The document outlines rules for graphing and linear regression. It discusses key aspects of graphing such as properly labeling axes, selecting appropriate scales, and drawing lines smoothly through data points. It then covers linear regression, defining the linear equation and describing how to calculate the slope and y-intercept by minimizing the sum of the deviations between observed and calculated data points. Formulas for determining the slope and y-intercept using linear regression are provided.
Simple Regression presentation is a
partial fulfillment to the requirement in PA 297 Research for Public Administrators, presented by Atty. Gayam , Dr. Cabling and Mr. Cagampang
Developed a system that generates a 3D face using a video of single frontal face as the input
Implemented Principal Component Analysis and Active Shape Model (ASM) for facial feature point detection
Applied the POSIT algorithm to estimate the head pose for realistic shading
Modified the classic ASM and critically appraised my approach by comparing it with alternative approaches
Simple Regression presentation is a
partial fulfillment to the requirement in PA 297 Research for Public Administrators, presented by Atty. Gayam , Dr. Cabling and Mr. Cagampang
Developed a system that generates a 3D face using a video of single frontal face as the input
Implemented Principal Component Analysis and Active Shape Model (ASM) for facial feature point detection
Applied the POSIT algorithm to estimate the head pose for realistic shading
Modified the classic ASM and critically appraised my approach by comparing it with alternative approaches
We present recent result on the numerical analysis of Quasi Monte-Carlo quadrature methods, applied to forward and inverse uncertainty quantification for elliptic and parabolic PDEs. Particular attention will be placed on Higher
-Order QMC, the stable and efficient generation of
interlaced polynomial lattice rules, and the numerical analysis of multilevel QMC Finite Element discretizations with applications to computational uncertainty quantification.
We present recent result on the numerical analysis of Quasi Monte-Carlo quadrature methods, applied to forward and inverse uncertainty quantification for elliptic and parabolic PDEs. Particular attention will be placed on Higher
-Order QMC, the stable and efficient generation of
interlaced polynomial lattice rules, and the numerical analysis of multilevel QMC Finite Element discretizations with applications to computational uncertainty quantification.
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Lord Jeshua Christos describes in more detail what He has purchsed for you and what you are an heir to as a Child of God.
Love Lord Jeshua Christos
משׁיח Hebrew Messiah Lord Jeshua Christos 14 hy
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Email: christosjeshua@gmail.com
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Web address: http://christosjeshua.wixsite.com/jeshuachristos
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Image segmentation is a computer vision task that involves dividing an image into multiple segments or regions, where each segment corresponds to a distinct object, region, or feature within the image. The goal of image segmentation is to simplify and analyze an image by partitioning it into meaningful and semantically relevant parts. This is a crucial step in various applications, including object recognition, medical imaging, autonomous driving, and more.
Key points about image segmentation:
Semantic Segmentation: This type of segmentation assigns each pixel in an image to a specific class, essentially labeling each pixel with the object or region it belongs to. It's commonly used for object detection and scene understanding.
Instance Segmentation: Here, individual instances of objects are separated and labeled separately. This is especially useful when multiple objects of the same class are present in the image.
Boundary Detection: Some segmentation methods focus on identifying the boundaries that separate different objects or regions in an image.
Methods: Image segmentation can be achieved through various techniques, including traditional methods like thresholding, clustering, and region growing, as well as more advanced techniques involving deep learning, such as using convolutional neural networks (CNNs) and fully convolutional networks (FCNs).
Challenges: Image segmentation can be challenging due to variations in lighting, color, texture, and object shape. Overlapping objects and unclear boundaries further complicate the task.
Applications: Image segmentation is used in diverse fields. For example, in medical imaging, it helps identify organs or abnormalities. In autonomous vehicles, it aids in identifying pedestrians, other vehicles, and obstacles.
Evaluation: Measuring the accuracy of segmentation methods can be complex. Metrics like Intersection over Union (IoU) and Dice coefficient are often used to compare segmented results to ground truth.
