List all cyclic subgroups of U_24.SolutionThe group consists of.pdf
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List all cyclic subgroups of U_24. Solution The group consists of the elements [i] belongsto [ Z/24Z] such that (i , 24) = 1. Thus the group is of order 8, consisting of the elements [1]; [5]; [7]; [11]; [13]; [17]; [19]; [23]. The exponent of a group G is the smallest positive number n such that gn = e for all g 2 G. For each of the 8 elements [i] 2 U24, one checks by direct computation thati2 1 mod 24. This shows the exponent is 2. it is abelian, and of exponent 2, so has no element of order 8 (= jU24j), hence is not cyclic..
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![List all cyclic subgroups of U_24.
Solution
The group consists of the elements [i] belongsto [ Z/24Z] such that (i , 24) = 1. Thus the group is
of order 8, consisting of the elements [1]; [5]; [7]; [11]; [13]; [17]; [19]; [23].
The exponent of a group G is the smallest positive number n such that gn = e for
all g 2 G. For each of the 8 elements [i] 2 U24, one checks by direct computation thati2 1 mod
24. This shows the exponent is 2.
it is abelian, and of exponent 2, so has no element of
order 8 (= jU24j), hence is not cyclic.](https://image.slidesharecdn.com/listallcyclicsubgroupsofu24-230904012603-8924fc20/85/List-all-cyclic-subgroups-of-U_24-SolutionThe-group-consists-of-pdf-1-320.jpg)
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most sense?a single layer of phospholipids with the polar heads facing outside the celltwo back-
to-back phospholipid layers with the non-polar tails facing out on both sidesa single layer of
phospholipids with the polar heads facing inside the celltwo back-to-back phospholipid layers
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Solution
Two back-to-back phospholipid layers with the polar heads facing out on both sides
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Solution
Ques-1:
Autosomal recessive trait is defined as a pattern of inheritance of an allele that is recessive over
an autosome as well as two copies of this allele is essential for an expression of phenotype. In
autosomal recessive inheritance, both the sons and daughters have a chance to be affected. Most
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The java.util.* packages provide support for the event model, collections framework, date and
time facilities, and contain various utility classes. The java.util.zip and java.util.jar packages
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It is our intention to produce interference fringes by illuminating some sort of arrangement
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value valB. Make all the modifications necessary to display BOTH values -- valA and valB -- for
each ListItem in the GUI window. Supply a second set of values for valB, to complement the
given valA.
Then, without modifying the compareTo() method you just created in part c., change how the
ListItems are sorted, such that they are sorted from highest to lowest according to the value of
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Solution
import java.awt.BasicStroke;
import java.awt.Color;
import java.awt.Component;
import java.awt.Font;
import java.awt.FontMetrics;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.Stroke;
import java.awt.geom.Ellipse2D;
import java.awt.geom.Rectangle2D;
import javax.swing.Icon;
public class ListItem implements Comparable
{
private int valA;
private int valB;
public ListItem (int valA,int B)
{
this.valA = valA;
}
public int getValA ()
{
return valA;
}
public int getValB()
{
return valB;
}
public Icon createIcon (int diameter) {
return new ItemIcon (this, diameter);
}
@Override
public int compareTo (ListItem o) {
// TODO Auto-generated method stub
return 0;
}
public class ItemIcon implements Icon
{
private final static int DEFAULT_THICKNESS = 5;
private final static int DEFAULT_DIAMETER = 50;
private int width = DEFAULT_DIAMETER;
private int height = DEFAULT_DIAMETER;
private int thickness = DEFAULT_THICKNESS;
private int borderSize = DEFAULT_THICKNESS;
private int valA;
private int valB;
public ItemIcon (int valA,int valB) {
this.valA = valA;
this.valB=valB;
}
public ItemIcon (int valA, int valB,int diameter)
{
this.valA=valA;
this.valB=valB;
setDiameter (diameter);
}
public ItemIcon (ListItem item) {
this.valA = item.getValA();
this.valB=item.getValB();
}
public ItemIcon (ListItem item, int diameter) {
this (item);
setDiameter (diameter);
}
public void setDiameter (int diameter) {
width = diameter;
height = diameter;
}
public int getIconHeight() {
return height + 2 * DEFAULT_THICKNESS;
}
public int getIconWidth() {
return width + 2 * DEFAULT_THICKNESS;
}
public void paintIcon (Component comp, Graphics g, int x, int y)
{
// get graphics context
Graphics2D g2 = (Graphics2D) g;
// set stroke size and color
g2.setPaint (Color.BLACK);
Stroke s = new BasicStroke (thickness);
g2.setStroke (s);
// draw white-filled circle with red border
Ellipse2D e1 = new Ellipse2D.Float (x + borderSize, y + borderSize, height, width);
g2.draw (e1);
g2.setPaint (Color.WHITE);
g2.fill (e1);
g2.setPaint (Color.RED);
// draw text
Font f = new Font (\"Arial\", Font.BOLD, 14);
FontMetrics fm = g.getFontMetrics (f);
String str = ((Integer) valA).toString();
Rectangle2D rect = fm.getStringBounds (str, g2);
int textHeight = (int) rect.getHeight();
int textWidth = (int) rect.getWidth();
int panelHeight= getIconHeight();
int panelWidth = getIconWidth();
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int offsetX = (panelWidth - textWidth) / 2;
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where a is associated with the interaction between molecules and b takes into account the finite
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YES
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extrinsic proteins or they may be traversing the membrane when they are said to be the intrinsic
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Explain your answer. Sickle cell disease in an inherited human condition of the blood cells.
Those who receive the allele from both parents (homozygous) has a fill blown sickle cell disease.
A person who only has one copy of the allele (heterozygous) may experiences stickling of their
blood cells under certain, stressful conditions. Give a simple explanation for the heterozygous
individual having a mild form of the disease. Why do individuals who inherit an A type allele
and a B type allele have blood type AB while a person who inherits an A type allele and an O
type allele have blood type A? How are pedigrees useful in counseling couples regarding the
likelihood they can pass a genetic disorder on to their offspring? A female child has the
autosomal dominant condition of polydactyl. The mother has the normal number of fingers and
toes while the father has six fingers on each hand. The couple has a second child, a boy, who has
the normal number of fingers and toes. Give the genotype of each parent and each child. Draw a
pedigree demonstrating polydactyl in the above family. What is the likely hood that if they have
a third child they will be born with polydactyl?
Solution
Ques-1:
Autosomal recessive trait is defined as a pattern of inheritance of an allele that is recessive over
an autosome as well as two copies of this allele is essential for an expression of phenotype. In
autosomal recessive inheritance, both the sons and daughters have a chance to be affected. Most
human genetic disorders are due to autosomal recessive inheritance. An individual can acquire
these conditions from parents that do not display symptoms because most of the parents carry
mutated alleles so that they act as \"carriers\" for the mutated gene resulting in autosomal
recessive disorder in the offspring
For example, Tay-Sachs disease is a rare autosomal recessive genetic disorder cuased by
mutations potentially affects nervous system in humans. Tay-Sachs disease is an autosomal
recessive genetic disease, which results in progressive degeneration of nerve cells that causes
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I know how to construct the Node class but still having trouble with the Linked List. I\'m
assuming it is called a Linked List, keyword being List. Do I import from java.util.* for the
list/array list? Which one do I use: List or ArrayList? If I am importing one of those types of
lists, do I add the LinkedList nodes to the list while having having references to link each other?
Solution
The java.util.* packages provide support for the event model, collections framework, date and
time facilities, and contain various utility classes. The java.util.zip and java.util.jar packages
provide support for ZIP file format and Java Archive (JAR) file formats respectively, and the
java.util.prefs package implements the Preferences API.
You need to only import java.util.* because it contains the all interfaces and classes.
Consider the below code for LinkedList:.
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It is our intention to produce interference fringes by illuminating some sort of arrangement
(Young\'s Experiment, a thin film, the Michelson Interferometer, etc.) with light at a mean
wavelength of 500 nm, having a linewidth of 2.5 times 10^-3 nm. At approximately what optical
path length difference can you expect the fringes to vanish?
Solution
Path length dt= 2/d=(500X10-9)2 /(2.5x10-3x10-9)
Path length =0.1 m.
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HELP
Modify the code so that the ListItem contains two values, instead of just one. Name the new
value valB. Make all the modifications necessary to display BOTH values -- valA and valB -- for
each ListItem in the GUI window. Supply a second set of values for valB, to complement the
given valA.
Then, without modifying the compareTo() method you just created in part c., change how the
ListItems are sorted, such that they are sorted from highest to lowest according to the value of
the new field, valB.
