b1 . x b2 . x2 b3 . x3
(2)
Then the corresponding slope equations are
y a1 2a2 .x 3a3 . x2 (3)
and y b1 2b2 .x 3b3 . x2 (4)

The corresponding curvature equations are
y 2a2 6a3 . x (5)
and y 2b2 6b3 . x (6)
For segment AB, we have
at A, with x 0, and y 10, and y -1, we
have
a0 10 from (1)
(6.5)
a1 -1 from (3 )
(6.6)
at B, with x 10, and y 20, we have
a0 10a1 100a2 1000a3 20 (7)
from (1)

For segment BC, we have
at B, with x 10, and y 20, we have
b0 10b1 100b2 1000b3 20 (8)
from (2)
at C, with x 20, y 30, and y 1, we have
b0 20b1 400b2 8000b3 30 (9)
from (1)
b1 2(20)b2 3(400)b3 1 (10) from
(2)

Matching slopes at B (x 10) from segments AB
and BC, from Eqs. (3 4), we get
a1 2(10)a2 3(100)a3 b1 2(10)b2 3(100)b3
(11)
Matching curvatures at B (x 10) from segments
AB and BC, from Eqs. (5 6), we get
2a2 6(10)a3 2b2 6(10)b3 (12)

Equations (6.5 to 12) represent 8 simultaneous
equations in the 8 unknowns a0, a1, a2, a3, b0,
b1, b2, and b3.
The first two explicitly known (because xA 0),
and hence we actually have only 6 equations in 6
unknowns.
They can be solved one way or another (even
longhand in this case), and the results are
a0 10
a1 -1
a2 0.35
a3 -0.015
b0 -10
b1 5
b2 -0.25
b3 0.005

To solve this linear system you may use the fact
that x inv(a).b
25

Thus, equations to the two segments AB and BC,
satisfying the boundary conditions are
y 10 x 0.35x2 0.015x3 (13)
y -10 5 x - 0.25x2 0.005x3 (14)
With the equations now available, the two curve
segments may be plotted. The end slopes are shown
as broken lines.
The corresponding slope equations for AB and BC
are
y -1 0.7x 0.045x2 (15)
y 5 0.5x 0.015x2 (16)
The corresponding curvature equations for AB and
BC are
y 0.7 0.09x (17)
y -0.5x 0.03x (18)

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By substitution of appropriate x values in Eqs.
(13) and Eqs. (15), it may be confirmed that all
the 6 specified boundary conditions Eqs. (7 to
12) are satisfied.
The slope and curvature matching conditions at B
(at x 10), Eqs. (11 12), can also be
confirmed from Eqs. (15 16) and Eqs. (17 18),
with the common slope being 1.5, and the common
curvature being 0.2.

28
Assignment 2

1. Tabulate five points for each segment to plot
the two-segment cubic spline passing through
A(5,5), B(15,10), and C(20,80).
Assume
Slopes at A and C as -1 and 1
Points P(10,12) and Q(25,2).

29
Bezier Curves

Bezier Curve Characteristics
In the late 1960s a French engineer, Pierre
Bezier developed a method of fitting a polynomial
curve of degree (n 1) to follow (not pass
through) a set of n control points.
The first curve (in the following figure)
controlled by 3 points is a quadratic.

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31

The second and third curves (in the previous
figure) are both cubics, but while in the second,
the 2 intermediate points (2 and 3) are on the
same side of the straight line connecting the
first and last points (1 and 4), in the second
figure they are on opposite sides of the line.

32
The characteristic features of these curves are

The curve passes through only the first and the
last point.
At the 2 ends the curve is tangential to the
chords joining the first and second point and the
last and penultimate point.
The intermediate control points, through which
the curve does not pass in general, are effective
in pulling the curve in directions and in
magnitudes we desire.
The shape of the curve is that of a convex hull
(a shipping term) within the lines bounding it,
and bulging towards the intersections of the
lines.
Bezier curves may be made to close or even cross
over, by appropriate choice of the control points.