Data Annotation: Creating ground truth annotations for segmentation can be labor-intensive, as each pixel must be labeled. This has led to the development of datasets and tools to facilitate annotation.
Semantic Segmentation Networks: Deep learning architectures like U-Net, Mask R-CNN, and Deeplab have significantly improved the accuracy of image segmentation by effectively learning complex patterns and features.
Image segmentation plays a fundamental role in understanding and processing images, enabling computers to "see" and interpret visual information in ways that mimic human perception.
Image segmentation is a computer vision task that involves dividing an image into meaningful and distinct segments or regions. The goal is to partition an image into segments that represent different objects or areas of interest within the image. Image segmentation plays a crucial role in various applications, such as object detection, medical imaging, autonomous vehicles, and more.
Q1Perform the two basic operations of multiplication and divisio.docxamrit47
Q1
Perform the two basic operations of multiplication and division to a complex number in both rectangular and polar form, to demonstrate the different techniques.
· Dividing complex numbers in rectangular and polar forms.
· Converting complex numbers between polar and rectangular forms and vice versa.
Q2
Calculate the mean, standard deviation and variance for a set of ungrouped data
· Completing a tabular approach to processing ungrouped data.
Q3
Calculate the mean, standard deviation and variance for a set of grouped data
· Completing a tabular approach to processing grouped data having selected an appropriate group size.
Q4
Sketch the graph of a sinusoidal trig function and use it to explain and describe amplitude, period and frequency.
· Calculate various features and coordinates of a waveform and sketch a plot accordingly.
· Explain basic elements of a waveform.
Q5
Use two of the compound angle formulae and verify their results.
· Simplify trigonometric terms and calculate complete values using compound formulae.
Q6
Find the differential coefficient for three different functions to demonstrate the use of function of a function and the product and quotient rules
· Use the chain, product and quotient rule to solve given differentiation tasks.
Q7
Use integral calculus to solve two simple engineering problems involving the definite and indefinite integral.
· Complete 3 tasks; one to practise integration with no definite integrals, the second to use definite integrals, the third to plot a graph and identify the area that relates to the definite integrals with a calculated answer for the area within such.
Q8
Use the laws of logarithms to reduce an engineering law of the type y = axn to a straight line form, then using logarithmic graph paper, plot the graph and obtain the values for the constants a and n.
· See Task.
Q9
Use complex numbers to solve a parallel arrangement of impedances giving the answer in both Cartesian and polar form
· See Task.
Q10
Use differential calculus to find the maximum/minimum for an engineering problem.
· See Task.
Q11
Using a graphical technique determine the single wave resulting from a combination of two waves of the same frequency and then verify the result using trig formulae.
· See Task.
Q12
Use numerical integration and integral calculus to analyse the results of a complex engineering problem
· See Task.
Level of Detail in
Solution
s: Need to show work leading to final answer
Need
Question 1
(a) Find:
(4 + i2)
(1 + i3)
Use the rules for multiplication and division of complex numbers in rectangular form.
(b) Convert the answer in rectangular form to polar form
(c) Repeat Q1a by first converting the complex numbers to polar form and then using the rules for multiplication and division of complex numbers in polar form.
(d) Convert the answer in polar form to rectangular form.
Question 2
The following data within the working area consists of measurements of resistor values from a producti ...
Bayesian Estimation for Missing Values in Latin Square Designinventionjournals
International Journal of Mathematics and Statistics Invention (IJMSI) is an international journal intended for professionals and researchers in all fields of computer science and electronics. IJMSI publishes research articles and reviews within the whole field Mathematics and Statistics, new teaching methods, assessment, validation and the impact of new technologies and it will continue to provide information on the latest trends and developments in this ever-expanding subject. The publications of papers are selected through double peer reviewed to ensure originality, relevance, and readability. The articles published in our journal can be accessed online.
