A plant with a very extensive root system would be able to better abs.pdf
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A plant with a very extensive root system would be able to better absorb and store nutrients. True False. Solution Ans. True. Roots absorb water and inorganic minerals (nutrients like NH4+, phosphate, Ca2+, Zn2+, etc.). The rate of water and mineral uptake is directly proportional to the surface area of roots. So, as plants grow larger, they extend their roots for greater uptake of water and minerals. A plant with relatively smaller roots absorb water and nutrients less efficiently that the plant with extensive root system. The roots may also be involved in storage- greater is root surface area, greater is the storage..
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In Drosophila, gray body color is dominant to ebony body color, while.pdf
In Drosophila, gray body color is dominant to ebony body color, while long wings are dominant
to vestigial wings. Assuming that the P_1 individuals are homozygous, work the following
crosses through the F_2 generation, and determine the phenotypic each generation. Enter your
answers in fraction form (ex: 1/4) in the blanks provided by the list of phenotypes: gray, long
gray, vestigial ebony, long ebony, vestigial
Solution
In Drosophila, gray color is dominant to ebony body color, while long wings are dominant to
vestigial wings. Assuming that the P1 individuals are homozygous, work the following crosses
through the F2 generation, and determine the phenotypic ratios for each generation
E = gray; e = ebony
V = long; v = vestigial
It is a dihybrid cross, in which phenotypic ratio should be 9:3:3:1
P1: ----> EEVV (homozygous dominant) *eevv (homozygous recessive)
F1 generation offspring are all are EeVv representing ---> heterozygous for both traits (genotype
and
dominant phenotypes); therefore, all gray body & and long wings can be seen as phenotypic
appearance.
P1 dihybrid cross: Punnet square
* ev ev ev ev
EV EeVv EeVv EeVv EeVv
EV EeVv EeVv EeVv EeVv
EV EeVv EeVv EeVv EeVv
EV EeVv EeVv EeVv EeVv
F2 –generation -dihybrid cross:
* EV Ev eV ev
EV EVEV EVEv EVeV EVeV
Ev EEVv EEvv EeVv Eevv
eV EeVV EeVv eeVV eeVv
ev EVev Evev eVev evev
Genotypic ratios ----> 1/16 EEVV; 1/8 EEVvl; 1/16 EEVV; 1/8 EeVV; 1/4 EeVv; 1/8 Eevv;
1/16 eeVV ;1/18 eeVv; 1/16 eevv
Phenotypic ratio ---->
9/16 gray body and long wings
3/16 gray body and vestigial wings
3/16 ebony body and long wings
1/16 ebony, vestigial
Staphylococcus aureus.
I need help with this maze gui that I wrote in java, I am trying to .pdf
I need help with this maze gui that I wrote in java, I am trying to find the shortest path but I dont
know how to implement the FindshortestPath Method.
using a BREADTH FIRST SEARCH and the stradgey below to ONLY FIND THE SHORTEST
PATH, the method \"FindshortestPath(); is the only part you need to add code too.
I was given this as a stragedy to solve the findshortestpathMethod():
Create two LinkedLists of MazeMap objects. One will be the list of map objects for cells at
distance x from the center. The other will be the list of map objects for cells at distance x + 1.
Initialize the list at distance x to contain a MazeMap object which refers to the center cell of the
maze. Since this map object refers to the center cell, it does not need to refer to any other
MazeMap object, so its MazeMap reference can be set to null. Set the center cell’s visited field
to true to indicate that this cell has already been added to the map.
Set up a loop to run until the maze entrance is mapped. This loop does the following:
Set up a loop to process each map object from the list at distance x. This loop does the following:
i.Get the cell from the map object
ii.If this cell is the entrance cell, set the entrance and break out of the loop!
iii.Check each direction from this cell and for an accessible neighbor cell
Get a reference to that neighbor cell
If the neighbor cell has not already been visited
Set the cell’s visited field to true so its only processed once
Create a MazeMap object that refers to this neighbor cell
Make this MazeMap object refer to the current map object being processed – ie the cell one step
closer to center
Add this MazeMap object to the list at distance x + 1
List at distance x has been processed. The list at distance x + 1 is now complete, so make the list
at distance x refer to list at distance x + 1. Then create a new empty list for the list at distance x +
1.
At this point, the entrance points at the map object which refers to cell 1,1. That map object also
refers to a map object which is one step closer to the center. Write a loop that traverses this trail
of references until it reaches the end (a null reference). For each map object:
Get the cell referred to.
Call the drawCircle method with a color, the cell’s row and column, and SMALL.
Call the delay method with a SHORT delay. This slows down the drawing process so you can
watch the path being displayed.
Advance to the next map object in the list.
here is the code so far*********
package mazepackage;
import java.util.LinkedList;
import java.util.Random;
import javafx.scene.paint.Color;
public class Maze {
private int N; // dimension of maze
private MazeCell[][] maze;
private Random rand = new Random();
// Used to signal the maze has been solved
private boolean done;
// time in milliseconds (from currentTimeMillis()) when we can draw again
// used to control the frame rate
private long nextDraw = -1;
private MazeApp mf;
// Define constants for the circle size
private final double BI.
G and g are dominant and recessive alleles, respectively, for a gene.pdf
G and g are dominant and recessive alleles, respectively, for a gene. If a mating of a gg female
with a Gg male resulted in offspring that all have the recessive phenotype, this would most likely
be an example of
Solution
Answer:
Maternal inheritance
The mitochon dria, like the chloroplasts, are self-replicating organelles which contain their own
genes and have a limited number of characters which are independent of the nucleus. They are
transmitted mainly through the female line and mutations in their genes show the same pattern or
maternal inheritance..
Explain how particles of different sizes are cleared from different a.pdf
Explain how particles of different sizes are cleared from different areas of the pulmonary
system. Include at least 3 mechanisms
Solution
Different particles are deposited is removed from the respiration tract(pulmonary system).
Different parts of pulmonary system are involved in different particle size removal:
1. Upper Respiratory Tract. Soluble material may be cleared from the nasal passages through
mucociliary transport. Some material also can be cleared via absorptive procedures into the
blood.
Insoluble material that deposit in the nasal passage clear in particular via mucociliary delivery.
The mucous usually flows in the direction of the nasopharynx.
The epithelium within the maximum anterior (the front) a part of the nasal passages does now
not comprise cilia. The mucus flows slowly toward the entrance to the nose and is moved by
means of traction associated with actions of extra distal cilia.
Particles depositing within the nonciliated front part of the nasal passages are notion to be
moved alongside at a charge of approximately 1-2 mm/h. Because clearance through the
indicated mechanism may additionally absorb to 12 h, deposited particles are commonly extra
unexpectedly eliminated with the aid of sneezing, wiping, or nose blowing.
Insoluble substances that deposit inside the oral passages (mouth, throat) may be translocated to
the gastrointestinal tract by way of swallowing. Some particles inhaled via the nose also can
become being swallowed.
2. Tracheobronchial Region. When soluble debris deposit inside the tracheobronchial region,
diffusion of material that dissolved among cells and organic tactics can also lead to absorption
into the blood. Intact soluble debris may also be cleared routinely thru the mucociliary shipping
pathway.
The mucociliary escalator is the primary clearance pathway in the tracheobronchial place for
insoluble debris. The escalator consists of the ciliated epithelium shifting a mucous layer on
pinnacle of it. This layer consists of a segment of low viscosity (hypophase) wherein the cilia
beat, and an covering gel segment of high viscosity (epiphase) concept to be moved via ciliary
motion toward the pharynx.
Clearance of particles depositing in the tracheobronchial region may additionally occur through
phagocytosis (consumption) by airway macrophages.
Another clearance mechanism, which is considered less important, is penetration of insoluble
debris of submicronic size thru the epithelium. Some debris may fit thru the epithelium, getting
into the peribronchial area.
Cough can be an efficient clearance mechanism, however clearance of debris thru this mode is
probably to be restricted to the upper generations of the conducting airlines.
The mucous transport rate varies in one of a kind areas of the tracheobronchial place. In healthy
nonsmokers, average tracheal mucous delivery fees range from about 4 to 6mm/min. For
insoluble debris, the delivery rate seems no longer to depend upon particle shape, length, and
co.