These features of Bezier curves make them
exceptionally useful in the design of imaginative
and innovative geometric designs for cars,
planes, architectural shapes, etc.
Most CAD packages and many utility packages that
include a graphics module provide a means for the
interactive development of Bezier curves.

Example of the Bezier curve is the MS-Windows
Paintbrush (or Paint) module. It has an icon,
which shows a general snake shaped curve, by
clicking on which a 4-point Bezier curve may be
drawn, as follows
a. Click on the icon
b. Click on the display space, at the location of
Point 1 and Point 4. A straight line is shown
connecting these 2 points.

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Move to Control point 2, the location to which we
wish to pull the curve and click the left button.
This can be done by moving the cursor to the
desired position, as displayed in the coordinates
field on the screen. Alternatively, it may be
done by trial and error, pressing the left button
and continuing to move the mouse, with the curve
also moving and changing shape, until the curve
shape is visually satisfactory.
d. Move the mouse to the location Control
Point 3 and click to complete the curve.

37
Equations for Bezier CurvesUsing Parametric Form

Let the four control points be
(x1, y1), (x2, y2), (x3, y3) and (x4, y4).
Then
x x(u) x1(1-u)3 3x2 u(1-u)2 3x3 u2(1-u)
x4 u3
y y(u) y1(1-u)3 3y2 u(1-u)2 3y3 u2(1-u)
y4 u3

38
Example

Generate a Bezier curve to the 4 control points
(60,20), (80,100), 150,90) and (180,50).

39
Solution

x 60(1-u)3 3(80)u(1-u)2 3(150)u2(1-u)
(180)u3
? x 60 60u 150u2 90u3
y 20(1-u)3 3(100)u(1-u)2 3(90)u2(1-u)
50u3
? y 20 240u - 270u2 60u3
Now we will be able to generate as many points as
are necessary to define the curve for plotting or
other purposes, by assigning various values of
parameter u ranging from 0 to 1.
The following table shows the computed values for
u from 0 to 1, in increments of 0.1. An actual
plot would require points at much smaller u
intervals so that no two successive values are
more than one pixel dimension apart.

40
Assignment 3 Write a program that generates u
automatically and calculates x and y.
41

The four control points are shown in the
following figure, with broken lines joining them.
It is clear that the first and last (third) lines
are tangential to the curve.

42
Concatenation of Multiple Bezier Curve Segments

Bezier Curve segments also may be concatenated
like cubic spline segments but the conditions at
the junction of the Bezier segments are easier to
satisfy and further, changes in one segment
affect only the adjacent segment and not all the
segments as in cubic splines.
If to a Bezier segments 'A' it is required to
attach another segment 'B', all that is needed to
be done is the following
Make the last point of the curve A the first
point of curve B, that is, B1 A4. That is
xB1 xA4 and yB1 yA4

2. Make the slope of the first chord B1-B2 of
curve B the same as the slope of the last chord
A3-A4 of curve A.
This condition, combined with the first,
implies that B2 may be anywhere along the
extension of the line A3-A4. This is easily
accomplished by setting B2 away from B1 (A4) by
a constant p times the x and y offsets by which
A4 is away from A3.
That is
xB2 xB1 p(xA4 xA3 ) xA4( 1 p) p .
xA3
yB2 yB1 p(yA4 yA3 ) yA4( 1 p) p .
yA3

The following figure illustrates the location of
B2 with the same slope as A3-A4, x and y standing
for the respective projections of A3-A4.
The two stipulations fix the starting point (B1)
of curve B and the direction of B1-B2, leaving
the location of B2 along B1-B2, as well as the
location of B3, for the user to manipulate to get
the desired curve.
Usually the ending point (B4) is fixed from other
considerations.
Obviously the actual curve will depend upon the
exact location of B2, even with the same slope as
A3-A4.