The aim of this presentation is to revise the functional regression models with scalar response (Linear, Nonlinear and Semilinear) and the extension to the more general case where the response belongs to the exponential family (binomial, poisson, gamma, ...). This extension allows to develop new functional classification methods based on this regression models. Some examples along with code implementation in R are provided during the talk. Lecturer: Manuel Febrero Bande, Univ. de Santiago de Compostela, Spain.
Tenser Product of Representation for the Group CnIJERA Editor
The main objective of this paper is to compute the tenser product of representation for the group Cn. Also
algorithms designed and implemented in the construction of the main program designated for the determination
of the tenser product of representation for the group Cn including a flow-diagram of the main program. Some
algorithms are followed by simple examples for illustration.
1. Henry R. Kang (1/2010)
General Chemistry
Lecture 6
Graphing
2. Henry R. Kang (7/2008)
Outlines
• Logarithm and Exponent
Definitions and rules
• Graphing
Graphing rules
Line drawing
• Linear Regression
Deviation of equations
Goodness of data fitting
Draw the least-square line
4. Henry R. Kang (7/2008)
Definition of Exponent
• Exponent is expressed as the power of a base
value.
Y = BX
.
B is the base and X is the power.
Any positive number greater than 1 can be used as
the base.
The power X can be a number or a more complex
expression such as X = a/4 or X = (a+b)/2.
Commonly used bases are 10 and e = 2.7182818…
(e is an irrational number).
Computer uses binary system; the base is 2.
7. Henry R. Kang (7/2008)
Definition of Logarithm
• Logarithm is the inverse of the exponent.
• log(base)Y = X.
Two popular bases are 10 and e.
• Common (or Briggsian) logarithm uses the base value of 10.
log10Y = X or log Y = X.
Some values of common logarithm:
log(1) = 0; log(10) = 1; log(100) = 2; etc.
• Natural (or hyperbolic) logarithm uses the base value of
e = 2.7182818…
logeY = X or lnY = X.
• Common and natural logarithms can be inter-converted.
lnY = 2.30259 × logY or logY = lnY / 2.30259.
8. Henry R. Kang (7/2008)
Rules of Logarithms
• The following rules apply to both common and natural
logarithms:
• log(A×B) = logA + logB.
log(5×7) = log(5) + log(7) = 0.698970004 + 0.84509804 = 1.544068044
log(35) = 1.544068044
• log(A/B) = logA – logB.
log(5/7) = log(5) – log(7) = 0.698970004 - 0.84509804 = -0.146128036
log(5/7) = log(0.714285714) = -0.146128036
• log(An
) = n logA.
log(73
) = 3 log(7) = 2.53529412
log(73
) = log(343) = 2.53529412
• log(A-n
) = -n logA.
• log(A1/n
) = (1/n) logA.
10. Henry R. Kang (7/2008)
Advantages of Graphing
• A graph or figure is a very powerful means
of delivering information.
The information is very compactly
represented.
The relationship between parameters is clearly
shown.
The general characteristics of the parameters
can be derived.
• A picture is worth a thousand words.
11. Henry R. Kang (7/2008)
Graphing Rules – Label & Size
• Graph should be neatly presented,
easily readable and properly titled.
Each axis should be clearly labeled with
the name of the parameter and
the unit.
Scales should be selected so that the
actual graph covers at least 50% of the
space available.
12. Henry R. Kang (7/2008)
Example – Correct Size
0
200
400
600
800
1000
1200
1400
0 10 20 30 40 50 60 70 80
Tem perature (C)
Pressure(mmH2O)
13. Henry R. Kang (7/2008)
Example – Incorrect Size
0
200
400
600
800
1000
1200
1400
0 10 20 30 40 50 60 70 80
Temperature (C)
Pressure(mmH2O)
14. Henry R. Kang (7/2008)
Graphing Rules - Axis
• An axis scale does not need to
start at “zero”
To avoid data points cluster in a
narrow range
Exception
If the extrapolation of the line is
required to the x-axis or y-axis
intercept.