Egineering Ethics Planned obsolescence is the practice of producing.pdf
Egineering Ethics: Planned obsolescence is the practice of producing goods designed to be
replaced, either because the goods \"go out of style\", or because the goods are designed to fail,
are designed to self-destruct, there is a lack of replacement parts, are made of less durable
materials than are readily available, etc. Give an example of planned obsolescence in
engineering. Then, argue that planned obsolescence is either morally acceptable or morally
unacceptable. (You can use one the normative or metaethical theories we\'ve discussed to
support your position. You can also appeal to the arguments of Plato, Micheal Davis, George
Harris, Harry Frankfurt, or the like to support your position.)
Solution
planned obsolence is a process of launching a product in the market with a limited life with an
ability to increase demand for its future version. The process have two different aspects in
effecting the environment.
1) Planned obsolescene by innovation firms
lets take an example of evolution of soft storage devices. the floppy disc to storage SD card. we
can observe that the during all these years the storage capacity is increased and the material used
is decreased. such types of evolution in the engineering are helpful to the humans as well as
environment.
2: By production firms:
there are many producers to make more money introduce products which are non -repaire able or
the parts with very low life. in such firms the updated product is launched with a periodic
intervel so that the competition in the market goes down.
ethically it is wrong because the consumer is being cheated. Taiwo K. Aladeojebi in his Journal
have discussed where the customers are being fooled by the producers. due to non-repaire abilty
of the product the wastage in the country increases due to which the environment is being
spoiled..
Endospores do not stain easily Perhaps you have seen them as unstain.pdf
Endospores do not stain easily: Perhaps you have seen them as unstained white objects inside
Bacillus species in other staining procedures. If they are visible as unstained objects in other
stains, of what use is the endospore stain?
Solution
A staining procedure is performed not to always provide a color to the desired target, but it aims
to generate a visible contrast between objects which can help them to visibly identify as
different. Although most of the bio-polymers/organisms uptake colors by easy diffusion or harsh
chemical-based methods such as acid-fuschin stain, some bio-molecules/organisms do not uptake
stains at all. However, since an organism always resides on a background, it remains necessary to
either stain the target object or the background in order to clearly identify the target as a separate
entity. This is termed as negative staining.
According to the question, the endospores do not uptake any stain and appear as opaque white
objects under microscope. However, although they do not uptake any stain, a staining procedure
is still performed in order to stain the background. This kind of staining procedure where the
target remains unstained and only the background is stained to provide a visible contrast is
termed as negative staining.
Hence, though endospores do not uptake any stain, a staining procedure is still performed..
Describe how the concepts of leadership and management differ from e.pdf
Describe how the concepts of leadership and management differ from each other. In what areas
do they overlap? Explain how the goals of management and leadership may sometimes overlap.
As a nurse leader, do you believe you can expand your influence to create change by taking
advantage of this overlap? Explain your answer.
Solution
Answer:
Nursing leadership: it is the method of creating a specific method of change followed by
establishing a direction to align patients for their empowerment in getting good health using clear
vision and motivation
Nursing management: it differs from the leadership competencies & management is mainly
associated with regulating order and consistency in implementing budget, staffing protocols
finally solving complexity of the situation when implementing an ethical based decision
The management & leadership goals of nurses in healthcare sector may overlap sometimes with
respect to nursing education, patient oriented communication-collaboration with other health
agencies, informatics design & information technology to encourage quality health care and
safety, intensive –based, evidence –based nursing policy, culture that foster regional and local
partnerships for nursing education, clinical research. Both leadership & management of nursing
skills initiative have specific impact on clinical nursing practice, for example, when conducting
clinical public health research, informatics & technology based electronic system useful to
record patient health previous health data including familial genetic –inherited disease
information. This type of information –based and evidence- based nursing practice promote
efficient health care delivery to the public. The innovative methods of initiative is mainly useful
early prevention of communicable chronic illness by identifying particular disease causing
organisms therefore \"major overlapping between leadership competency & management in
nursing\" can act as an initiative to deliver good patient outcomes. The advantage of this overlap
is going to enable identification of communicable diseases in the communities caused by
microbial species finally implementing other preventive approaches of the chronic illness include
that initial screening, predisposition of the disease an also its growth and development (example
Polio). Finally, the patients result in reduced death rate (low mortality) and we can reduce the
disease propagation in the communities by alerting the people with this disease..
Contemporary historyFor each year since 1945, select an event that.pdf
Contemporary history
For each year since 1945, select an event that had a major importance for world affairs.
Solution
The 1956 Suez Crisis is one of the most important and controversial events in British history
since the Second World War. Not only did Suez result in deep political and public division in
Britain, it also caused international uproar. It has come to be regarded as the end of Britain\'s role
as one of the world powers and as the beginning of the end for the British Empire. In future
British foreign policy would be conducted in concurrence with American diplomatic support.
This special online exhibition has been developed to mark the fiftieth anniversary of the Crisis.
The exhibition draws from the Bodleian\'s rich holdings of modern political papers to provide an
\'insider\'s viewpoint\' of Suez from politicians, diplomats, civil servants and leading public
figures. It includes items from a number of collections which have been acquired within the last
decade or so.
The Suez Crisis erupts on July 26, 1956 when Egyptian President Abdul Nasser nationalises the
Suez Canal Company – which had been run by the French with the British government as the
largest single shareholder.
On October 29, 1956, Israeli armed forces pushed into Egypt toward the Suez Canal after
Egyptian president Gamal Abdel Nasser (1918-70) nationalized the canal in July of that same
year, initiating the Suez Crisis. The Israelis soon were joined by French and British forces, which
nearly brought the Soviet Union into the conflict, and damaged their relationships with the
United States. In the end, the British, French and Israeli governments withdrew their troops in
late 1956 and early 1957.
The 120-mile Suez Canal, which connects the Mediterranean Sea and the Red Sea, took 10 years
to construct and opened in 1869. The canal was developed by Frenchman Ferdinand de Lesseps,
who in the 1880s made an unsuccessful attempt to develop the Panama Canal..
Recommended
In Drosophila, gray body color is dominant to ebony body color, while.pdf
In Drosophila, gray body color is dominant to ebony body color, while long wings are dominant
to vestigial wings. Assuming that the P_1 individuals are homozygous, work the following
crosses through the F_2 generation, and determine the phenotypic each generation. Enter your
answers in fraction form (ex: 1/4) in the blanks provided by the list of phenotypes: gray, long
gray, vestigial ebony, long ebony, vestigial
Solution
In Drosophila, gray color is dominant to ebony body color, while long wings are dominant to
vestigial wings. Assuming that the P1 individuals are homozygous, work the following crosses
through the F2 generation, and determine the phenotypic ratios for each generation
E = gray; e = ebony
V = long; v = vestigial
It is a dihybrid cross, in which phenotypic ratio should be 9:3:3:1
P1: ----> EEVV (homozygous dominant) *eevv (homozygous recessive)
F1 generation offspring are all are EeVv representing ---> heterozygous for both traits (genotype
and
dominant phenotypes); therefore, all gray body & and long wings can be seen as phenotypic
appearance.
P1 dihybrid cross: Punnet square
* ev ev ev ev
EV EeVv EeVv EeVv EeVv
EV EeVv EeVv EeVv EeVv
EV EeVv EeVv EeVv EeVv
EV EeVv EeVv EeVv EeVv
F2 –generation -dihybrid cross:
* EV Ev eV ev
EV EVEV EVEv EVeV EVeV
Ev EEVv EEvv EeVv Eevv
eV EeVV EeVv eeVV eeVv
ev EVev Evev eVev evev
Genotypic ratios ----> 1/16 EEVV; 1/8 EEVvl; 1/16 EEVV; 1/8 EeVV; 1/4 EeVv; 1/8 Eevv;
1/16 eeVV ;1/18 eeVv; 1/16 eevv
Phenotypic ratio ---->
9/16 gray body and long wings
3/16 gray body and vestigial wings
3/16 ebony body and long wings
1/16 ebony, vestigial
Staphylococcus aureus.
I need help with this maze gui that I wrote in java, I am trying to .pdf
I need help with this maze gui that I wrote in java, I am trying to find the shortest path but I dont
know how to implement the FindshortestPath Method.
using a BREADTH FIRST SEARCH and the stradgey below to ONLY FIND THE SHORTEST
PATH, the method \"FindshortestPath(); is the only part you need to add code too.