Solution
import java.awt.BasicStroke;
import java.awt.Color;
import java.awt.Component;
import java.awt.Font;
import java.awt.FontMetrics;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.Stroke;
import java.awt.geom.Ellipse2D;
import java.awt.geom.Rectangle2D;
import javax.swing.Icon;
public class ListItem implements Comparable
{
private int valA;
private int valB;
public ListItem (int valA,int B)
{
this.valA = valA;
}
public int getValA ()
{
return valA;
}
public int getValB()
{
return valB;
}
public Icon createIcon (int diameter) {
return new ItemIcon (this, diameter);
}
@Override
public int compareTo (ListItem o) {
// TODO Auto-generated method stub
return 0;
}
public class ItemIcon implements Icon
{
private final static int DEFAULT_THICKNESS = 5;
private final static int DEFAULT_DIAMETER = 50;
private int width = DEFAULT_DIAMETER;
private int height = DEFAULT_DIAMETER;
private int thickness = DEFAULT_THICKNESS;
private int borderSize = DEFAULT_THICKNESS;
private int valA;
private int valB;
public ItemIcon (int valA,int valB) {
this.valA = valA;
this.valB=valB;
}
public ItemIcon (int valA, int valB,int diameter)
{
this.valA=valA;
this.valB=valB;
setDiameter (diameter);
}
public ItemIcon (ListItem item) {
this.valA = item.getValA();
this.valB=item.getValB();
}
public ItemIcon (ListItem item, int diameter) {
this (item);
setDiameter (diameter);
}
public void setDiameter (int diameter) {
width = diameter;
height = diameter;
}
public int getIconHeight() {
return height + 2 * DEFAULT_THICKNESS;
}
public int getIconWidth() {
return width + 2 * DEFAULT_THICKNESS;
}
public void paintIcon (Component comp, Graphics g, int x, int y)
{
// get graphics context
Graphics2D g2 = (Graphics2D) g;
// set stroke size and color
g2.setPaint (Color.BLACK);
Stroke s = new BasicStroke (thickness);
g2.setStroke (s);
// draw white-filled circle with red border
Ellipse2D e1 = new Ellipse2D.Float (x + borderSize, y + borderSize, height, width);
g2.draw (e1);
g2.setPaint (Color.WHITE);
g2.fill (e1);
g2.setPaint (Color.RED);
// draw text
Font f = new Font (\"Arial\", Font.BOLD, 14);
FontMetrics fm = g.getFontMetrics (f);
String str = ((Integer) valA).toString();
Rectangle2D rect = fm.getStringBounds (str, g2);
int textHeight = (int) rect.getHeight();
int textWidth = (int) rect.getWidth();
int panelHeight= getIconHeight();
int panelWidth = getIconWidth();
// Center text horizontally and vertically
int offsetX = (panelWidth - textWidth) / 2;
int offsetY =.
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Find the second virial coefficient for the Dieterici equation of state
p = RT/V-b exp (-a/RTV)
Solution
where a is associated with the interaction between molecules and b takes into account the finite
size of the molecules, similarly to the Van der Waals equation.
the viral coefficient are
Tc= a/(4Rb)
Pc= a/(4c2b2)
Vc=2b.
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Find the composite phi and velocity field. Find the velocity vector at (x, y) = (0, 0) and (1, 1)
Find the pressure difference between the two points in part (b) (assume gravity is in the -Z
direction.
Solution
1) calculate deflection of point A , in the Y direction
2) use another method to solve this problem, one method MUST be castigliano\'s method
Note: the material is aluminum.. P is in Newtons(N) and L\'s,D\'s,and T\'s are in millimeters
(mm)
please show all work in detail on how to get part 1&2 thanks.
EnvironmentalCivil Engineering QuestionWhen chemicals are added t.pdf
Environmental/Civil Engineering Question
When chemicals are added to water for coagulation, they are flash-mixed. Is there an advantage
to this or would prolonged slow stirring accomplish the same purpose?
Solution
YES
there is advantage of adding the chemical
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Did Carl Linnaeus incorporate evolutionary thinking in the development of his Systema Naturae?
Solution
Answer
No Carl Linnaeus did not incorporated evolutionary thinking in the development of his Systema
Naturae.
He published this book in 1735
In this book Linnaeus presented an outline of the sexual system of classification of plant
kingdom.
He classified plant kingdom into 24 classes on the basis of presence and absence of stamens and
their number, without taking into account any evolution relationship.
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Do you think wood is mechanically homogeneous (i.e. as strong in all directions? Why or why
not?
Solution
Wood is mechanically heterogeneous because the composition of wood are of solid,some liquid
and little gases when heated.There are differences in the grain that show the material is not
uniformly mixed. This allows certain areas of the sample of wood to have different properties
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Design a database to store details about U.S. presidents and their terms in office. Also, record
details of their date and place of birth, gender, and political party affiliation (e.g., Caluthumpian
Progress Party). You are required to record the sequence of presidents so the predecessor and
successor of any president can be identified. How will you model the case of Grover Cleveland
who served nonconsecutive terms as president? Is it feasible that political party affiliation may
change? If so, how will you handle it? In populating your solution include the following
information (note this is not the format you will store it in the database):
Chester
Arthur
Born 10/05/1829
Whig until 01/01/1854
Republican after that
21st president
Grover
Cleveland
Born 03/18/1837
Democrat his entire career
22nd President
24th President
Benjamin
Harrison
Born 08/20/1833
Whig until 01/01/1856
Republican after that
23rd President
Chester
Arthur
Born 10/05/1829
Whig until 01/01/1854
Republican after that
21st president
Grover
Cleveland
Born 03/18/1837
Democrat his entire career
22nd President
24th President
Benjamin
Harrison
Born 08/20/1833
Whig until 01/01/1856
Republican after that
23rd President
Solution
Create table President(
PresidentID integer(6) primary key,
Name varchar (30)
Place varchar (20),
Dateofbirth date,
Gender varchar(1) check in (m,f),
Ruling_startdate date,
Ruling_end date date,
Affiliation_party varchar(20),
Terms varchar(30));
We will design all the attributes under one table or define separate table for president personal
details and party details.
We can separate the tables now.
President table:
Create table president(
PresidentId integer(6) primary key,
Name varchar(30) foreign key,
Dateofbirth date,
Gender varchar(1) check in (m,f)
Affiliation_party varchar(20),
Terms varchar(20));
Create table pres_det(
presidentID integer(6) primarykey,
Place_of_elect varchar(20),
Birthdate date,
Ruling_startdate date,
Ruling_enddate date,
);
CREATE SEQUENCE seq_presidentID
MINVALUE 10000
START WITH 10000
INCREMENT BY 100
Cycle.
If we execute this sequence president id is set to default value and increment by 100 for every
president value.
DECLARE @maxTerm INT
DECLARE @minTerm INT
SELECT @maxTerm = MAX(term_id), @minTerm = MIN(term_id) FROM President AS p
JOIN President AS p
ON p. presidentId = pres_det.presidentId
WHERE p.Name = \'name\'
Then predecessor is
SELECT p.Name FROM pres_det AS t
JOIN President AS p
ON p.presidentId = t.presidentId
WHERE p.term = @minTerm - 1
To find the successor is
SELECT p.Name FROM pres_det AS t
JOIN President AS p
ON p.presidentId = t.presidentId
WHERE p.term = @maxTerm +1.
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Chapter 1 - A View of Life and its Chemical Basis. 1. Explain the basic characteristics that are
common to all living things. 2. Describe the levels of organization of life. 3. Distinguish among
the three domains of life and kingdoms. 4. Explain the process of natural selection. 5. Identify
the components of the scientific method. 6. Contrast ionic and covalent bonds. 7. Explain how
the properties of water make life possible. 8. Recognize the importance of functional groups in
determining the chemical properties of an organic molecule. 9. Summarize the categories of the
main macromolecules and provide examples of their diverse biological functions: carbohydrates,
lipids, proteins and nucleic acids.
Solution
please split the questions and ask. it was too lengthy to answer so all are in short.
1. Explain the basic characteristics that are common to all living things.
The fundamental characteristics of all living beings are growth, reproduction. All are made up of
cells, they adapt and respond to the environment. They all obtain and utilize energy. Though all
these are in common, they have different cellular organization.
2. Describe the levels of organization of life.
The various levels of organization of life include organelles, cells, tissues, organs, organ systems,
organisms, populations, communities, ecosystem and biosphere. They are in the chronological
order starting from the simplest form to the most complex one.
3. Distinguish among the three domains of life and kingdoms.
Three domain system introduced by carl woese divides cellular forms into three domains- archae,
bacteria and eukaryotes. It mainly separates archae from prokaryotes based on their 16sRNA
differences. This classification is above the kingdom level.
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are placed into 3 kingdoms plantae, fungi and animalia. Protista are a separate kingdom
consisting of diversified algae and others which does not come under other groups
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Charles Darwin proposed the natural selection in his theory of evolution. Natural selection is a
process by which the organisms which are adapted to the environmental conditions and survive
in them. Shortly said as “survival of the fittest”. They transfer the genetic traits to further
generations making them more dominant while those who cannot survive get eliminated.
5. Identify the components of the scientific method.
Scientific method is a planned method which helps scientists in answering or solving a problem.