15. Henry R. Kang (7/2008)
Data Cluster - Incorrect
0
1
2
3
4
5
6
7
8
9
10
0 5 10 15 20
Tem perature (C)
Volume(mL)
16. Henry R. Kang (7/2008)
Expand the Scale
0
1
2
3
4
5
6
7
8
9
10
16 16.5 17 17.5 18 18.5 19
Tem perature (C)
Volume(mL)
17. Henry R. Kang (7/2008)
Extrapolation
0
1
2
3
4
5
6
7
8
9
-10 -5 0 5 10 15 20
X
Y
18. Henry R. Kang (7/2008)
Graphing Rules - Divisions
• Scale on the graph
paper should have
divisions that are
easily “divided by
the eye”
1, 2, 5, or 10
Not 3, 6, 7, or 11
0
1
2
3
4
5
6
7
8
9
0 5 10 15 20
0 2 4 6 8or
3 60 9 12Not
2
0
4
6
8
10
12
5
10
15
20
25
30
10
20
30
40
50
60
00
19. Henry R. Kang (7/2008)
Graphing Rules – Table of Data
•A table of data may be
provided (This rule is not
always obeyed)
No individual coordinates of
data points should appear on
the graph.
20. Henry R. Kang (7/2008)
Example
0
1
2
3
4
5
6
7
8
9
10
16 16.5 17 17.5 18 18.5 19
Temperature (C)
Volume(mL)
T(°c) V
(mL)
17.0 6.58
16.8 5.76
17.8 7.71
18.5 8.84
18.2 8.47
17.5 7.04
16.5 5.25
(17.0,
6.58)
21. Henry R. Kang (7/2008)
Graphing Rules - Resolution
• Ideally, one should be able to read
all significant figures of the data
from its position on the graph
paper
Often, the significant figure of the
data is higher than the resolution of
the graph paper.
22. Henry R. Kang (7/2008)
Graphing Rules – Data Symbols
• A data point should be circled (or other
shapes such as square or triangle)
surrounding it.
• If more than one data set, each set
should have its own shape (or symbol)
to represent the data points.
You can use different color for different
data set, if color display or print is
available.
23. Henry R. Kang (7/2008)
Example
0
2
4
6
8
10
12
14
0 5 10 15
X
Y
Series1
Series2
Series3
24. Henry R. Kang (7/2008)
Graphing Rules - Drawing
• The curve or straight line is drawn smoothly
among the points, rather than connecting dots
(piece-wise linearization).
The curve or line should represent the best average of
the data.
Roughly about equal number of points above and below the
curve or line.
The curve or line does not have to touch any of the data
points.
Use a clear straightedge for drawing lines.
Use French curves for drawing curves.
25. Henry R. Kang (7/2008)
Incorrect Way of Line Drawing
0
1
2
3
4
5
6
7
8
9
10
16 16.5 17 17.5 18 18.5 19
Tem perature (C)
Volume(mL)
26. Henry R. Kang (7/2008)
Correct Way of Line Drawing
0
1
2
3
4
5
6
7
8
9
10
16 16.5 17 17.5 18 18.5 19
Tem perature (C)
Volume(mL)
27. Henry R. Kang (7/2008)
Graphing Rules – Data Points
•Data point should not exist
on the axis line.
This rule is not always obeyed.
You still plot the point if it lies
on the axis line.
29. Henry R. Kang (7/2008)
Linear Regression - Equation
• Linear regression is used to find the
straight line that fits the data best.
• General equation for a line is
Y = m X + b
X is the independent variable,
Y is the dependent variable,
m is the slope of the line, and
b is the y-axis intercept
30. Henry R. Kang (7/2008)
Linear Regression - Deviation
• Let yn = the observed values
• and ýn= the calculated value from the linear
equation
• The deviation
dn = ýn– yn (n = 1, 2, 3, - - -, N)
N is the number of data sets.