I was given this as a stragedy to solve the findshortestpathMethod():
Create two LinkedLists of MazeMap objects. One will be the list of map objects for cells at
distance x from the center. The other will be the list of map objects for cells at distance x + 1.
Initialize the list at distance x to contain a MazeMap object which refers to the center cell of the
maze. Since this map object refers to the center cell, it does not need to refer to any other
MazeMap object, so its MazeMap reference can be set to null. Set the center cell’s visited field
to true to indicate that this cell has already been added to the map.
Set up a loop to run until the maze entrance is mapped. This loop does the following:
Set up a loop to process each map object from the list at distance x. This loop does the following:
i.Get the cell from the map object
ii.If this cell is the entrance cell, set the entrance and break out of the loop!
iii.Check each direction from this cell and for an accessible neighbor cell
Get a reference to that neighbor cell
If the neighbor cell has not already been visited
Set the cell’s visited field to true so its only processed once
Create a MazeMap object that refers to this neighbor cell
Make this MazeMap object refer to the current map object being processed – ie the cell one step
closer to center
Add this MazeMap object to the list at distance x + 1
List at distance x has been processed. The list at distance x + 1 is now complete, so make the list
at distance x refer to list at distance x + 1. Then create a new empty list for the list at distance x +
1.
At this point, the entrance points at the map object which refers to cell 1,1. That map object also
refers to a map object which is one step closer to the center. Write a loop that traverses this trail
of references until it reaches the end (a null reference). For each map object:
Get the cell referred to.
Call the drawCircle method with a color, the cell’s row and column, and SMALL.
Call the delay method with a SHORT delay. This slows down the drawing process so you can
watch the path being displayed.
Advance to the next map object in the list.
here is the code so far*********
package mazepackage;
import java.util.LinkedList;
import java.util.Random;
import javafx.scene.paint.Color;
public class Maze {
private int N; // dimension of maze
private MazeCell[][] maze;
private Random rand = new Random();
// Used to signal the maze has been solved
private boolean done;
// time in milliseconds (from currentTimeMillis()) when we can draw again
// used to control the frame rate
private long nextDraw = -1;
private MazeApp mf;
// Define constants for the circle size
private final double BI.
G and g are dominant and recessive alleles, respectively, for a gene.pdf
G and g are dominant and recessive alleles, respectively, for a gene. If a mating of a gg female
with a Gg male resulted in offspring that all have the recessive phenotype, this would most likely
be an example of
Solution
Answer:
Maternal inheritance
The mitochon dria, like the chloroplasts, are self-replicating organelles which contain their own
genes and have a limited number of characters which are independent of the nucleus. They are
transmitted mainly through the female line and mutations in their genes show the same pattern or
maternal inheritance..
Explain how particles of different sizes are cleared from different a.pdf
Explain how particles of different sizes are cleared from different areas of the pulmonary
system. Include at least 3 mechanisms
Solution
Different particles are deposited is removed from the respiration tract(pulmonary system).
Different parts of pulmonary system are involved in different particle size removal:
1. Upper Respiratory Tract. Soluble material may be cleared from the nasal passages through
mucociliary transport. Some material also can be cleared via absorptive procedures into the
blood.
Insoluble material that deposit in the nasal passage clear in particular via mucociliary delivery.
The mucous usually flows in the direction of the nasopharynx.
The epithelium within the maximum anterior (the front) a part of the nasal passages does now
not comprise cilia. The mucus flows slowly toward the entrance to the nose and is moved by
means of traction associated with actions of extra distal cilia.
Particles depositing within the nonciliated front part of the nasal passages are notion to be
moved alongside at a charge of approximately 1-2 mm/h. Because clearance through the
indicated mechanism may additionally absorb to 12 h, deposited particles are commonly extra
unexpectedly eliminated with the aid of sneezing, wiping, or nose blowing.
Insoluble substances that deposit inside the oral passages (mouth, throat) may be translocated to
the gastrointestinal tract by way of swallowing. Some particles inhaled via the nose also can
become being swallowed.
2. Tracheobronchial Region. When soluble debris deposit inside the tracheobronchial region,
diffusion of material that dissolved among cells and organic tactics can also lead to absorption
into the blood. Intact soluble debris may also be cleared routinely thru the mucociliary shipping
pathway.
The mucociliary escalator is the primary clearance pathway in the tracheobronchial place for
insoluble debris. The escalator consists of the ciliated epithelium shifting a mucous layer on
pinnacle of it. This layer consists of a segment of low viscosity (hypophase) wherein the cilia
beat, and an covering gel segment of high viscosity (epiphase) concept to be moved via ciliary
motion toward the pharynx.
Clearance of particles depositing in the tracheobronchial region may additionally occur through
phagocytosis (consumption) by airway macrophages.
Another clearance mechanism, which is considered less important, is penetration of insoluble
debris of submicronic size thru the epithelium. Some debris may fit thru the epithelium, getting
into the peribronchial area.
Cough can be an efficient clearance mechanism, however clearance of debris thru this mode is
probably to be restricted to the upper generations of the conducting airlines.
The mucous transport rate varies in one of a kind areas of the tracheobronchial place. In healthy
nonsmokers, average tracheal mucous delivery fees range from about 4 to 6mm/min. For
insoluble debris, the delivery rate seems no longer to depend upon particle shape, length, and
co.
Egineering Ethics Planned obsolescence is the practice of producing.pdf
Egineering Ethics: Planned obsolescence is the practice of producing goods designed to be
replaced, either because the goods \"go out of style\", or because the goods are designed to fail,
are designed to self-destruct, there is a lack of replacement parts, are made of less durable
materials than are readily available, etc. Give an example of planned obsolescence in
engineering. Then, argue that planned obsolescence is either morally acceptable or morally
unacceptable. (You can use one the normative or metaethical theories we\'ve discussed to
support your position. You can also appeal to the arguments of Plato, Micheal Davis, George
Harris, Harry Frankfurt, or the like to support your position.)
Solution
planned obsolence is a process of launching a product in the market with a limited life with an
ability to increase demand for its future version. The process have two different aspects in
effecting the environment.
1) Planned obsolescene by innovation firms
lets take an example of evolution of soft storage devices. the floppy disc to storage SD card. we
can observe that the during all these years the storage capacity is increased and the material used
is decreased. such types of evolution in the engineering are helpful to the humans as well as
environment.
2: By production firms:
there are many producers to make more money introduce products which are non -repaire able or
the parts with very low life. in such firms the updated product is launched with a periodic
intervel so that the competition in the market goes down.
ethically it is wrong because the consumer is being cheated. Taiwo K. Aladeojebi in his Journal
have discussed where the customers are being fooled by the producers. due to non-repaire abilty
of the product the wastage in the country increases due to which the environment is being
spoiled..
Endospores do not stain easily Perhaps you have seen them as unstain.pdf
Endospores do not stain easily: Perhaps you have seen them as unstained white objects inside
Bacillus species in other staining procedures. If they are visible as unstained objects in other
stains, of what use is the endospore stain?
Solution
A staining procedure is performed not to always provide a color to the desired target, but it aims
to generate a visible contrast between objects which can help them to visibly identify as
different. Although most of the bio-polymers/organisms uptake colors by easy diffusion or harsh
chemical-based methods such as acid-fuschin stain, some bio-molecules/organisms do not uptake
stains at all. However, since an organism always resides on a background, it remains necessary to
either stain the target object or the background in order to clearly identify the target as a separate
entity. This is termed as negative staining.
According to the question, the endospores do not uptake any stain and appear as opaque white
objects under microscope. However, although they do not uptake any stain, a staining procedure
is still performed in order to stain the background. This kind of staining procedure where the
target remains unstained and only the background is stained to provide a visible contrast is
termed as negative staining.
Hence, though endospores do not uptake any stain, a staining procedure is still performed..
Describe how the concepts of leadership and management differ from e.pdf
Describe how the concepts of leadership and management differ from each other. In what areas
do they overlap? Explain how the goals of management and leadership may sometimes overlap.
As a nurse leader, do you believe you can expand your influence to create change by taking
advantage of this overlap? Explain your answer.