It consists of 6 components. 1. Purpose or question of research, 2. Research to find more
information on the question, 3. Hypothesis after getting preliminary information, 4. Experimental
design of the question, 5. Analysis of the results, 6. Conclusion of the experiment.
6. Contrast ionic and covalent bonds.
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between two non-metals .
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CHOOSE THE CORRECT ANSWER:
1) PVC Corporation is considering an investment proposal in which a technology investment of
$10,000 would be required. The investment would provide cash inflows of $2,000 per year for
six years. The technology investment will have no salvage value at the end of six years. If the
company\'s discount rate is 10%, the investment\'s net present value is closest to (Ignore income
taxes.): (See the time value of money BELOW THE PAGE.)
A. $1,289
B. $(1,289)
C. $3,066
D. $2,000
2) RST Corporation has provided the following data concerning an investment project that it is
considering:
Initial investment $160,000
Annual cash flow $54,000 per year
Salvage value at the end of the project $11,000
Expected life of the project years 4 years
Discount rate 10%
The net present value of the project is closest to: (See the time value of money BELOW THE
PAGE)
A. $175
B. ($23,694)
C. $3,662
D. $11,175
3) An increase in the expected salvage value at the end of a capital budgeting project will
A. Increase the cash outflows of the project.
B. Increase the present value of cash inflows of the project
C. Decrease the cash inflows of the project
D. Decrease the net present value of a project
HERE IS THE Time Value of Money Factor for the Present Value of a Sum
PV =FV * TVM Factor
Present Value of a Sum = FV * 1/ (1 + i) n
2%
3%
4%
5%
6%
8%
10%
12%
1
0.9804
0.9709
0.9615
0.9524
0.9434
0.9259
0.9091
0.8929
2
0.9612
0.9426
0.9246
0.9070
0.8900
0.8573
0.8264
0.7972
3
0.9423
0.9151
0.8890
0.8638
0.8396
0.7938
0.7513
0.7118
4
0.9238
0.8885
0.8548
0.8227
0.7921
0.7350
0.6830
0.6355
5
0.9057
0.8626
0.8219
0.7835
0.7473
0.6806
0.6209
0.5674
6
0.8880
0.8375
0.7903
0.7462
0.7050
0.6302
0.5645
0.5066
7
0.8706
0.8131
0.7599
0.7107
0.6651
0.5835
0.5132
0.4523
8
0.8535
0.7894
0.7307
0.6768
0.6274
0.5403
0.4665
0.4039
9
0.8368
0.7664
0.7026
0.6446
0.5919
0.5002
0.4241
0.3606
10
0.8203
0.7441
0.6756
0.6139
0.5584
0.4632
0.3855
0.3220
20
0.6730
0.5537
0.4564
0.3769
0.3118
0.2145
0.1486
0.1037
25
0.6095
0.4776
0.3751
0.2953
0.2330
0.1460
0.0923
0.0588
Time Value of Money Factor for the Present Value of an Annuity
PV = Annuity * TVM Factor
Present Value of an Annuity = X*((1-(1/(1+i)^n))/i)
2%
3%
4%
5%
6%
8%
10%
12%
1
0.9804
0.9709
0.9615
0.9524
0.9434
0.9259
0.9091
0.8929
2
1.9416
1.9135
1.8861
1.8594
1.8334
1.7833
1.7355
1.6901
3
2.8839
2.8286
2.7751
2.7232
2.6730
2.5771
2.4869
2.4018
4
3.8077
3.7171
3.6299
3.5460
3.4651
3.3121
3.1699
3.0373
5
4.7135
4.5797
4.4518
4.3295
4.2124
3.9927
3.7908
3.6048
6
5.6014
5.4172
5.2421
5.0757
4.9173
4.6229
4.3553
4.1114
7
6.4720
6.2303
6.0021
5.7864
5.5824
5.2064
4.8684
4.5638
8
7.3255
7.0197
6.7327
6.4632
6.2098
5.7466
5.3349
4.9676
9
8.1622
7.7861
7.4353
7.1078
6.8017
6.2469
5.7590
5.3282
10
8.9826
8.5302
8.1109
7.7217
7.3601
6.7101
6.1446
5.6502
20
16.3514
14.8775
13.5903
12.4622
11.4699
9.8181
8.5136
7.4694
25
19.5235
17.4131
15.6221
14.0939
12.7834
10.6748
9.0770
7.8431
HERE IS THE Time Value of Money Factor for the Pre.
Civil Infrastructure Problems Why the public is not concerned about t.pdf
Civil Infrastructure Problems Why the public is not concerned about the condition of civil
infrastructure while it affects them personally? Why the public do not comment more to their
elected representatives?
Solution
Infrastructure is the backbone of a country and has serious impact on progress of that country
and happiness of its people. The basic infrastructure comprises of transportation, communication,
sewage, water and electric systems. Further elaborating, it covers roads, highways, bridges,
telecommunication, ports, airports and railways.
People use the infrastructure daily, in some form or other and are affected by it. Poor
infrastructure does generate discontent and is reflected in people movements protesting and
demanding rapid improvements. These protests are more visible in poor and developing
countries where even the basic infrastructure of housing and allied services itself is deficient or
non-existent. The protests are now visible in developed countries also where aging infrastructure
because of lack of funds is deteriorating rapidly. In fact, it was a major point of agenda in recent
presidential elections in USA. The administration is understood to be planning on a major
overhaul of infrastructure costing trillion of dollars. In countries like India the Prime Minister
personally monitors the overall progress of improvement of infrastructure as it was a major
election promise on which he was elected.
So, it is not correct to say that people do not complain of poor infrastructure,though it may be
muted in some countries, cities , states and provinces .People do complain to their elected
representative and that is why it is everywhere a major point of promises in the election agenda.
These infrastructure projects tend to be high-cost investments; though they are vital to a
country\'s economic development and prosperity. Therefore, their improvements take time and is
generally spread over several years..
Cell Structure and FunctionLabel each of the arrows in the followi.pdf
Cell Structure and Function
Label each of the arrows in the following slide image: AE Onion Root Tip: 1000X
Solution
A) Cells are in anaphase
The cells are arranged such that the centromeres have split at this point and the chromatids have
been separated by migrating to the opposite poles.
B) Cells here are present in the early prophase of mitosis, where the centriole can now be seen to
move towards the opposite poles of the cell.
C) cell plate
D) not clear.y distinguishable.
C Language ProblemThere are more than one wrong things I think.P.pdf
C Language Problem
There are more than one wrong things I think.
Please Explain Is the following case structure correct? If incorrect - what\'s wrong? double name;
switch(name){ease 1: break: case 2: break case 3: break: default:}
Solution
Change datatype of \"name\" to int or change case values to double type
Default keyword should be followed by colon(:).
Briefly discuss the adaptive changes seen in the chordates over the o.pdf
Briefly discuss the adaptive changes seen in the chordates over the other animal phyla?
Solution
Animals in the phylum Chordata reveal four key features that appear at some point all through
their development.
a. Notochord - The chordates are called for the notochord: a flexible, rod-shaped design that is
found within the embryonic stage of most chordates and also in the adult stage of some chordate
species. It is found between the intestinal pipe and the nerve cord, providing skeletal help
through along the body. In some chordates, the notochord acts while the primary axial support of
the human body throughout the animal\'s lifetime. In vertebrates, the notochord is provide during
embryonic development, of which time it induces the development of the neural pipe which acts
as an assistance for the developing embryonic body. The notochord, however, is replaced by the
vertebral column (spine) in most adult vertebrates.
b.Dorsal Hollow Nerve Cord - The dorsal hollow nerve cord derives from ectoderm that sheets
into a worthless pipe all through development. In chordates, it is found dorsally (at the top of the
animal) to the notochord. In contrast to the chordates, other animal phyla are known by strong
nerve cables which are positioned either ventrally or laterally. The nerve cord within most
chordate embryos develops into the brain and spinal cord, which include the main anxious
system.
c.Pharyngeal Slits - Pharyngeal slits are openings in the pharynx (the place just posterior to the
mouth) that expand to the surface environment. In organisms that are now living in aquatic
surroundings, pharyngeal slits allow for the quit of water that enters the mouth all through
feeding. Some invertebrate chordates use the pharyngeal slits to filtration food out of the water
that enters the mouth. In vertebrate fishes, the pharyngeal slits develop into gill arches, the bony
or cartilaginous gill supports. In most terrestrial animals, including mammals and birds,
pharyngeal slits are providing just during embryonic development. In these animals, the
pharyngeal slits develop into the jaw and inner ear bones.
d.Post-anal Tail - The post-anal tail is a posterior elongation of the human body, increasing
beyond the anus. The trail contains skeletal things and muscles, which give a way to obtain
locomotion in aquatic species. In some terrestrial vertebrates, the trail also assists with stability,
courting, and signaling when risk is near. In individuals and other apes, the post-anal tail is
provide during embryonic development, but is vestigial as an adult..