Figure 3.5(b) shows four different curves for
four values of p. (The same principles may be
extended to curves with more than 4 control
points).

46
ExampleMultiple Segment Bezier Curve

In this figure Bezier curves have to be fitted
between the points (60,20), (180,50) and
(300,80), with a hill between the first pair, and
a continuous valley and a second smaller hill
within the second pair.
Fig
Let us further assume that by trial and error,
the user has picked control points 2 and 3 for
the first curve (say A) as (80,100) and (150,90).
This segment has already been developed in
Section 3.4.3, and shown as in Fig. 3.6.

xB1 xA4 180
yB1 yA4 50
Hence B1 (180,50)
xB2 180( 11 ) 150 210,
yB2 50(11) 90 10
Hence B2 (210, 10)
The end point B4 is given as (300,80).
Now let us say that, by trial and error, we
choose Point B3 as (250,150), raising the curve
from a valley, which has been pulled down to y2
of 10, to a hilly location above the end point
with given y4 of 80.

Applying the same process as in the previous
example to this segment B, we get
x 180(1-u)3 3(210) . u . (1-u)2 3(250) . u2
(1-u) 300 . u3
180 90u 30 u2
y 50(1-u)3 3(10). u . (1-u)2 3(150) . u2 .
(1-u) 80 . u3
50 120u 540u2 390u3
It is interesting, but not terribly worrisome,
that the x equation does not have an x3 term.
Now sufficient number of points may be computed
for display of the second segment. This is left
as an assignment 4.

49
Assignment 5Exercise 3.2

Tabulate 5 points for each segment to plot the
3-segment A-B-C Bezier Curve with control points
as follows
A1(5,12)
A2(10,5)
A3(15,24)
A4(18,17)
C1(29,14)
C2(31,20)
C3(35,22)
C4(40,15)
Choose p of 0.5.
Hint Get B3 and B4 by continuity conditions from
C1 and C2.

50
REGRESSION ANALYSIS

In computer graphic representation of a curve, if
we have the equation to the curve, or appropriate
data to determine the curve exactly, then it is
best to include the curve in the plot routine and
let it compute as many pixel positions as it is
necessary to plot.
However, in real life, most natural phenomena
cannot be reduced to simple equations. We may
have a large number of observations either from
nature (such as the weather) or from the
laboratory experiments (such as from controlled
chemical reactions), which lead to a number of
(x,y) data.

Plotting of these raw data will indeed be the
best means of understanding and interpreting the
findings in qualitative terms. On the other
hand, frequently it becomes necessary to
interpret these data in scientific terms, trying
to find meaning and logic behind these
variations, and to be able to interpolate missing
data within the observed range, and/or
extrapolate (predict) data outside the observed
range of data.
This is done by a further statistical and
mathematical treatment called Regression
Analysis, leading to algebraic equations to fit
the observed data and predict the interpolated or
extrapolated behaviour. In such cases too,
plotting of both the observed and predicted
information would be the best means of analysis
and further decisions.

52
Least Squares Fit

Let us consider n sets of (x,y) observations.
When plotted, they may be almost on a straight
line, so much so that we want (at least as a
first approximation) to fit a straight line
passing through them, as in the following figure,
and then use the equation of the fitted straight
line for interpolation and extrapolation.
Fig

The best-fit straight line is obtained when the
sum of the squares of the deviations of the
points (marked as e1 and e2 for points P1 and P2
in the inset of the figure) from the straight
line is the minimum, and hence the fit is called
the least Squares Fit.
In the equation y m . x c for the best-fit
straight line, the slope m and intercept c are
given by
m n ? x y ( ? x ) ( ? y
) / D
and
c (? y ) (? x2 ) ( ? x ) ( ? x y ) / D
where D n ? x2 - (?x)2
The ? symbol indicating summation for i
1,2,...,n, of quantities following the symbol.
Once the equation is determined, the interpolated
or extrapolated value of x within or outside the
range may be computed from y m . x c. In
actual plotting, only the line will be plotted,
but none of the data points.