Solution
Answer:
Nursing leadership: it is the method of creating a specific method of change followed by
establishing a direction to align patients for their empowerment in getting good health using clear
vision and motivation
Nursing management: it differs from the leadership competencies & management is mainly
associated with regulating order and consistency in implementing budget, staffing protocols
finally solving complexity of the situation when implementing an ethical based decision
The management & leadership goals of nurses in healthcare sector may overlap sometimes with
respect to nursing education, patient oriented communication-collaboration with other health
agencies, informatics design & information technology to encourage quality health care and
safety, intensive –based, evidence –based nursing policy, culture that foster regional and local
partnerships for nursing education, clinical research. Both leadership & management of nursing
skills initiative have specific impact on clinical nursing practice, for example, when conducting
clinical public health research, informatics & technology based electronic system useful to
record patient health previous health data including familial genetic –inherited disease
information. This type of information –based and evidence- based nursing practice promote
efficient health care delivery to the public. The innovative methods of initiative is mainly useful
early prevention of communicable chronic illness by identifying particular disease causing
organisms therefore \"major overlapping between leadership competency & management in
nursing\" can act as an initiative to deliver good patient outcomes. The advantage of this overlap
is going to enable identification of communicable diseases in the communities caused by
microbial species finally implementing other preventive approaches of the chronic illness include
that initial screening, predisposition of the disease an also its growth and development (example
Polio). Finally, the patients result in reduced death rate (low mortality) and we can reduce the
disease propagation in the communities by alerting the people with this disease..
Contemporary historyFor each year since 1945, select an event that.pdf
Contemporary history
For each year since 1945, select an event that had a major importance for world affairs.
Solution
The 1956 Suez Crisis is one of the most important and controversial events in British history
since the Second World War. Not only did Suez result in deep political and public division in
Britain, it also caused international uproar. It has come to be regarded as the end of Britain\'s role
as one of the world powers and as the beginning of the end for the British Empire. In future
British foreign policy would be conducted in concurrence with American diplomatic support.
This special online exhibition has been developed to mark the fiftieth anniversary of the Crisis.
The exhibition draws from the Bodleian\'s rich holdings of modern political papers to provide an
\'insider\'s viewpoint\' of Suez from politicians, diplomats, civil servants and leading public
figures. It includes items from a number of collections which have been acquired within the last
decade or so.
The Suez Crisis erupts on July 26, 1956 when Egyptian President Abdul Nasser nationalises the
Suez Canal Company – which had been run by the French with the British government as the
largest single shareholder.
On October 29, 1956, Israeli armed forces pushed into Egypt toward the Suez Canal after
Egyptian president Gamal Abdel Nasser (1918-70) nationalized the canal in July of that same
year, initiating the Suez Crisis. The Israelis soon were joined by French and British forces, which
nearly brought the Soviet Union into the conflict, and damaged their relationships with the
United States. In the end, the British, French and Israeli governments withdrew their troops in
late 1956 and early 1957.
The 120-mile Suez Canal, which connects the Mediterranean Sea and the Red Sea, took 10 years
to construct and opened in 1869. The canal was developed by Frenchman Ferdinand de Lesseps,
who in the 1880s made an unsuccessful attempt to develop the Panama Canal..
A variety of media and culture conditions can be used to culture and .pdf
A variety of media and culture conditions can be used to culture and identify Listeria
monocytogenes. For each, characterize it as selective or differential.
Solution
Listeria monocytogenes is basically a soil-borne micro-organism. It has been found in various
sources such as water, agricultural products, and in animals. Listeria monocytogenes has been
known as etiologic agent of listeriosis, causing meningitis, encephalitis, septicemia and
endocarditis in humans.
various selective and differential methods were developed to isolate the Listeria monocytogenes
species as follows
Selective
Incubation at 30 degree celsius
Adding antibiotics such as nalidixic acid
Adding high concentarion of salts such as lithium chloride and sodium chloride
Adding antifungals such as amphotericin B
Differential
Adding pH indicator to detect ferementaion of rhamanose to acid
Detecting beta heamolysis.
A program consists of 4 musical numbers and 3 speeches. In how many .pdf
A program consists of 4 musical numbers and 3 speeches. In how many ways can the program be
arranged so that it begins and ends with a musical number?
Solution
There are 4 musical numbers and 3 speeches
AS begin and end musical numbers 2 musical numbers selected from 4 and put in the ends
This can be done in 2(4C2) = 12 ways
The remaining are 2 musical and 3 speeches. This can be arranged in 5! ways
Hence total no of ways = 12 (5!) = 1440.
A Moving to another question will save this response. Question 4 Iden.pdf
A Moving to another question will save this response. Question 4 Identify a problem with water
that is classified as hard a scaling especially when exposed to heat b.scaling especially when
exposed to cold c inability to lather soap (onto bubbles) d soap very lathery/bubbly hard to rinse
soap off Of b&d; AMoving to another question will save this response
Solution
Option e is correct. Hard water is scaling when exposd to heat and it is not able to give lather to
the soap..
property, plant, and equipment are categorized as property, pla.pdf
property, plant, and equipment are categorized as
property, plant, and equipment are categorized as
Solution
Answer:
Property, Plant and Equipment are categorized as \"Fixed Assets\" in the Asset Section of the
Balance Sheet.
Property, Plant and Equipment are Non Current Assets which includes Building, Plant, Vehicle,
Land etc. Some of these Fixed Assets are subject to Depreciation, while some are not subject to
Depreciation which are Land etc..
Write a program in Matlab using Newtons method to solve this syste.pdf
Write a program in Matlab using Newton\'s method to solve this system of equations for x, y,
and z...
Solution
f=inline(\'[-10*p(:,1)-10*p(:,2); 28*p(:,1)-p(:,2)-p(:,1).*p(:,3); p(:,1).*p(:,2)-(8*p(:,3))/2]\',\'p\');
tol = 0.4e-6; %error tolerance
MaxIter = 10;
format long
x = sym(\'x\') ;
y = sym(\'y\') ;
z = sym(\'z\') ;
F = f([x,y,z]);
J = jacobian(F);
invJ = inv(J);
s = zeros(MaxIter,3);
s(1,:) = [1 1 1]; % first set of values
dsnorm = inf;
iter = 1;
while dsnorm>tol && itertol && itertol && iter.
Why is it that bacterial ribosomes can begin translation before mRNA.pdf
Why is it that bacterial ribosomes can begin translation before mRNA synthesis is completed,
but eukaryotic ribosomes cannot?
a. mRNA splicing occurs very quickly in prokaryotic cells.
b. In prokaryotes, transcription and translation take place in the cytoplasm.
c. Prokaryotic mRNAs are elongated differently from eukaryotic mRNAs.
d. Prokaryotic mRNAs are synthesized with ribosomes attached.
Solution
Answer is b. In prokaryotic organisms there is no nuclei, so both transcription and translation
occur in the same place that is Cytosol. Moreover, no post transcriptional modifications like
caping and Tailing. Due to this the half life of Prokaryotic m-RNA is too small. Due to these two
reasons, in prokaryotic organisms, translation start, before completion of transcription.
Write a C++ program with a function defined to give the person a .pdf
Write a C++ program with a function defined to give the person a 20% raise?
Please explain. Thanks
Write a C++ program with a function defined to give the person a 20% raise?
Please explain. Thanks
Please explain. Thanks
Solution
#include
#include
#define SIZE 5 // we are defining the size of entries / variables to 5
class emp
{
float basic,da,it,netsal; // salary , da and netsalary are mentioned in float
char name[20],num[10];
public:
void getdata(); // function to get employee data
void net_sal(); // function to get employee salary
void sal_hike(); // fuction to display hike based on net salary
void dispdata(); // to dislay data
};
void emp::getdata() // enter basic employee details
{
cout<<\"\ Enter employee number: \" ;
cin>>name;
cout<<\" Enter employee name: \" ;
cin>>num;
cout<<\"Enter employee basic salary in Rs: \" ;
cin>>basic;
}
void emp::net_sal() // calculate net salary
{
da=((0.52)*basic );
float gsal=da+basic;
it=((0.3)*gsal);
netsal=gsal-it;
}
void emp::sal_hike() // compute salary hike
int hike=net_sal + 0.2*net_sal;
}
void emp::dispdata() // to display data
{
cout
<<\"\ Employee number: \"<>n;
for(int i=0;i.
Which standards organization defined the Ethernet standardA.IEEE.pdf
Which standards organization defined the Ethernet standard?