Argue in favor or against the following statements (include examples.pdf
Argue in favor or against the following statements (include examples, rationale, explanations,
etc.):
1. Plants migrate
2. Plants defend themselves
3. Plants are good mothers.
4. Grasses are the most important food for humans.
Solution
Plant migration
The Dispersal of the seeds is continuous process, by which plant species survives and spreads.
Migration of plants is a result of the dispersal of the seeds directly to wider areas of earth,also a
phenomenon constant in plant kingdom.Angiosperms species covered the whole earth migrating
to the different countries along the different routes. Due to global warming,the plants are heading
for hills for better survival. A study shows that 171 forest species in the Western Europe, among
most of spp are shifting toward their a favourable locations to cooler spots for better
survival.Plants are migrate toward higher altitudes only because of the climate change, most of
world plant species diversity are found in the tropical areas, a very few studies, have examine
either tropical mountain plant species are they affected by change in climate to same extent as
the temperate plant species. new study has determined, major changes occurred in last two
[2]centuries.If we compare the migration of the plant communities on Chimborazo volcano in the
Ecuador with a historical data from the 1802, the Aarhus University researchers found that
average upslope shifted more than [> 500 metres].and The entire boundary of vegetation have
been moved upwards from the 4,600 metres to the 5,200 metres. The main region for this
dramatically shift is only the climate change last 210 years
2.Plants defend themselves
Plant can not run away from danger like most of the animal do,so that they have developed there
own weapon & aroma & release toxic chemical in order to protect themselves,
Plant are able to move anywhere, so that every single cell defend itself against the all attack from
the invading pathogens, for example- fungi and bacteria.
first line defense in plant
The first line of defense in plants is intact & impenetrable barrier that composed of the bark &
waxy cuticle. Both are protect plants against the herbivores. And Other adaptations against
herbivores are include the, thorns the modified branches of plant, hard shells and the
spines.[modified leaves]first line defense occur when the plant cells can detect the presence of
the foreign invader like(flagella of the bacteria or chitin which found in fungi cell walls) and they
alert the surrounding cells through releasing molecules that give massage to other cells so other
cell ready to fight invader their defences. example, the alerted infected plant cells secrete
molecules are harmful to invadors or plant cell build up their more cell walls for extra protection
form invader.
second line of defense is hypersensitive response (HR).infected plant cells around the infection
site will kill [itself]themselves to stop pathogen to spreading infection throughout the plant. This
pr.
27. What is a prosthetic group Identify prosthetic groups in the fol.pdf
27. What is a prosthetic group? Identify prosthetic groups in the following conjugated proteins:
A prosthetic group is a part of the protein that is not composed of amino acids. Lipoproteins
Carbohydrates Hemoproteins Lipids Flavin nucleotides Metalloproteins Glycoproteins
Phosphate groups Fe, Ca, Copper, Zinc Phosphoproteins Flavoproteins Heme (iron porphyrin)
Solution
part-a: Hemoproteins contain hemoglobin, in which \"iron (Fe2+) is the prosthetic group useful
to carry oxygen
Part-b: Lipoproteins contain lipid moieties with double & triple bonds as prosthetic groups
Part-c: Metalloproteins such as cobalamin contain \"cobalt\" as prosthetic groups in their
structure of intrinsic factor human digestive system
Part-c: Flavoproteins contain flavin as prosthetic group
Glycoproteins is conjugate protein that contains \"glycosidic linkage\" as prosthetic groups.
What is the role of sequence numbers in the rdt protocolWhat is t.pdf
What is the role of sequence numbers in the rdt protocol?
What is the role of timers in the rdt protocol?
Solution
The internet network layer provides only best effort service with no guarantee that packets arrive
at their destination. Also, since each packet is routed individually it is possible that packets are
received out of order. For connection-oriented service provided by TCP, it is necessary to have a
reliable data transfer (RDT) protocol to ensure delivery of all packets and to enable the receiver
to deliver the packets in order to its application layer.
A simple alternating bit RDT protocol can be designed using some basic tools. This protocol is
also known as a stop-and-wait protocol: after sending each packet the sender stops and waits for
feedback from the receiver indicating that the packet has been received.
Stop-and-wait RDT protocols have poor performance in a long-distance connection. At best, the
sender can only transmit one packet per round-trip time. For a 1000 mile connection this
amounts to approximately 1 packet (about 1500 bytes) every 20 ms. That results in a pathetic 75
KB per second rate.
To improve transmission rates, a realistic RDT protocol must use pipelining. This allows the
sender to have a large number of packets \"in the pipeline\". This phrase refers to packets that
have been sent but whose receipt has not yet verified by the receiver..
Which of the following statements about mitosis is true O a. Mitosis.pdf
Which of the following statements about mitosis is true? O a. Mitosis produces two daughter
cells identical to the mother cell. b Mitosis is the division of the nuclear component (DNA)
which is then followed by the division of the cytoplasm (cytokinesis). C.A mitotically dividing
cell copies its DNA, separates these copies at opposite poles and then divides. d All of the above
O e. A and B only
Solution
1. Answer: a. Mitosis produces two daughter cells identical to the mother cell.
Explanation- The chromosomes undergo duplication before mitosis. After this, identical
chromosomes reach daughter cells.
2. Answer: b. During metaphase, chromosomes align at the metaphase plate, the equator of the
mitotic spindle
Explanation- The chromosomes align at the centre of the mitotic spindle and the two
kinetochores of each chromosome get attached to the microtubules from opposite poles of the
spindle.
3. Interphase- D. Cell growth. Replication of DNA.
Prophase- G. The chromosomes condense, each with two sister chromatids. Mitotic spindle
begins to form.
Prometaphase- C. The nuclear envelope breaks down and the chromosomes attach to the mitotic
spindle at their kinetochores.
Metaphse- A. Chromosomes align at the equator of the mitotic spindle. For each chromosome,
the sister chromatids are attached at the kinetochore to microtubules from opposing poles.
Anaphase- E. The sister chromatids are pulled apart to opposite poles of the mitotic spindle.
Telophase- F. A nuclear membrane forms around each set of chromosomes at the mitotic
spindle\'s poles. Mitosis ends.
Cytokinesis- B. Cell pinches in until cytoplasm is divided in two (in animal cells). Results in two
daughter cells that each contain a nucleus.
Explanation- Mitosis comprises of four basic phases, which according to their order are-
prophase, metaphase, anaphase, and telophase. Prometaphase is the late phase of prophase.
Cytokinesis is the division of the cytoplasm. Interphase is the preparatory stage of mitosis..
Which of the following factors would NOT contribute to allopatric spe.pdf
Which of the following factors would NOT contribute to allopatric speciation? The isolated
population is exposed to different selection pressures than the ancestral population Different
mutations begin to manifest in the gene pools of the separated populations The separated
population is small, and genetic drift occurs. There is high gene flow between the two
populations A population becomes geographically isolated from the parent population.
Solution
Ans: There is high gene flow between the tow population is not contribute to allopatric
speciation.
In allopatric speciation, a population splits into two geographically isolated populations due to
formation of a barrier between portions of a population.The isolated populations then experience
differentiating genotypic and phenotypic divergence as a result of different selective pressures in
their differing environments. Additionally, genetic drift will occur differently, eventually
differentiating the population’s genotypes and phenotypes. The separated populations will also
experience different mutations that may persist differently in their respective environments. The
each population have evolved differently to an extent that they remain isolated because changes
are too great to enable genetic mixing through sexual reproduction. Allopatric speciation is not
necessarily precluded if some individuals from one group cross the barrier and mate with
members of the other group. In other words, allopatric speciation can occur if gene flow between
groups is greatly reduced, but not entirely eliminated..
What would be the exact Materials and Method for this experiment2.pdf
What would be the exact Materials and Method for this experiment?
2. Material and methods
2.1. Subjects
One hundred obese children (58 girls and 42 boys) were studied
with a mean age (±SD) of 11.4 ± 2.7 (range 5–17 y). Obesity was de-
fined as a body mass index (BMI) above the 97th percentile for age
and sex according to the relevant national curves [28]. The participants
were recruited as part of the pediatric medical consultations of the Pre-
vention and Management Pediatric Obesity Network of Besançon
(France).
The study was conducted in accordance with the Declaration of Hel-
sinki and approved by the local ethics committee (Comité de Protection
des Personnes, Est-II, France, n°2015-A00763-46). All participants and
their parents or legal guardians were fully informed of the experimental
procedures and gave written informed consent.
2.2. Experimental procedures and measurements
After a medical exam performed by a pediatrician to ensure that the
participant had the ability to complete the whole protocol, the included
subjects underwent different assessments on the same day: i) anthro-
pometric characteristics and maturation status; ii) blood sample and
blood pressure; iii) physical activity level; iv) sleep-disordered
breathing.