One measure of Goodness of Outfit of the straight
line to the points is the Coefficient of
Correlation given by
When all the points are actually on a straight
line, all the deviation e will be zero, and r
will equal 1, representing a perfect fit. The
closer r is to 1, the better is the
representation of the group of points by the
fitted straight line.

In the B.C. (Before Computer) era, many
professionals who wanted to get a best fit
straight line through a number of scattered
points on regular graph paper, and, by eye
judgment, moved a transparent plastic scale over
the points to locate a give and take line
(meaning they left as much gap between points on
one side of the straight edge as on the other).
Most of the time, this quick and dirty method
served as well as any mathematical procedure, at
least for practical application!
Similar procedures and equations are available
for parabolic and higher degree fits, but those
get quite a bit more involved than justified in
this introductory work.

56
ExampleStraight Line Fit

Fit straight line to the five points with
observed (x,y) coordinates (2, 7.1) (3, 9.2) (4,
10.8) (6, 15.2) and (7, 16.7) and estimate the
values at x 5 and x 8.

57
Solution

Let us first look at the five points, with a
straight line from the first to the last point
for judging the deviations, as in the following
figure.
Fig

They are almost on a straight line, but not
quite, especially when we see them against the
straight line drawn from the first to the last
point (which will almost certainly be not the
best-fit line!)
To fit the straight line, we compute the various
quantities required and tabulate them in the
following table.

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? m 1.948
c 3.230
giving the best-fit equation as
y 1.948x 3.230
The Coefficient of Correlation is
r 0.999
which is not bad at all!
From the developed equation, we get
for x 5 y 1.948
3.230 12.970
and for x 8 y 1.948(8)
3.230 18.814

The best fit-line is shown plotted in the right
side of the previous figure, solid between P1 and
P5 and broken beyond this range.
The visual slope (of nearly 45º) is not the true
slope because of the difference in the x and y
scales.
The best-fit line in (b) may appear to be the
same as the check line in (a) shown connecting P1
and P5. But in actuality, the slope and intercept
of the check line could be shown to be 1.920 and
3.26 as against the 1.948 and 3.23 of the
best-fit line.
Exercise Try y values of 7, 9, 11, 15 and 17 for
the given x-values and confirm that the least
square equations (slide 53) lead to the
equation y 2x 3, with an r of 1.0, a perfect
fit.

62
Curves Transformable to Straight Lines

Certain standard curves as shown in the following
figure may be transformed to straight-line
equations by substitution of the variables x
and/or y and possibly the constant also, with
some appropriate functions of the variables. Then
regression analysis may be applied to get the
best-fit straight line to the transformed forms
and reverse functions applied to the variables
and constants to replace the given points with
the best-fit functions.
Fig

First, the scattered points available are
examined visually, and the one among the three
curves given, which would approximate them best,
is selected.
It must be remembered that the transformations
work only with reference to the specific origins
O indicated. The coordinates of the points may be
given with reference to some other origin O'.
Hence, before regression analysis equations can
be applied, the origin and axes appropriate to
the selected curve must be located, and the
coordinates modified by the shifts amounts.
Occasionally, trial and error may be necessary
and each time the coefficient of correlation r
must be found the final choice of course will be
the one with the r closest to 1.

A) y m/x c The equation is sketched in Fig
(a). Vertically the two arms of the curve
converge to the y axis. The broken line depicts
the asymptote at y c to which both branches of
the curve converge. The origin O has to be
located with this in mind and if the curves do
not extend far enough or only one branch is
available some trial and error may be needed
(especially for y-axis) to choose the best fit
with the highest r.
Let x 1/x. Then the equation reads, y mx
c, the standard form. Now regression analysis x
and y gives m and c.