A.IEEE
B.ISO
C.ANSI
D.DIX
Solution
Answer: D) DIX
DIX stands for Digital Equopment Company, Intel and Xerox. DIX teamed up for the first
standard for Ethernet standard.
IEEE also created Ethernet standards but they were not the frst.
ISO defined ISO reference model not the Ethernet standard
ANSI associted with ASCII codes and standards of others..
When the pH of a solution changes from 2 to 6 a) That solution beco.pdf
When the pH of a solution changes from 2 to 6: a) That solution becomes more acidic due to an
increase in the proton concentration. b) That solution contains 4X fewer protons. c) That solution
becomes 1000X more basic. d) That solution is 4X more acidic. e) None of the above
Solution
Answer :
Answer: The correct answer is B (That solution contains 4X fewer protons).
When the pH of a solution changes from 2 to 6, the [H+] hydronium ions decrease and [OH-]
will increase.
Which is the correct answer Consider the characteristics of moss an.pdf
Which is the correct answer? Consider the characteristics of moss and fern life cycles. Which of
the following sets of statements is true? The gametophyte is the dominant stage of the life cycle
in both mosses and ferns. Mosses require moisture for sperm to reach the egg, but ferns do not
In mosses, the gametophyte is the dominant stage of the life cycle; in ferns, the sporophyte is the
dominant stage of the life cycle. Ferns require moisture for sperm to reach the egg, but mosses
do not The gametophyte is the dominant stage of the life cycle in both mosses and ferns. In both
mosses and ferns, moisture is required for sperm to reach the egg. In mosses, the sporophyte is
the dominant stage of the life cycle; in ferns, the dominant stage of the life cycle. In both mosses
and ferns moisture is reach for spent to reach the egg. In moses, the gametophyte is the
dominant state of the life cycle, if ferns, the sprophyte is the dominant stage of the life cycle. In
both mosses and fens, moisture is required for sperm to reach to egg.
Solution
5. In mosses, the gametophyte is the dominant stage of the life cycle, in ferns, the sporophyte is
the dominant stage of the life cycle. In both mosses and ferns, moisture is required for the sperm
to reach the egg..
Unequal crossing over is responsible for creating paracentric inversi.pdf
Unequal crossing over is responsible for creating paracentric inversion duplications and
deletions pericentric inversions translocations
Solution
unequal crossing overs are responsible for creating translocation as one part would break and
join the chromatid while the other one would remain intact..
Using java script for an html page how do you add a name to a locati.pdf
Using java script for an html page how do you add a name to a location storage then it is
automatically added to a table on the webpage
Solution
Answer:
There are 2 methods for storing data : -
1) localStorage ,
2) sessionStorage
if we want the data to be stored forever then use localStorage if not then use sessionStorage .
Example :
localStorage.name = \"elixir black\".
True False When the null hypothesis is not rejected, there is no p.pdf
True/ False: When the null hypothesis is not rejected, there is no possibility of making a Type I
error. (
Solution
Type I error : It is the probability of rejecting true null hypothesis.
Type II error : It is the probability of fail to reject the false null hypothesis.
That is When the null hypothesis is not rejected, there is no possibility of making a Type II error.
Hence, given statement is false..
Thesus the rat is in a labyrinth. Initially, he is in a chamber with.pdf
Thesus the rat is in a labyrinth. Initially, he is in a chamber with two openings. Thesus always
chooses one of these two possible directions with equal probability.
If he goes right then he needs to walk three minutes only to come back inside to its initial
position.
If he goes to the left, then he needs to walk for one minute before finding a fork in the path. With
probability 1/3 he will choose the upwards path to his right, which leads to the exit, after another
minute walk.
With probability 2/3 he will choose the other way, which after four more minutes walking, will
lead to his departing point. What is the expected time that Theseus will remain in the labyrinth?
Solution
EXPECTED(X) = X1*P(X1)+ X2*P(X2).......+Xn*P(Xn)
probability of right = 1/3
probability of left= 2/3
expected minute = 1*1/3 + 4*2/3 = 9/3 = 3 minute..
summarizing the case #3 on facebook freedom of speech and giving add.pdf
summarizing the case #3 on facebook freedom of speech and giving additional source to support
your answer
Solution
We will complete our review and update that the our User Operations team uses to evaluate
reports of irreverence of our Community Standards around hate speech. To check that these
guidelines reflect best practices, we will solicit feedback from legal experts and others, including
representatives of the women\'s coalition and other groups that have historically faced
recognition and understanding of the difference between one thing and another. A few months
ago we began testing a new requirement that the creator of any content containing cruel and
insensitive humor include his or her authentic identity for the content to remain on Facebook. As
a result, if an individual decides to publicly share cruel and insensitive content, users can hold
the author accountable and directly object to the content.
the Anti-Defamation League’s Anti-Cyberhate working group and other international working
groups that we currently work with on these issues to include representatives of the women’s
coalition to identify how to balance considerations of free expression, to undertake research on
the effect of online hate speech on the online experiences of members of groups that have
historically faced discrimination in society, and to evaluate progress on our collective objectives.
..
So Far I have these two classes but I need help with my persontest c.pdf
So Far I have these two classes but I need help with my persontest class:
The next step is to add method calls for the accessor methods after each input:
//Reading and updating values
System.out.print(\"\ Enter Person Name: \");
person1.setPersonName(sc.nextLine());
System.out.println(\"You entered: \" + person1.getPersonName());
Call the accessor function each time you ask for input, after capturing that input. Then repeat for
a second Person, only this time call the constructor that takes the student id as a parameter.
7 import java.util.Scanner;
8
9 class PersonTest
10 {
11 //Main method
12 public static void main(String args[])
13 {
14 Scanner sc = new Scanner(System.in);
15
16 //Creating object
17 Persons_Information person1 = new Persons_Information();
18
19 //Reading and updating values
20 System.out.print(\"\ Enter Person Name: \");
21 person1.setPersonName(sc.nextLine());
22
23 System.out.print(\"\ Enter Current Address: \");
24 person1.setCurrentAdress(sc.nextLine());
25
26 System.out.print(\"\ Enter Permanent Address: \");
27 person1.setpermanentAdress(sc.nextLine());
28
29 System.out.print(\"\ Enter ID number: \");
30 person1.setIdNumber(sc.nextInt());
31
32 sc.nextLine();
33
34 System.out.print(\"\ Enter Birth Date: \");
35 person1.setBirthDate(sc.nextLine());
36
37
38 System.out.print(\"\ Enter Person Age: \");
39 person1.setPersonAge(sc.nextInt());
40
41
42 System.out.print(\"\ Enter Entry Year: \");
43 person1.setEntryYear(sc.nextInt());
44
45
46 System.out.print(\"\ Enter Total Years: \");
47 person1.setTotalYears(sc.nextInt());
48
49 //Printing person 1 details
50 System.out.println(\"\ Person 1: \ \" + person1.toString());
51 }
52 }
Solution
Persons_Information.java
public class Persons_Information {
//Declaring instance variables
private String personName;
private String currentAdress;
private String permanentAdress;
private int idNumber;
private String birthDate;
private int personAge;
private int entryYear;
private int totalYears;
//Zero argumented constructor
public Persons_Information()
{
}
//Parameterized constructor
public Persons_Information(int idNumber) {
this.idNumber = idNumber;
}
//Getters and setters
public String getPersonName() {
return personName;
}
public void setPersonName(String personName) {
this.personName = personName;
}
public String getCurrentAdress() {
return currentAdress;
}
public void setCurrentAdress(String currentAdress) {
this.currentAdress = currentAdress;
}
public String getPermanentAdress() {
return permanentAdress;
}
public void setPermanentAdress(String permanentAdress) {
this.permanentAdress = permanentAdress;
}
public int getIdNumber() {
return idNumber;
}
public void setIdNumber(int idNumber) {
this.idNumber = idNumber;
}
public String getBirthDate() {
return birthDate;
}
public void setBirthDate(String birthDate) {
this.birthDate = birthDate;
}
public int getPersonAge() {
return personAge;
}
public void setPersonAge(int personAge) {
this.personAge = personAge;
}
public int getEntryYe.
Shock Case Study (15 pts)HPI. Mrs. K is a 22 yo college student, .pdf
Shock Case Study (15 pts)
HPI. Mrs. K is a 22 y/o college student, rushed to the ED 35 minutes after sustaining multiple
stab wounds to the chest and abdomen by an unidentified assailant. A witness called 911.