Solution
2. Material and methods:
Assessment of anthro-pometric characteristics and maturation status in childhood who have
obesity followed by examination of regular blood sample and monitoring blood pressure;
monitoring of physical activity level as per regular ime scale, finally asking them for any
presence of sleep-disordered breathing
2.1. Subjects - obese children --> One hundred obese children (58 girls and 42 boys) were
studied with a mean age (±SD) of 11.4 ± 2.7 (range 5–17 y). Obesity was de-fined as a body
mass index (BMI) above the 97th percentile for age and sex according to the relevant national
curves.
What are the most challenges of wireless networksSolutionThe .pdf
What are the most challenges of wireless networks?
Solution
The most challenges in wireless networks are:
1. Security: Mobility of users increases the security threats in a wireless network. The present
wireless networks utilize authentication and data encryption methods on the interface to provide
security to the users.
2. Connectivity issues: The connectivity issues can be addressed in many ways such as, adding
access points, dividing networks to sustain proper access, limiting the bandwidth of the guest
network and implementing management and policy software. These issues may occur from the
addition of applications to the network.
3. power and energy: A mobile device is generally useful, small in size, and committed to carry
out a certain set of functions. When a device is permitted to move freely, it would usually be
tough to receive a continuous supply of power. To preserve energy, a mobile device should be
able to operate in a successful and well-organized manner..
What is particle What is a rigid body What is conservative .pdf
What is particle? What is a \"rigid body\"? What is \"conservative force\"? Give an example of
a \"non-conservative\" force? The below relationship embodies what general principle of
physical science? T_1 + V_1 = T_2 + V_2 What are the three ways in which a body can
move? What is \"kinetics\"? What is \"kinematics\"? What is \"power\"? What is \"work\"?
What significance does the date 07 December 1941 have in American history?
Solution
2.rigid body is a body which does not under any deformation on application of load
3.conservative force is a force under which internal energy of system remains conserved
4.friction force is an example of non conservative force
6.a body can move either by pure translatory motion or pure rotary motion or with both rotary
plus translatory
8.kinematics is the branch of mechanics which deals with motion of body without considering
force
9.power is defined as rate of doing work
10.work is defined force multiplied fy displacement in the direction of applied load.
What are some of the obstacles or barriers to implementing EBP (evid.pdf
What are some of the obstacles or barriers to implementing EBP (evidenced based practices) in
nursing? Provide a rationale for your answer. Since there are numerous topics on the issue, it is
not appropriate to repeat one that has already been mentioned unless providing new information.
Solution
Ans:
Obstacles or barriers to implementing EBP (evidenced based practices) in nursing includes
(1) Evidence might have a limited role in decision-making processes; (2) aspects other than
quality of care steer the evidence-based practice agenda; (3) some health care providers benefit
less from evidence-based practice than others and (4) there is a lack of competences to put the
evidence-based principles in practice.
Policy makers might consider health care system characteristics from and strategies developed or
suggested by others to respond to country-specific obstacles. Examples include but are not
limited to; (a) providing incentives for patient-centred care coordination and patient
communication, (b) supporting practitioners interested in applying research-related activities, (c)
considering direct access systems and interprofessional learning to respond to the demand for
autonomous decision-making from satellite professional groups, (d) systematically involving
allied health professionals in important governmental advisory boards, (e) considering
pharmaceutical companies perceived as \'the enemy\' an ally in filling in research gaps, (f)
embedding the evaluation of evidence-based knowledge and skills in examinations (g) moving
from (in)formative learning to transformative learning and (h) organizing high quality catch-up
programs for those who missed out on evidence-based medicine in their curriculum..
Type II hypersensitivity occurs when a.An immediate allergic or ana.pdf
Type II hypersensitivity occurs when: a.An immediate allergic or anaphylactic type of reaction
occur b. Antibodies arc formed against antigens on cell surfaces c. Antigen-antibody complexes
arc deposited in tissues d. Reactions do not require antibody production Which of the following
is an example of type III hypersensitivity: a. Anaphylaxis b. SLE- Systemic Lupus
Erythematosus c. Transfusions reactions d. Transplant rejection. Which of the following is
capable of establishing the MAC- membrane attack complex, effect cell lysis? a. Mast cells b.
MHC I proteins c. MHC II proteins d. Complement proteins In Myasthenia Gravis a.
Acetylcholine receptors on neurons arc bound by antibodies b. Acetylcholine receptors on the
motor end plate arc bound by antibodies c. Acetylcholine vesicles arc bound antibodies d. An
autoimmune response is not involved. Cytotoxic T cells (CD8^+) react to cells that haw: a.
Foreign MHC class II proteins on their surfaces b. B cell receptors c. Foreign MHC class I
proteins on their surfaces d. Soluble antigens
Solution
13. b
The type II hypersensitivity occurs when, the antibodies produced by the immune response bind
to antigens on the patient\'s own cell surfaces. This causes a B cell response, because here
antibodies are formked against the foreign antigen.
14. b
SLE- Systemic lupus erythematosus, it is due to Nuclear antigens. It occurs when accumulation
of immune complexes (antigen-antibody) takes place. Type III hypersensitivity causes
inflammatory response and attraction of leukocytes.
15 d. complement proteins
Transmembrane channels formed by membrane-attack complex (MAC) leads to cell lysis and
death by disrupting the cell membrane of target cells. It involves proteins C5 through C9
16 b.
Antibody block or destroy nicotinic acetylcholine receptors at the junction between the nerve and
muscle.
17. c. Foreign MHC class I protein on thier surface.
cytotoxic T cell or CD8+ T-cell or killer T cell are able to recognize a specific antigen. TCR to
bind to the class I MHC molecule,.
The sum of two integers is 19. The larger integer is 5 less than twi.pdf
The sum of two integers is 19. The larger integer is 5 less than twice the smaller. Find the two
integers.
Solution
Let the larger integer be X and smaller integer be Y
Given sum of two integers is 19 i.e., X+Y=19
The larger integer is 5 less than twice the smaller i.e., X=2Y-5
X+Y=19
By plugging X=2Y-5 in X+Y=19 we get
2Y-5+Y=19
3Y=24
Y=24/3=8
X+Y=19
X+8=19
X=19-8=11
Therefore the two integers are 11,8.
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Did Carl Linnaeus incorporate evolutionary thinking in the developme.pdf
Did Carl Linnaeus incorporate evolutionary thinking in the development of his Systema Naturae?
Solution
Answer
No Carl Linnaeus did not incorporated evolutionary thinking in the development of his Systema
Naturae.
He published this book in 1735
In this book Linnaeus presented an outline of the sexual system of classification of plant
kingdom.
He classified plant kingdom into 24 classes on the basis of presence and absence of stamens and
their number, without taking into account any evolution relationship.
Evolutionary thinking was incorporated into classifications after the publication of Darwin\'s
Origin of Species (1859).
Do you think wood is mechanically homogeneous (i.e. as strong in all.pdf
Do you think wood is mechanically homogeneous (i.e. as strong in all directions? Why or why
not?
Solution
Wood is mechanically heterogeneous because the composition of wood are of solid,some liquid
and little gases when heated.There are differences in the grain that show the material is not
uniformly mixed. This allows certain areas of the sample of wood to have different properties
than other areas..
Design a database to store details about U.S. presidents and their t.pdf
Design a database to store details about U.S. presidents and their terms in office. Also, record
details of their date and place of birth, gender, and political party affiliation (e.g., Caluthumpian
Progress Party). You are required to record the sequence of presidents so the predecessor and
successor of any president can be identified. How will you model the case of Grover Cleveland
who served nonconsecutive terms as president? Is it feasible that political party affiliation may
change? If so, how will you handle it? In populating your solution include the following
information (note this is not the format you will store it in the database):
Chester
Arthur
Born 10/05/1829
Whig until 01/01/1854
Republican after that
21st president
Grover
Cleveland
Born 03/18/1837
Democrat his entire career
22nd President
24th President
Benjamin
Harrison
Born 08/20/1833
Whig until 01/01/1856
Republican after that
23rd President
Chester
Arthur
Born 10/05/1829
Whig until 01/01/1854
Republican after that
21st president
Grover
Cleveland
Born 03/18/1837
Democrat his entire career
22nd President
24th President
Benjamin
Harrison
Born 08/20/1833
Whig until 01/01/1856
Republican after that
23rd President
Solution
Create table President(
PresidentID integer(6) primary key,
Name varchar (30)
Place varchar (20),
Dateofbirth date,
Gender varchar(1) check in (m,f),
Ruling_startdate date,
Ruling_end date date,
Affiliation_party varchar(20),
Terms varchar(30));
We will design all the attributes under one table or define separate table for president personal
details and party details.
We can separate the tables now.
President table:
Create table president(
PresidentId integer(6) primary key,
Name varchar(30) foreign key,
Dateofbirth date,
Gender varchar(1) check in (m,f)
Affiliation_party varchar(20),
Terms varchar(20));
Create table pres_det(
presidentID integer(6) primarykey,
Place_of_elect varchar(20),
Birthdate date,
Ruling_startdate date,
Ruling_enddate date,
);
CREATE SEQUENCE seq_presidentID
MINVALUE 10000
START WITH 10000
INCREMENT BY 100
Cycle.