B) y c . ax , The equation is sketched in Fig.
(b). The broken line shows the intercept of the
curve with the y-axis. In case there is an
additive constant d on the right hand side, trial
and error for d are inevitable, to gather results
on various best-fit lines, and pick the one with
the highest r.
When we take logarithm of both sides, we get
Log y log c
x. log a
Let y log y, c log c, and m log a.
Then the equations reads, y m.x c, the
standard form to which Least Square Equation may
be applied.
After the best-fit line has been found connecting
the x and y, we get m and c. Then, we can
simply revert back to (x,y) coordinates, by
taking c exp (c), and, a exp (m). Note that
for this analysis, y cannot take values lt 0
unless clt0, in which case, signs may reversed on
both sides to proceed further.
This technique by manual plotting of log y
against x on Semi-Log graph paper, and eye
fitting a best-fit straight line, used to be and
continuous to be known as the Semi-Log Plot
method. See the following figure.

(c) y c . xm The equation is sketched in figure
c.
The full line shows the curve when m is odd, and
the broken line shows the curves ( for xlt0 ) when
m is even. The origin O must be chosen at the
horizontal tangent position, if available within
the range of points.
When we take logarithms of both sides, we get
log y
log c m. log x
Let y log y, c log c. and x log x.
Then the equation reads y m. x c, the
standard form to which least square equation may
be applied.
After the best-fit line has been found connecting
x and y, we get m and c.
Then, we can simply revert back to ( x, y )
coordinates by taking c exp (c)
Note that when m is not an integer, both x and y
cannot take values lt0, and that the broken line
and solid line portions for xlt0 in Fig. (c) are
simply reflections of the curve in the first
quadrant, about the horizontal axis and about the
origin respectively.
This technique by manual plotting of log y on
Log-Log graph paper and eye fitting a best-fit
straight line, used to be known as the Log-Log
Plot method. See the previous figure (b).

Example This sample of data obtained in a study
of the relationship between the number of years
that applicants for certain foreign service jobs
have studied English in high school or college
and the grades which they received in a
proficiency test in that language

No. of years (x) Grade in Test (y) x2 x.y
3 57 9 171
4 78 16 312
4 72 16 288
2 58 4 116
5 89 25 445
3 63 9 189
4 73 16 292
5 84 25 420
3 75 9 225
2 48 4 96
35 697 133 2,554

The goal is to find one line which fits the data.
Normal Equations
?y na b?x n no. of pairs
?xy a?x b ?x2
697 10 a 35 b
2,554 35 a 133 b

Two methods to find a and b
(1)
1st 7 ? 4,879 70 a 245 b
2nd 2 ? 5,108 70 a 266 b
? (2nd 1st) 229 0 21 b
? b 10.90 ? a 31.55
? y 31.55 10.90 x

(2)
1st 697 10 a 35 b
? a (679 35b)/10
2nd ? 2,554 35 (679 35b)/10 133 b
? b 10.90 ? a 32.55

Where the slope of the line is given by m (Dx
over Dy or the rise over the run) and the
intercept of the y axis is given as b. For n
data pairs, the equations used to find the slope
m and intercept b are
1.) b ( Sy Sx2 - Sx Sxy ) / ( n Sx2
(Sx)2 )
2.) m ( N Sxy SySx ) / (n Sx2 (Sx) 2 )

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Assignment 6.1 (deadline Monday 13th Jan)

The following area data on the IQs of 25
students, the number of hours they studies for a
certain achievement test, and their scores on the
test

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(No Transcript)
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1. Use the computer to find the least squares
line which will enable us to predict a student
score on the test in terms of his/her IQ. And
draw the line.
2. Use the computer to find the least squares
line which will enable us to predict a students
score on the test in terms of the numbers of
hours he/she studied for the test. And draw the
line.
3. Use the computer to predict how many hours a
student will study for the test given his/her IQ.
Draw the line.

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Assignment 6.2 (deadline Monday 13th Jan)