Paramedics arriving at the scene found the victim to be in severe acute distress.
Vital signs were as follows: HR 128 (baseline 80), BP 80/55 (baseline 115/80), RR 37 and
labored. Chest auscultation revealed decreased breath sounds in the R lung consistent with
basilar atelectasis (ie. collapsed R lung). Pupils were equal, round, reactive to light, and
accommodation. Her LOC was reported as “awake, slightly confused, and complaining of severe
chest and abdominal pain.” Pedal pulses were absent, radial pulses were weak, and carotid pulses
were palpable. The patient was immediately started on IV Lactated Ringer’s solution at a rate of
150 mL/hr.
An ECG monitor placed at the scene of the attack revealed that the patient has developed sinus
tachycardia. She was tachypneic, became short of breath with conversation and reported her
heart was pounding out of her chest. She appeared to be very anxious and continued to c/o pain.
Her skin was cool and nail beds were pale but not cyanotic. Skin turgor was poor. Peripheral
pulses were absent with the exception of a thread, brachial pulse. Capillary refill time was 7-8
seconds. Doppler ultrasound had been required to obtain an accurate BP reading. The patient’s
skin was cool and clammy. There was a significant amount of blood on her dress and on the
pavement where she was lying.
Question 2. What is the pathophysiologic sequence of events for shock? (2pts)
Question 3. What type of shock does this patient seem to have? What is your rationale? (2 pts)
Question 4. Does this patient need a blood transfusion? Provide rationale for your answer. (2pts)
During transport to the hospital, vital signs were reassessed: HR 138, BP 75/50, RR 38 with
confusion. Patient was diagnosed with hypovolemic shock and IV fluids were doubled. Oxygen
was started at 3L/min by nasal cannula. ER physicians chose not to start a central venous line.
An indwelling foley catheter was inserted with return of 180mL of amber colored urine. Urine
output measured over the next hour was 14mL. Patient was taken to the OR for surgical
correction of lacerations to the right lung, liver and pancreas. In total, patient received 1L of
Lactated Ringers.
Table 1.
Class
Parameter
I
II
III
IV
Blood loss (ml)
<750
750–1500
1500–2000
>2000
Blood loss (%)
<15%
15–30%
30–40%
>40%
Pulse rate (beats/min)
<100
>100
>120
>140
Blood pressure
Normal
Decreased
Decreased
Decreased
Respiratory rate (breaths/min)
14–20
20–30
30–40
>35
Urine output (ml/hour)
>30
20–30
5–15
Negligible
CNS symptoms
Normal
Anxious
Confused
Lethargic
Class
Parameter
I
II
III
IV
Blood loss (ml)
<750
750–1500
1500–2000
>2000
Blood loss (%)
<15%
15–30%
30–40%
>40%
Pulse rate (beats/min)
<100
>100
>120
>140
Blood pressure
Normal
Decreased
Decreased
Decreased
Respiratory rate (breaths/min.
results obtained for the different tests and the characteristics of .pdf
results obtained for the different tests and the characteristics of E. coli, S. epidermidis, and S.
cerevisiae including the results of the test
Solution
1. Staphylococcus Epidermidis: Its develops on sheep blood agar : White colour colonies are
appears , circular and raised.
2. Entamoeba coli - Cyst and Trophozoite.
3. Saccharomyces cerevisiae is a kind of yeast. It is conceivably the most functional of yeast,
having been influential to baking, winemaking and brewing since early times. It is suspect that it
was initially isolated from the grapes skin. It is one of the most concentrated studied eukaryotic
duplicate organisms in cell biology and molecular. It is the microorganism beyond the most
ordinary type of fermentation..
Q1. What limits the number of different wavelengths that can be used.pdf
Q1. What limits the number of different wavelengths that can be used in a Dense Wavelength
Division Multiplexing (DWDM) System? Provide strong arguments for you answer. Explain any
terms used.
Solution
Dense wavelength division multiplexing(DWDM) refers originally to optical signals multiplexed
within the 1550 nm band so as to leverage the capabilities (and cost) of erbium doped fiber
amplifiers (EDFAs), which are effective for wavelengths between approximately 1525–1565 nm
(C band), or 1570–1610 nm (L band). EDFAs were originally developed to replace
SONET/SDHoptical-electrical-optical (OEO) regenerators, which they have made practically
obsolete. EDFAs can amplify any optical signal in their operating range, regardless of the
modulated bit rate. In terms of multi-wavelength signals, so long as the EDFA has enough pump
energy available to it, it can amplify as many optical signals as can be multiplexed into its
amplification band (though signal densities are limited by choice of modulation format). EDFAs
therefore allow a single-channel optical link to be upgraded in bit rate by replacing only
equipment at the ends of the link, while retaining the existing EDFA or series of EDFAs through
a long haul route. Furthermore, single-wavelength links using EDFAs can similarly be upgraded
to WDM links at reasonable cost. The EDFA\'s cost is thus leveraged across as many channels as
can be multiplexed into the 1550 nm band..
11. Internal validity refers to a. The extent to which a researcher.pdf
11. Internal validity refers to: a. The extent to which a researcher is able to draw causal
conclusions from a study. b. The extent to which the researcher can determine that changes to the
IV cause changes to the DV. c. The extent to which results of a study can be generalized. d. All
of the above. e. More than one of the above.
Solution
The correct answer is d.
BÀI TẬP BỔ TRỢ TIẾNG ANH GLOBAL SUCCESS LỚP 3 - CẢ NĂM (CÓ FILE NGHE VÀ ĐÁP Á...
BÀI TẬP BỔ TRỢ TIẾNG ANH GLOBAL SUCCESS LỚP 3 - CẢ NĂM (CÓ FILE NGHE VÀ ĐÁP Á...Nguyen Thanh Tu Collection
https://app.box.com/s/hqnndn05v4q5a4k4jd597rkdbda0fniiDigital Tools and AI for Teaching Learning and Research
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A variety of media and culture conditions can be used to culture and .pdf
A variety of media and culture conditions can be used to culture and identify Listeria
monocytogenes. For each, characterize it as selective or differential.
Solution
Listeria monocytogenes is basically a soil-borne micro-organism. It has been found in various
sources such as water, agricultural products, and in animals. Listeria monocytogenes has been
known as etiologic agent of listeriosis, causing meningitis, encephalitis, septicemia and
endocarditis in humans.
various selective and differential methods were developed to isolate the Listeria monocytogenes
species as follows
Selective
Incubation at 30 degree celsius
Adding antibiotics such as nalidixic acid
Adding high concentarion of salts such as lithium chloride and sodium chloride
Adding antifungals such as amphotericin B
Differential
Adding pH indicator to detect ferementaion of rhamanose to acid
Detecting beta heamolysis.
A program consists of 4 musical numbers and 3 speeches. In how many .pdf
A program consists of 4 musical numbers and 3 speeches. In how many ways can the program be
arranged so that it begins and ends with a musical number?
Solution
There are 4 musical numbers and 3 speeches
AS begin and end musical numbers 2 musical numbers selected from 4 and put in the ends
This can be done in 2(4C2) = 12 ways
The remaining are 2 musical and 3 speeches. This can be arranged in 5! ways
Hence total no of ways = 12 (5!) = 1440.
A Moving to another question will save this response. Question 4 Iden.pdf
A Moving to another question will save this response. Question 4 Identify a problem with water
that is classified as hard a scaling especially when exposed to heat b.scaling especially when
exposed to cold c inability to lather soap (onto bubbles) d soap very lathery/bubbly hard to rinse
soap off Of b&d; AMoving to another question will save this response
Solution
Option e is correct. Hard water is scaling when exposd to heat and it is not able to give lather to
the soap..
property, plant, and equipment are categorized as property, pla.pdf
property, plant, and equipment are categorized as
property, plant, and equipment are categorized as
Solution
Answer:
Property, Plant and Equipment are categorized as \"Fixed Assets\" in the Asset Section of the
Balance Sheet.
Property, Plant and Equipment are Non Current Assets which includes Building, Plant, Vehicle,
Land etc. Some of these Fixed Assets are subject to Depreciation, while some are not subject to
Depreciation which are Land etc..