If we execute this sequence president id is set to default value and increment by 100 for every
president value.
DECLARE @maxTerm INT
DECLARE @minTerm INT
SELECT @maxTerm = MAX(term_id), @minTerm = MIN(term_id) FROM President AS p
JOIN President AS p
ON p. presidentId = pres_det.presidentId
WHERE p.Name = \'name\'
Then predecessor is
SELECT p.Name FROM pres_det AS t
JOIN President AS p
ON p.presidentId = t.presidentId
WHERE p.term = @minTerm - 1
To find the successor is
SELECT p.Name FROM pres_det AS t
JOIN President AS p
ON p.presidentId = t.presidentId
WHERE p.term = @maxTerm +1.
Chapter 1 - A View of Life and its Chemical Basis. 1. Explain the.pdf
Chapter 1 - A View of Life and its Chemical Basis. 1. Explain the basic characteristics that are
common to all living things. 2. Describe the levels of organization of life. 3. Distinguish among
the three domains of life and kingdoms. 4. Explain the process of natural selection. 5. Identify
the components of the scientific method. 6. Contrast ionic and covalent bonds. 7. Explain how
the properties of water make life possible. 8. Recognize the importance of functional groups in
determining the chemical properties of an organic molecule. 9. Summarize the categories of the
main macromolecules and provide examples of their diverse biological functions: carbohydrates,
lipids, proteins and nucleic acids.
Solution
please split the questions and ask. it was too lengthy to answer so all are in short.
1. Explain the basic characteristics that are common to all living things.
The fundamental characteristics of all living beings are growth, reproduction. All are made up of
cells, they adapt and respond to the environment. They all obtain and utilize energy. Though all
these are in common, they have different cellular organization.
2. Describe the levels of organization of life.
The various levels of organization of life include organelles, cells, tissues, organs, organ systems,
organisms, populations, communities, ecosystem and biosphere. They are in the chronological
order starting from the simplest form to the most complex one.
3. Distinguish among the three domains of life and kingdoms.
Three domain system introduced by carl woese divides cellular forms into three domains- archae,
bacteria and eukaryotes. It mainly separates archae from prokaryotes based on their 16sRNA
differences. This classification is above the kingdom level.
The five kingdom system has prokaryotes without membranes placed under Monera, eukaryotes
are placed into 3 kingdoms plantae, fungi and animalia. Protista are a separate kingdom
consisting of diversified algae and others which does not come under other groups
4. Explain the process of natural selection
Charles Darwin proposed the natural selection in his theory of evolution. Natural selection is a
process by which the organisms which are adapted to the environmental conditions and survive
in them. Shortly said as “survival of the fittest”. They transfer the genetic traits to further
generations making them more dominant while those who cannot survive get eliminated.
5. Identify the components of the scientific method.
Scientific method is a planned method which helps scientists in answering or solving a problem.
It consists of 6 components. 1. Purpose or question of research, 2. Research to find more
information on the question, 3. Hypothesis after getting preliminary information, 4. Experimental
design of the question, 5. Analysis of the results, 6. Conclusion of the experiment.
6. Contrast ionic and covalent bonds.
Covalent bonds are less polar than ionic bonds.Covalent bond is a chemical bond formed
between two non-metals .
CHOOSE THE CORRECT ANSWER1) PVC Corporation is considering an inv.pdf
CHOOSE THE CORRECT ANSWER:
1) PVC Corporation is considering an investment proposal in which a technology investment of
$10,000 would be required. The investment would provide cash inflows of $2,000 per year for
six years. The technology investment will have no salvage value at the end of six years. If the
company\'s discount rate is 10%, the investment\'s net present value is closest to (Ignore income
taxes.): (See the time value of money BELOW THE PAGE.)
A. $1,289
B. $(1,289)
C. $3,066
D. $2,000
2) RST Corporation has provided the following data concerning an investment project that it is
considering:
Initial investment $160,000
Annual cash flow $54,000 per year
Salvage value at the end of the project $11,000
Expected life of the project years 4 years
Discount rate 10%
The net present value of the project is closest to: (See the time value of money BELOW THE
PAGE)
A. $175
B. ($23,694)
C. $3,662
D. $11,175
3) An increase in the expected salvage value at the end of a capital budgeting project will
A. Increase the cash outflows of the project.
B. Increase the present value of cash inflows of the project
C. Decrease the cash inflows of the project
D. Decrease the net present value of a project
HERE IS THE Time Value of Money Factor for the Present Value of a Sum
PV =FV * TVM Factor
Present Value of a Sum = FV * 1/ (1 + i) n
2%
3%
4%
5%
6%
8%
10%
12%
1
0.9804
0.9709
0.9615
0.9524
0.9434
0.9259
0.9091
0.8929
2
0.9612
0.9426
0.9246
0.9070
0.8900
0.8573
0.8264
0.7972
3
0.9423
0.9151
0.8890
0.8638
0.8396
0.7938
0.7513
0.7118
4
0.9238
0.8885
0.8548
0.8227
0.7921
0.7350
0.6830
0.6355
5
0.9057
0.8626
0.8219
0.7835
0.7473
0.6806
0.6209
0.5674
6
0.8880
0.8375
0.7903
0.7462
0.7050
0.6302
0.5645
0.5066
7
0.8706
0.8131
0.7599
0.7107
0.6651
0.5835
0.5132
0.4523
8
0.8535
0.7894
0.7307
0.6768
0.6274
0.5403
0.4665
0.4039
9
0.8368
0.7664
0.7026
0.6446
0.5919
0.5002
0.4241
0.3606
10
0.8203
0.7441
0.6756
0.6139
0.5584
0.4632
0.3855
0.3220
20
0.6730
0.5537
0.4564
0.3769
0.3118
0.2145
0.1486
0.1037
25
0.6095
0.4776
0.3751
0.2953
0.2330
0.1460
0.0923
0.0588
Time Value of Money Factor for the Present Value of an Annuity
PV = Annuity * TVM Factor
Present Value of an Annuity = X*((1-(1/(1+i)^n))/i)
2%
3%
4%
5%
6%
8%
10%
12%
1
0.9804
0.9709
0.9615
0.9524
0.9434
0.9259
0.9091
0.8929
2
1.9416
1.9135
1.8861
1.8594
1.8334
1.7833
1.7355
1.6901
3
2.8839
2.8286
2.7751
2.7232
2.6730
2.5771
2.4869
2.4018
4
3.8077
3.7171
3.6299
3.5460
3.4651
3.3121
3.1699
3.0373
5
4.7135
4.5797
4.4518
4.3295
4.2124
3.9927
3.7908
3.6048
6
5.6014
5.4172
5.2421
5.0757
4.9173
4.6229
4.3553
4.1114
7
6.4720
6.2303
6.0021
5.7864
5.5824
5.2064
4.8684
4.5638
8
7.3255
7.0197
6.7327
6.4632
6.2098
5.7466
5.3349
4.9676
9
8.1622
7.7861
7.4353
7.1078
6.8017
6.2469
5.7590
5.3282
10
8.9826
8.5302
8.1109
7.7217
7.3601
6.7101
6.1446
5.6502
20
16.3514
14.8775
13.5903
12.4622
11.4699
9.8181
8.5136
7.4694
25
19.5235
17.4131
15.6221
14.0939
12.7834
10.6748
9.0770
7.8431
HERE IS THE Time Value of Money Factor for the Pre.
Civil Infrastructure Problems Why the public is not concerned about t.pdf
Civil Infrastructure Problems Why the public is not concerned about the condition of civil
infrastructure while it affects them personally? Why the public do not comment more to their
elected representatives?
Solution
Infrastructure is the backbone of a country and has serious impact on progress of that country
and happiness of its people. The basic infrastructure comprises of transportation, communication,
sewage, water and electric systems. Further elaborating, it covers roads, highways, bridges,
telecommunication, ports, airports and railways.
People use the infrastructure daily, in some form or other and are affected by it. Poor
infrastructure does generate discontent and is reflected in people movements protesting and
demanding rapid improvements. These protests are more visible in poor and developing
countries where even the basic infrastructure of housing and allied services itself is deficient or
non-existent. The protests are now visible in developed countries also where aging infrastructure
because of lack of funds is deteriorating rapidly. In fact, it was a major point of agenda in recent
presidential elections in USA. The administration is understood to be planning on a major
overhaul of infrastructure costing trillion of dollars. In countries like India the Prime Minister
personally monitors the overall progress of improvement of infrastructure as it was a major
election promise on which he was elected.
So, it is not correct to say that people do not complain of poor infrastructure,though it may be
muted in some countries, cities , states and provinces .People do complain to their elected
representative and that is why it is everywhere a major point of promises in the election agenda.
These infrastructure projects tend to be high-cost investments; though they are vital to a
country\'s economic development and prosperity. Therefore, their improvements take time and is
generally spread over several years..