• The best result is obtained by minimizing the
deviation (or the square of the deviation)
∑ dn
2
= (y1–ý1)2
+ (y2–ý2)2
+ - - - + (yN–ýN)
31. Henry R. Kang (7/2008)
Linear Regression – Minimize Deviation
• Calculate ýn from the linear equation, we have
ýn = m xn + b
• The deviation becomes
dn = ýn– yn = m xn+ b – yn
• The square of deviation is
dn
2
= (m xn+ b – yn )2
= m2
xn
2
+ b2
+ yn
2
+ 2mbxn – 2mxnyn – 2byn
• The overall deviation is
∑ dn
2
= ∑ (m xn+ b – yn )2
∀ ∑dn
2
can be minimized by taking the partial derivatives
with respect to m and b, respectively.
32. Henry R. Kang (7/2008)
Linear Regression - Formulas
• Minimize ∑dn
2
by taking the partial derivatives
with respect to m (slope) and b (intercept)
∂(∑ dn
2
)/ ∂m = ∑(2mxn
2
+ 2bxn – 2xnyn) = 0
∂(∑ dn
2
)/ ∂b = ∑(2b + 2mxn – 2yn) = 0
• This set of equations can be solved for m and b.
N (∑xnyn ) – (∑xn) (∑ yn)
N (∑xn
2
) – (∑xn)2
N (∑xn
2
) – (∑xn)2
(∑xn
2
)(∑yn) – (∑xn)(∑xn yn)
m =
b =
where N is the number of data sets
33. Henry R. Kang (7/2008)
Linear Regression - Example
N (∑xn yn)– (∑xn) (∑ yn)
[N(∑xn
2
) – (∑xn)2
]1/2
[N(∑yn
2
)– (∑yn )2
]1/2r =
x
(C)
y
(liter)
xy x2
y2
1 0.0 20.0 0.0 0.0 400.
2 10.0 22.0 220. 100. 484.
3 20.0 23.0 460. 400. 529.
Sum 30.0
(∑xn)
65.0
(∑ yn)
680.
(∑xn yn)
500.
(∑xn
2
)
1413
(∑yn
2
)
b =
m =
N (∑xn
2
) – (∑xn)2
N (∑xn yn)– (∑xn) (∑ yn)
=
3×680.– 30.0×65.0
3×500.– 30.02
=
2040.– 1950.
1500.– 900.
=
90.0
600.
=0.150
N (∑xn
2
) – (∑xn)2
(∑xn
2
)(∑ yn) – (∑xn)(∑xn yn)
=
500.×65.0 – 30.0×680.
3×500.– 30.02
=
12100.
600.
= 20.2
=
=
3×680.– 30.0×65.0
(3×500.– 30.02
)1/2
(3×1413– 65.02
)1/2
90.0
r =
(600.)1/2
(14)1/2
0.982
r2
= 0.964
34. Henry R. Kang (7/2008)
Linear Regression - Goodness
• The “goodness” of the data fitting is expressed by
the regression coefficient r2
• If r2
= 1, perfect fit
• If r2
> 0.95, excellent fit
• If r2
> 0.90, good fit
• If r2
> 0.80, reasonable fit
• If r2
= 0, completely unrelated
N (∑xn yn)– (∑xn) (∑ yn)
[N (∑xn
2
) – (∑xn)2
]1/2
[N (∑yn
2
) – (∑yn )2
]1/2
r =
35. Henry R. Kang (7/2008)
Draw the Least-Square Line
• Once the slope m and intercept b are calculated from a
set of data (x and y), the best line can be drawn to fit the
data.
• A line, y = mx + b, is defined by two points.
• The least computational cost to find these two points is
Set x = 0, then y = b, giving the first point (0, b)
Set y = 0, then x = -b/m, giving the second point (-b/m, 0)
Put these two points in the graph, then draw a straight line
connecting these two points.
This line is the least-square line to fit the given data set (x and y) best
with a minimum error between the calculated and measured y values.