Write a program in Matlab using Newtons method to solve this syste.pdf
Write a program in Matlab using Newton\'s method to solve this system of equations for x, y,
and z...
Solution
f=inline(\'[-10*p(:,1)-10*p(:,2); 28*p(:,1)-p(:,2)-p(:,1).*p(:,3); p(:,1).*p(:,2)-(8*p(:,3))/2]\',\'p\');
tol = 0.4e-6; %error tolerance
MaxIter = 10;
format long
x = sym(\'x\') ;
y = sym(\'y\') ;
z = sym(\'z\') ;
F = f([x,y,z]);
J = jacobian(F);
invJ = inv(J);
s = zeros(MaxIter,3);
s(1,:) = [1 1 1]; % first set of values
dsnorm = inf;
iter = 1;
while dsnorm>tol && itertol && itertol && iter.
Why is it that bacterial ribosomes can begin translation before mRNA.pdf
Why is it that bacterial ribosomes can begin translation before mRNA synthesis is completed,
but eukaryotic ribosomes cannot?
a. mRNA splicing occurs very quickly in prokaryotic cells.
b. In prokaryotes, transcription and translation take place in the cytoplasm.
c. Prokaryotic mRNAs are elongated differently from eukaryotic mRNAs.
d. Prokaryotic mRNAs are synthesized with ribosomes attached.
Solution
Answer is b. In prokaryotic organisms there is no nuclei, so both transcription and translation
occur in the same place that is Cytosol. Moreover, no post transcriptional modifications like
caping and Tailing. Due to this the half life of Prokaryotic m-RNA is too small. Due to these two
reasons, in prokaryotic organisms, translation start, before completion of transcription.
Write a C++ program with a function defined to give the person a .pdf
Write a C++ program with a function defined to give the person a 20% raise?
Please explain. Thanks
Write a C++ program with a function defined to give the person a 20% raise?
Please explain. Thanks
Please explain. Thanks
Solution
#include
#include
#define SIZE 5 // we are defining the size of entries / variables to 5
class emp
{
float basic,da,it,netsal; // salary , da and netsalary are mentioned in float
char name[20],num[10];
public:
void getdata(); // function to get employee data
void net_sal(); // function to get employee salary
void sal_hike(); // fuction to display hike based on net salary
void dispdata(); // to dislay data
};
void emp::getdata() // enter basic employee details
{
cout<<\"\ Enter employee number: \" ;
cin>>name;
cout<<\" Enter employee name: \" ;
cin>>num;
cout<<\"Enter employee basic salary in Rs: \" ;
cin>>basic;
}
void emp::net_sal() // calculate net salary
{
da=((0.52)*basic );
float gsal=da+basic;
it=((0.3)*gsal);
netsal=gsal-it;
}
void emp::sal_hike() // compute salary hike
int hike=net_sal + 0.2*net_sal;
}
void emp::dispdata() // to display data
{
cout
<<\"\ Employee number: \"<>n;
for(int i=0;i.
Which standards organization defined the Ethernet standardA.IEEE.pdf
Which standards organization defined the Ethernet standard?
A.IEEE
B.ISO
C.ANSI
D.DIX
Solution
Answer: D) DIX
DIX stands for Digital Equopment Company, Intel and Xerox. DIX teamed up for the first
standard for Ethernet standard.
IEEE also created Ethernet standards but they were not the frst.
ISO defined ISO reference model not the Ethernet standard
ANSI associted with ASCII codes and standards of others..
When the pH of a solution changes from 2 to 6 a) That solution beco.pdf
When the pH of a solution changes from 2 to 6: a) That solution becomes more acidic due to an
increase in the proton concentration. b) That solution contains 4X fewer protons. c) That solution
becomes 1000X more basic. d) That solution is 4X more acidic. e) None of the above
Solution
Answer :
Answer: The correct answer is B (That solution contains 4X fewer protons).
When the pH of a solution changes from 2 to 6, the [H+] hydronium ions decrease and [OH-]
will increase.
Which is the correct answer Consider the characteristics of moss an.pdf
Which is the correct answer? Consider the characteristics of moss and fern life cycles. Which of
the following sets of statements is true? The gametophyte is the dominant stage of the life cycle
in both mosses and ferns. Mosses require moisture for sperm to reach the egg, but ferns do not
In mosses, the gametophyte is the dominant stage of the life cycle; in ferns, the sporophyte is the
dominant stage of the life cycle. Ferns require moisture for sperm to reach the egg, but mosses
do not The gametophyte is the dominant stage of the life cycle in both mosses and ferns. In both
mosses and ferns, moisture is required for sperm to reach the egg. In mosses, the sporophyte is
the dominant stage of the life cycle; in ferns, the dominant stage of the life cycle. In both mosses
and ferns moisture is reach for spent to reach the egg. In moses, the gametophyte is the
dominant state of the life cycle, if ferns, the sprophyte is the dominant stage of the life cycle. In
both mosses and fens, moisture is required for sperm to reach to egg.
Solution
5. In mosses, the gametophyte is the dominant stage of the life cycle, in ferns, the sporophyte is
the dominant stage of the life cycle. In both mosses and ferns, moisture is required for the sperm
to reach the egg..
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Unequal crossing over is responsible for creating paracentric inversion duplications and
deletions pericentric inversions translocations
Solution
unequal crossing overs are responsible for creating translocation as one part would break and
join the chromatid while the other one would remain intact..
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Using java script for an html page how do you add a name to a location storage then it is
automatically added to a table on the webpage
Solution
Answer:
There are 2 methods for storing data : -
1) localStorage ,
2) sessionStorage
if we want the data to be stored forever then use localStorage if not then use sessionStorage .
Example :
localStorage.name = \"elixir black\".
True False When the null hypothesis is not rejected, there is no p.pdf
True/ False: When the null hypothesis is not rejected, there is no possibility of making a Type I
error. (
Solution
Type I error : It is the probability of rejecting true null hypothesis.
Type II error : It is the probability of fail to reject the false null hypothesis.
That is When the null hypothesis is not rejected, there is no possibility of making a Type II error.
Hence, given statement is false..
Thesus the rat is in a labyrinth. Initially, he is in a chamber with.pdf
Thesus the rat is in a labyrinth. Initially, he is in a chamber with two openings. Thesus always
chooses one of these two possible directions with equal probability.
If he goes right then he needs to walk three minutes only to come back inside to its initial
position.
If he goes to the left, then he needs to walk for one minute before finding a fork in the path. With
probability 1/3 he will choose the upwards path to his right, which leads to the exit, after another
minute walk.
With probability 2/3 he will choose the other way, which after four more minutes walking, will
lead to his departing point. What is the expected time that Theseus will remain in the labyrinth?
Solution
EXPECTED(X) = X1*P(X1)+ X2*P(X2).......+Xn*P(Xn)
probability of right = 1/3
probability of left= 2/3
expected minute = 1*1/3 + 4*2/3 = 9/3 = 3 minute..
summarizing the case #3 on facebook freedom of speech and giving add.pdf
summarizing the case #3 on facebook freedom of speech and giving additional source to support
your answer
Solution
We will complete our review and update that the our User Operations team uses to evaluate
reports of irreverence of our Community Standards around hate speech. To check that these
guidelines reflect best practices, we will solicit feedback from legal experts and others, including
representatives of the women\'s coalition and other groups that have historically faced
recognition and understanding of the difference between one thing and another. A few months
ago we began testing a new requirement that the creator of any content containing cruel and
insensitive humor include his or her authentic identity for the content to remain on Facebook. As
a result, if an individual decides to publicly share cruel and insensitive content, users can hold
the author accountable and directly object to the content.
the Anti-Defamation League’s Anti-Cyberhate working group and other international working
groups that we currently work with on these issues to include representatives of the women’s
coalition to identify how to balance considerations of free expression, to undertake research on
the effect of online hate speech on the online experiences of members of groups that have
historically faced discrimination in society, and to evaluate progress on our collective objectives.
..
So Far I have these two classes but I need help with my persontest c.pdf
So Far I have these two classes but I need help with my persontest class:
The next step is to add method calls for the accessor methods after each input:
//Reading and updating values
System.out.print(\"\ Enter Person Name: \");
person1.setPersonName(sc.nextLine());
System.out.println(\"You entered: \" + person1.getPersonName());
Call the accessor function each time you ask for input, after capturing that input. Then repeat for
a second Person, only this time call the constructor that takes the student id as a parameter.