Cell Structure and FunctionLabel each of the arrows in the followi.pdf
Cell Structure and Function
Label each of the arrows in the following slide image: AE Onion Root Tip: 1000X
Solution
A) Cells are in anaphase
The cells are arranged such that the centromeres have split at this point and the chromatids have
been separated by migrating to the opposite poles.
B) Cells here are present in the early prophase of mitosis, where the centriole can now be seen to
move towards the opposite poles of the cell.
C) cell plate
D) not clear.y distinguishable.
C Language ProblemThere are more than one wrong things I think.P.pdf
C Language Problem
There are more than one wrong things I think.
Please Explain Is the following case structure correct? If incorrect - what\'s wrong? double name;
switch(name){ease 1: break: case 2: break case 3: break: default:}
Solution
Change datatype of \"name\" to int or change case values to double type
Default keyword should be followed by colon(:).
Briefly discuss the adaptive changes seen in the chordates over the o.pdf
Briefly discuss the adaptive changes seen in the chordates over the other animal phyla?
Solution
Animals in the phylum Chordata reveal four key features that appear at some point all through
their development.
a. Notochord - The chordates are called for the notochord: a flexible, rod-shaped design that is
found within the embryonic stage of most chordates and also in the adult stage of some chordate
species. It is found between the intestinal pipe and the nerve cord, providing skeletal help
through along the body. In some chordates, the notochord acts while the primary axial support of
the human body throughout the animal\'s lifetime. In vertebrates, the notochord is provide during
embryonic development, of which time it induces the development of the neural pipe which acts
as an assistance for the developing embryonic body. The notochord, however, is replaced by the
vertebral column (spine) in most adult vertebrates.
b.Dorsal Hollow Nerve Cord - The dorsal hollow nerve cord derives from ectoderm that sheets
into a worthless pipe all through development. In chordates, it is found dorsally (at the top of the
animal) to the notochord. In contrast to the chordates, other animal phyla are known by strong
nerve cables which are positioned either ventrally or laterally. The nerve cord within most
chordate embryos develops into the brain and spinal cord, which include the main anxious
system.
c.Pharyngeal Slits - Pharyngeal slits are openings in the pharynx (the place just posterior to the
mouth) that expand to the surface environment. In organisms that are now living in aquatic
surroundings, pharyngeal slits allow for the quit of water that enters the mouth all through
feeding. Some invertebrate chordates use the pharyngeal slits to filtration food out of the water
that enters the mouth. In vertebrate fishes, the pharyngeal slits develop into gill arches, the bony
or cartilaginous gill supports. In most terrestrial animals, including mammals and birds,
pharyngeal slits are providing just during embryonic development. In these animals, the
pharyngeal slits develop into the jaw and inner ear bones.
d.Post-anal Tail - The post-anal tail is a posterior elongation of the human body, increasing
beyond the anus. The trail contains skeletal things and muscles, which give a way to obtain
locomotion in aquatic species. In some terrestrial vertebrates, the trail also assists with stability,
courting, and signaling when risk is near. In individuals and other apes, the post-anal tail is
provide during embryonic development, but is vestigial as an adult..
Argue in favor or against the following statements (include examples.pdf
Argue in favor or against the following statements (include examples, rationale, explanations,
etc.):
1. Plants migrate
2. Plants defend themselves
3. Plants are good mothers.
4. Grasses are the most important food for humans.
Solution
Plant migration
The Dispersal of the seeds is continuous process, by which plant species survives and spreads.
Migration of plants is a result of the dispersal of the seeds directly to wider areas of earth,also a
phenomenon constant in plant kingdom.Angiosperms species covered the whole earth migrating
to the different countries along the different routes. Due to global warming,the plants are heading
for hills for better survival. A study shows that 171 forest species in the Western Europe, among
most of spp are shifting toward their a favourable locations to cooler spots for better
survival.Plants are migrate toward higher altitudes only because of the climate change, most of
world plant species diversity are found in the tropical areas, a very few studies, have examine
either tropical mountain plant species are they affected by change in climate to same extent as
the temperate plant species. new study has determined, major changes occurred in last two
[2]centuries.If we compare the migration of the plant communities on Chimborazo volcano in the
Ecuador with a historical data from the 1802, the Aarhus University researchers found that
average upslope shifted more than [> 500 metres].and The entire boundary of vegetation have
been moved upwards from the 4,600 metres to the 5,200 metres. The main region for this
dramatically shift is only the climate change last 210 years
2.Plants defend themselves
Plant can not run away from danger like most of the animal do,so that they have developed there
own weapon & aroma & release toxic chemical in order to protect themselves,
Plant are able to move anywhere, so that every single cell defend itself against the all attack from
the invading pathogens, for example- fungi and bacteria.
first line defense in plant
The first line of defense in plants is intact & impenetrable barrier that composed of the bark &
waxy cuticle. Both are protect plants against the herbivores. And Other adaptations against
herbivores are include the, thorns the modified branches of plant, hard shells and the
spines.[modified leaves]first line defense occur when the plant cells can detect the presence of
the foreign invader like(flagella of the bacteria or chitin which found in fungi cell walls) and they
alert the surrounding cells through releasing molecules that give massage to other cells so other
cell ready to fight invader their defences. example, the alerted infected plant cells secrete
molecules are harmful to invadors or plant cell build up their more cell walls for extra protection
form invader.
second line of defense is hypersensitive response (HR).infected plant cells around the infection
site will kill [itself]themselves to stop pathogen to spreading infection throughout the plant. This
pr.
27. What is a prosthetic group Identify prosthetic groups in the fol.pdf
27. What is a prosthetic group? Identify prosthetic groups in the following conjugated proteins:
A prosthetic group is a part of the protein that is not composed of amino acids. Lipoproteins
Carbohydrates Hemoproteins Lipids Flavin nucleotides Metalloproteins Glycoproteins
Phosphate groups Fe, Ca, Copper, Zinc Phosphoproteins Flavoproteins Heme (iron porphyrin)
Solution
part-a: Hemoproteins contain hemoglobin, in which \"iron (Fe2+) is the prosthetic group useful
to carry oxygen
Part-b: Lipoproteins contain lipid moieties with double & triple bonds as prosthetic groups
Part-c: Metalloproteins such as cobalamin contain \"cobalt\" as prosthetic groups in their
structure of intrinsic factor human digestive system
Part-c: Flavoproteins contain flavin as prosthetic group
Glycoproteins is conjugate protein that contains \"glycosidic linkage\" as prosthetic groups.
What is the role of sequence numbers in the rdt protocolWhat is t.pdf
What is the role of sequence numbers in the rdt protocol?
What is the role of timers in the rdt protocol?
Solution
The internet network layer provides only best effort service with no guarantee that packets arrive
at their destination. Also, since each packet is routed individually it is possible that packets are
received out of order. For connection-oriented service provided by TCP, it is necessary to have a
reliable data transfer (RDT) protocol to ensure delivery of all packets and to enable the receiver
to deliver the packets in order to its application layer.
A simple alternating bit RDT protocol can be designed using some basic tools. This protocol is
also known as a stop-and-wait protocol: after sending each packet the sender stops and waits for
feedback from the receiver indicating that the packet has been received.
Stop-and-wait RDT protocols have poor performance in a long-distance connection. At best, the
sender can only transmit one packet per round-trip time. For a 1000 mile connection this
amounts to approximately 1 packet (about 1500 bytes) every 20 ms. That results in a pathetic 75
KB per second rate.
To improve transmission rates, a realistic RDT protocol must use pipelining. This allows the
sender to have a large number of packets \"in the pipeline\". This phrase refers to packets that
have been sent but whose receipt has not yet verified by the receiver..
Which of the following statements about mitosis is true O a. Mitosis.pdf
Which of the following statements about mitosis is true? O a. Mitosis produces two daughter
cells identical to the mother cell. b Mitosis is the division of the nuclear component (DNA)
which is then followed by the division of the cytoplasm (cytokinesis). C.A mitotically dividing
cell copies its DNA, separates these copies at opposite poles and then divides. d All of the above
O e. A and B only
Solution
1. Answer: a. Mitosis produces two daughter cells identical to the mother cell.
Explanation- The chromosomes undergo duplication before mitosis. After this, identical
chromosomes reach daughter cells.
2. Answer: b. During metaphase, chromosomes align at the metaphase plate, the equator of the
mitotic spindle
Explanation- The chromosomes align at the centre of the mitotic spindle and the two
kinetochores of each chromosome get attached to the microtubules from opposite poles of the
spindle.
3. Interphase- D. Cell growth. Replication of DNA.
Prophase- G. The chromosomes condense, each with two sister chromatids. Mitotic spindle
begins to form.
Prometaphase- C. The nuclear envelope breaks down and the chromosomes attach to the mitotic
spindle at their kinetochores.
Metaphse- A. Chromosomes align at the equator of the mitotic spindle. For each chromosome,
the sister chromatids are attached at the kinetochore to microtubules from opposing poles.
Anaphase- E. The sister chromatids are pulled apart to opposite poles of the mitotic spindle.
Telophase- F. A nuclear membrane forms around each set of chromosomes at the mitotic
spindle\'s poles. Mitosis ends.
Cytokinesis- B. Cell pinches in until cytoplasm is divided in two (in animal cells). Results in two
daughter cells that each contain a nucleus.
Explanation- Mitosis comprises of four basic phases, which according to their order are-
prophase, metaphase, anaphase, and telophase. Prometaphase is the late phase of prophase.
Cytokinesis is the division of the cytoplasm. Interphase is the preparatory stage of mitosis..
Which of the following factors would NOT contribute to allopatric spe.pdf
Which of the following factors would NOT contribute to allopatric speciation? The isolated
population is exposed to different selection pressures than the ancestral population Different
mutations begin to manifest in the gene pools of the separated populations The separated
population is small, and genetic drift occurs. There is high gene flow between the two
populations A population becomes geographically isolated from the parent population.
Solution
Ans: There is high gene flow between the tow population is not contribute to allopatric
speciation.
In allopatric speciation, a population splits into two geographically isolated populations due to
formation of a barrier between portions of a population.The isolated populations then experience
differentiating genotypic and phenotypic divergence as a result of different selective pressures in
their differing environments. Additionally, genetic drift will occur differently, eventually
differentiating the population’s genotypes and phenotypes. The separated populations will also
experience different mutations that may persist differently in their respective environments. The
each population have evolved differently to an extent that they remain isolated because changes
are too great to enable genetic mixing through sexual reproduction. Allopatric speciation is not
necessarily precluded if some individuals from one group cross the barrier and mate with
members of the other group. In other words, allopatric speciation can occur if gene flow between
groups is greatly reduced, but not entirely eliminated..
What would be the exact Materials and Method for this experiment2.pdf
What would be the exact Materials and Method for this experiment?
2. Material and methods
2.1. Subjects
One hundred obese children (58 girls and 42 boys) were studied
with a mean age (±SD) of 11.4 ± 2.7 (range 5–17 y). Obesity was de-
fined as a body mass index (BMI) above the 97th percentile for age
and sex according to the relevant national curves [28]. The participants
were recruited as part of the pediatric medical consultations of the Pre-
vention and Management Pediatric Obesity Network of Besançon
(France).
The study was conducted in accordance with the Declaration of Hel-
sinki and approved by the local ethics committee (Comité de Protection
des Personnes, Est-II, France, n°2015-A00763-46). All participants and
their parents or legal guardians were fully informed of the experimental
procedures and gave written informed consent.
2.2. Experimental procedures and measurements
After a medical exam performed by a pediatrician to ensure that the
participant had the ability to complete the whole protocol, the included
subjects underwent different assessments on the same day: i) anthro-
pometric characteristics and maturation status; ii) blood sample and
blood pressure; iii) physical activity level; iv) sleep-disordered
breathing.
Solution
2. Material and methods:
Assessment of anthro-pometric characteristics and maturation status in childhood who have
obesity followed by examination of regular blood sample and monitoring blood pressure;
monitoring of physical activity level as per regular ime scale, finally asking them for any
presence of sleep-disordered breathing
2.1. Subjects - obese children --> One hundred obese children (58 girls and 42 boys) were
studied with a mean age (±SD) of 11.4 ± 2.7 (range 5–17 y). Obesity was de-fined as a body
mass index (BMI) above the 97th percentile for age and sex according to the relevant national
curves.
What are the most challenges of wireless networksSolutionThe .pdf
What are the most challenges of wireless networks?
Solution
The most challenges in wireless networks are:
1. Security: Mobility of users increases the security threats in a wireless network. The present
wireless networks utilize authentication and data encryption methods on the interface to provide
security to the users.
2. Connectivity issues: The connectivity issues can be addressed in many ways such as, adding
access points, dividing networks to sustain proper access, limiting the bandwidth of the guest
network and implementing management and policy software. These issues may occur from the
addition of applications to the network.
3. power and energy: A mobile device is generally useful, small in size, and committed to carry
out a certain set of functions. When a device is permitted to move freely, it would usually be
tough to receive a continuous supply of power. To preserve energy, a mobile device should be
able to operate in a successful and well-organized manner..
What is particle What is a rigid body What is conservative .pdf
What is particle? What is a \"rigid body\"? What is \"conservative force\"? Give an example of
a \"non-conservative\" force? The below relationship embodies what general principle of
physical science? T_1 + V_1 = T_2 + V_2 What are the three ways in which a body can
move? What is \"kinetics\"? What is \"kinematics\"? What is \"power\"? What is \"work\"?
What significance does the date 07 December 1941 have in American history?
Solution
2.rigid body is a body which does not under any deformation on application of load
3.conservative force is a force under which internal energy of system remains conserved
4.friction force is an example of non conservative force
6.a body can move either by pure translatory motion or pure rotary motion or with both rotary
plus translatory
8.kinematics is the branch of mechanics which deals with motion of body without considering
force
9.power is defined as rate of doing work
10.work is defined force multiplied fy displacement in the direction of applied load.
What are some of the obstacles or barriers to implementing EBP (evid.pdf
What are some of the obstacles or barriers to implementing EBP (evidenced based practices) in
nursing? Provide a rationale for your answer. Since there are numerous topics on the issue, it is
not appropriate to repeat one that has already been mentioned unless providing new information.
Solution
Ans:
Obstacles or barriers to implementing EBP (evidenced based practices) in nursing includes
(1) Evidence might have a limited role in decision-making processes; (2) aspects other than
quality of care steer the evidence-based practice agenda; (3) some health care providers benefit
less from evidence-based practice than others and (4) there is a lack of competences to put the
evidence-based principles in practice.
Policy makers might consider health care system characteristics from and strategies developed or
suggested by others to respond to country-specific obstacles. Examples include but are not
limited to; (a) providing incentives for patient-centred care coordination and patient
communication, (b) supporting practitioners interested in applying research-related activities, (c)
considering direct access systems and interprofessional learning to respond to the demand for
autonomous decision-making from satellite professional groups, (d) systematically involving
allied health professionals in important governmental advisory boards, (e) considering
pharmaceutical companies perceived as \'the enemy\' an ally in filling in research gaps, (f)
embedding the evaluation of evidence-based knowledge and skills in examinations (g) moving
from (in)formative learning to transformative learning and (h) organizing high quality catch-up
programs for those who missed out on evidence-based medicine in their curriculum..
Type II hypersensitivity occurs when a.An immediate allergic or ana.pdf
Type II hypersensitivity occurs when: a.An immediate allergic or anaphylactic type of reaction
occur b. Antibodies arc formed against antigens on cell surfaces c. Antigen-antibody complexes
arc deposited in tissues d. Reactions do not require antibody production Which of the following
is an example of type III hypersensitivity: a. Anaphylaxis b. SLE- Systemic Lupus
Erythematosus c. Transfusions reactions d. Transplant rejection. Which of the following is
capable of establishing the MAC- membrane attack complex, effect cell lysis? a. Mast cells b.
MHC I proteins c. MHC II proteins d. Complement proteins In Myasthenia Gravis a.
Acetylcholine receptors on neurons arc bound by antibodies b. Acetylcholine receptors on the
motor end plate arc bound by antibodies c. Acetylcholine vesicles arc bound antibodies d. An
autoimmune response is not involved. Cytotoxic T cells (CD8^+) react to cells that haw: a.
Foreign MHC class II proteins on their surfaces b. B cell receptors c. Foreign MHC class I
proteins on their surfaces d. Soluble antigens
Solution
13. b
The type II hypersensitivity occurs when, the antibodies produced by the immune response bind
to antigens on the patient\'s own cell surfaces. This causes a B cell response, because here
antibodies are formked against the foreign antigen.
14. b
SLE- Systemic lupus erythematosus, it is due to Nuclear antigens. It occurs when accumulation
of immune complexes (antigen-antibody) takes place. Type III hypersensitivity causes
inflammatory response and attraction of leukocytes.
15 d. complement proteins
Transmembrane channels formed by membrane-attack complex (MAC) leads to cell lysis and
death by disrupting the cell membrane of target cells. It involves proteins C5 through C9
16 b.
Antibody block or destroy nicotinic acetylcholine receptors at the junction between the nerve and
muscle.
17. c. Foreign MHC class I protein on thier surface.
cytotoxic T cell or CD8+ T-cell or killer T cell are able to recognize a specific antigen. TCR to
bind to the class I MHC molecule,.
The sum of two integers is 19. The larger integer is 5 less than twi.pdf
The sum of two integers is 19. The larger integer is 5 less than twice the smaller. Find the two
integers.
Solution
Let the larger integer be X and smaller integer be Y
Given sum of two integers is 19 i.e., X+Y=19
The larger integer is 5 less than twice the smaller i.e., X=2Y-5
X+Y=19
By plugging X=2Y-5 in X+Y=19 we get
2Y-5+Y=19
3Y=24
Y=24/3=8
X+Y=19
X+8=19
X=19-8=11
Therefore the two integers are 11,8.
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