7 import java.util.Scanner;
8
9 class PersonTest
10 {
11 //Main method
12 public static void main(String args[])
13 {
14 Scanner sc = new Scanner(System.in);
15
16 //Creating object
17 Persons_Information person1 = new Persons_Information();
18
19 //Reading and updating values
20 System.out.print(\"\ Enter Person Name: \");
21 person1.setPersonName(sc.nextLine());
22
23 System.out.print(\"\ Enter Current Address: \");
24 person1.setCurrentAdress(sc.nextLine());
25
26 System.out.print(\"\ Enter Permanent Address: \");
27 person1.setpermanentAdress(sc.nextLine());
28
29 System.out.print(\"\ Enter ID number: \");
30 person1.setIdNumber(sc.nextInt());
31
32 sc.nextLine();
33
34 System.out.print(\"\ Enter Birth Date: \");
35 person1.setBirthDate(sc.nextLine());
36
37
38 System.out.print(\"\ Enter Person Age: \");
39 person1.setPersonAge(sc.nextInt());
40
41
42 System.out.print(\"\ Enter Entry Year: \");
43 person1.setEntryYear(sc.nextInt());
44
45
46 System.out.print(\"\ Enter Total Years: \");
47 person1.setTotalYears(sc.nextInt());
48
49 //Printing person 1 details
50 System.out.println(\"\ Person 1: \ \" + person1.toString());
51 }
52 }
Solution
Persons_Information.java
public class Persons_Information {
//Declaring instance variables
private String personName;
private String currentAdress;
private String permanentAdress;
private int idNumber;
private String birthDate;
private int personAge;
private int entryYear;
private int totalYears;
//Zero argumented constructor
public Persons_Information()
{
}
//Parameterized constructor
public Persons_Information(int idNumber) {
this.idNumber = idNumber;
}
//Getters and setters
public String getPersonName() {
return personName;
}
public void setPersonName(String personName) {
this.personName = personName;
}
public String getCurrentAdress() {
return currentAdress;
}
public void setCurrentAdress(String currentAdress) {
this.currentAdress = currentAdress;
}
public String getPermanentAdress() {
return permanentAdress;
}
public void setPermanentAdress(String permanentAdress) {
this.permanentAdress = permanentAdress;
}
public int getIdNumber() {
return idNumber;
}
public void setIdNumber(int idNumber) {
this.idNumber = idNumber;
}
public String getBirthDate() {
return birthDate;
}
public void setBirthDate(String birthDate) {
this.birthDate = birthDate;
}
public int getPersonAge() {
return personAge;
}
public void setPersonAge(int personAge) {
this.personAge = personAge;
}
public int getEntryYe.
Shock Case Study (15 pts)HPI. Mrs. K is a 22 yo college student, .pdf
Shock Case Study (15 pts)
HPI. Mrs. K is a 22 y/o college student, rushed to the ED 35 minutes after sustaining multiple
stab wounds to the chest and abdomen by an unidentified assailant. A witness called 911.
Paramedics arriving at the scene found the victim to be in severe acute distress.
Vital signs were as follows: HR 128 (baseline 80), BP 80/55 (baseline 115/80), RR 37 and
labored. Chest auscultation revealed decreased breath sounds in the R lung consistent with
basilar atelectasis (ie. collapsed R lung). Pupils were equal, round, reactive to light, and
accommodation. Her LOC was reported as “awake, slightly confused, and complaining of severe
chest and abdominal pain.” Pedal pulses were absent, radial pulses were weak, and carotid pulses
were palpable. The patient was immediately started on IV Lactated Ringer’s solution at a rate of
150 mL/hr.
An ECG monitor placed at the scene of the attack revealed that the patient has developed sinus
tachycardia. She was tachypneic, became short of breath with conversation and reported her
heart was pounding out of her chest. She appeared to be very anxious and continued to c/o pain.
Her skin was cool and nail beds were pale but not cyanotic. Skin turgor was poor. Peripheral
pulses were absent with the exception of a thread, brachial pulse. Capillary refill time was 7-8
seconds. Doppler ultrasound had been required to obtain an accurate BP reading. The patient’s
skin was cool and clammy. There was a significant amount of blood on her dress and on the
pavement where she was lying.
Question 2. What is the pathophysiologic sequence of events for shock? (2pts)
Question 3. What type of shock does this patient seem to have? What is your rationale? (2 pts)
Question 4. Does this patient need a blood transfusion? Provide rationale for your answer. (2pts)
During transport to the hospital, vital signs were reassessed: HR 138, BP 75/50, RR 38 with
confusion. Patient was diagnosed with hypovolemic shock and IV fluids were doubled. Oxygen
was started at 3L/min by nasal cannula. ER physicians chose not to start a central venous line.
An indwelling foley catheter was inserted with return of 180mL of amber colored urine. Urine
output measured over the next hour was 14mL. Patient was taken to the OR for surgical
correction of lacerations to the right lung, liver and pancreas. In total, patient received 1L of
Lactated Ringers.
Table 1.
Class
Parameter
I
II
III
IV
Blood loss (ml)
<750
750–1500
1500–2000
>2000
Blood loss (%)
<15%
15–30%
30–40%
>40%
Pulse rate (beats/min)
<100
>100
>120
>140
Blood pressure
Normal
Decreased
Decreased
Decreased
Respiratory rate (breaths/min)
14–20
20–30
30–40
>35
Urine output (ml/hour)
>30
20–30
5–15
Negligible
CNS symptoms
Normal
Anxious
Confused
Lethargic
Class
Parameter
I
II
III
IV
Blood loss (ml)
<750
750–1500
1500–2000
>2000
Blood loss (%)
<15%
15–30%
30–40%
>40%
Pulse rate (beats/min)
<100
>100
>120
>140
Blood pressure
Normal
Decreased
Decreased
Decreased
Respiratory rate (breaths/min.
results obtained for the different tests and the characteristics of .pdf
results obtained for the different tests and the characteristics of E. coli, S. epidermidis, and S.
cerevisiae including the results of the test
Solution
1. Staphylococcus Epidermidis: Its develops on sheep blood agar : White colour colonies are
appears , circular and raised.
2. Entamoeba coli - Cyst and Trophozoite.
3. Saccharomyces cerevisiae is a kind of yeast. It is conceivably the most functional of yeast,
having been influential to baking, winemaking and brewing since early times. It is suspect that it
was initially isolated from the grapes skin. It is one of the most concentrated studied eukaryotic
duplicate organisms in cell biology and molecular. It is the microorganism beyond the most
ordinary type of fermentation..
Q1. What limits the number of different wavelengths that can be used.pdf
Q1. What limits the number of different wavelengths that can be used in a Dense Wavelength
Division Multiplexing (DWDM) System? Provide strong arguments for you answer. Explain any
terms used.
Solution
Dense wavelength division multiplexing(DWDM) refers originally to optical signals multiplexed
within the 1550 nm band so as to leverage the capabilities (and cost) of erbium doped fiber
amplifiers (EDFAs), which are effective for wavelengths between approximately 1525–1565 nm
(C band), or 1570–1610 nm (L band). EDFAs were originally developed to replace
SONET/SDHoptical-electrical-optical (OEO) regenerators, which they have made practically
obsolete. EDFAs can amplify any optical signal in their operating range, regardless of the
modulated bit rate. In terms of multi-wavelength signals, so long as the EDFA has enough pump
energy available to it, it can amplify as many optical signals as can be multiplexed into its
amplification band (though signal densities are limited by choice of modulation format). EDFAs
therefore allow a single-channel optical link to be upgraded in bit rate by replacing only
equipment at the ends of the link, while retaining the existing EDFA or series of EDFAs through
a long haul route. Furthermore, single-wavelength links using EDFAs can similarly be upgraded
to WDM links at reasonable cost. The EDFA\'s cost is thus leveraged across as many channels as
can be multiplexed into the 1550 nm band..
11. Internal validity refers to a. The extent to which a researcher.pdf
11. Internal validity refers to: a. The extent to which a researcher is able to draw causal
conclusions from a study. b. The extent to which the researcher can determine that changes to the
IV cause changes to the DV. c. The extent to which results of a study can be generalized. d. All
of the above. e. More than one of the above.
Solution
The correct answer is d.
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