briefly write about the variability of natural systems and the signi.pdf
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Ecological systems are naturally variable in their physical and biological characteristics, and exist in a dynamic equilibrium with their biota. Systems with greater diversity of species are generally more stable and resistant to external disturbances. Keystone species significantly impact ecosystem functioning and overall community stability. Therefore, variability and stability are important for the proper functioning of ecological systems.
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Give background information concerning EMT and include what is known.pdf
Give background information concerning EMT and include what is known about EMT in normal
development as well as in neoplasms. What are strategies that could be used to overcome drug
resistance in cancer cells by targeting EMT (minimum four paragraphs).
Solution
Answer:
Cancer stem cell (CSC) hypothesis: This hypothesis elucidating that the CSC are the
subpopulation and able to exert higher resistance associated with explicit capability to self-renew
finally to differentiate into different offspring of cancer cells to generate a tumor. This
hypothesis is based on the cellular mechanisms exclusively developed inside the CSC to exert
resistance such pathways are mainly through Wnt signaling, Notch signaling. There are other
targets to regulate proliferation and differentiation through ABC transporter system and receptor
tyrosine kinases. These targets are the future stem cells anti-cancer therapies in regulating the
proliferation & differentiation of cancer cells as CSCs have a potential relationship with
epithelial-mesenchymal transition (EMT).
Metastasis has many steps. The main cancer cell morphological features that support invasion,
intravasation, and extravasation are epithelial-mesenchymal transition via cell -adhesion
molecules. Invasion is mainly due to presence of cell adhesion molecules such as integrins on the
cancer cells finally promote the movement of cancer cells to another region. This process is
going to connect adjacent cells to tumor cells finally undergo epithelial-mesenchymal transition
(EMT) result in formation of \"continuously dividing neoplasms\" with higher drug resistence to
cure compared to normal cells
Strategies that could be used to overcome drug resistance in cancer cells by targeting EMT:
\"A combination therapy of neutraceuticals & chemotherapeutic agents\" are most promising
strategies to overcome drug resistance in cancer cells by targeting EMTs, cancer stem cells
(CSC) &
Modulation of microRNAs do promote effective strategy to overcome drug resistance in cancer
cells
Oligonucletodie deliver: Another strategy could be used to overcome drug resistance in cancer
cells by targeting EMT. In this method, synthesized microRNAs are going to make in the form of
\"nanoparticle or microsomal\" formulation finally used to target microRNAs in cancer cells
result in \"complete destruction of pre-mRNA\" finally there will no translation mechanism to
synthesize \"cell survival proteins\" to proliferate.
Explain how the current economic recession differs from the depressio.pdf
Explain how the current economic recession differs from the depression in the 1930\'s.
Solution
The Great depression was worldwide. It was due to the collapse of the international financial
system It also resulted from the mutual adoption by many countries ( including USA) of high-
tariff policies, which were intended to keep out foreign goods in order to protect domestic
producers. The policies were called \"beggar thy neighbour\" strategies since they attempted to
\"export\" unemployment by improving one country\'s trade position and hence demand for its
goods at the expense of its trading partners And , of course, if each country keeps out foreign
goods, the volume of world trade declines, providing a contractionary influence on the world
economy
Almost every country suffered a deep recession in the 1930s, but some countries did better than
USA Sweden began an expansionary policy in the early 1930s and reduced its unemployment
relatively quicker Britain\'s economy suffered high unemployment but in 1931 it went off the
gold standard and the ensuing devaluation of the pound sterling set the stage for some
improvement. Germany grew rapidly after Hitler came to power and expanded government
spending. China escaped the recession until after 1931, because it had a floating exchange rate.
In 1938, real GNP in US rose above its 1929 level for the first time in the decade, but it was not
until 1942 , after the US formally entered world war II , that the unemployment rate finally fell
below 5%
The experience of the US in the early 1980s- the worst recession since the Great Depression,
casts doubt on the optimistism of recovery. During the 1980s, the US experienced the largest
sustained budget deficits in its peacetime history. Even so cutting spending or raising taxes was
not politically popular. Gradually, in the 1990s, the deficit began to be brought under control,
and toward the end of the decade the budget swung into surplus.This was due to raising the
prices and government spending.
By,
Nishant Bhatt.
Does yeast uses CopI and CopII formation for vesicle budding to grow.pdf
Does yeast uses CopI and CopII formation for vesicle budding to grow asexually?Is CopII starts
from the er then goes out the cell and CopI works for the molecules comes in the cell and goes to
nucleus?Does copII carries proteins away to make the budding?Please explain in details.
Solution
Yes, yeasts reproduce asexually by an asymmetric division process known as budding. Initially,
yeast forms a small protuberance which then grows to a full size and forms a bud. Then the
nucleus splits and migrates into the daughter cell. After the movement of nucleus, the entire bud
detaches from the parent cell by forming a constriction at the base. This process will repeat until
a chain of bud cells are formed..
DNA molecules consist of chemically linked sequences of the bases ad.pdf
DNA molecules consist of chemically linked sequences of the bases adenine, guanine, cytosine
and thymine, denoted A, G, C and T. A sequence of three bases is called a codon. A base may
appear more than once in a codon.
1) Represent all possible base sequences using a tree diagram.
2) How many different codons are there?
3) The bases A and G are purines, while C and T are pyrimidines. How many codons are there
whose first and third bases are purines and whose second base is a pyrimidine?
4) How many codons consist of three different bases?
Solution.
Data StructuresPLEASE USING THIS C++ PROGRAM BELOW, I NEED HEL.pdf
Data Structures
PLEASE USING THIS C++ PROGRAM BELOW, I NEED HELP ON IMPLEMENTING
BOTH QUEUE AND STACK TO PRINT \"I able\"
REVERSE/BACKWARDS(PALINDROME)
#include
#include
using namespace std;
struct st
{
int year;
st *next;
};
class queue
{
public:
int sz,i = 0;
char chr[100];
queue()
{
//constructor for initializing front and rear to NULL
front = NULL;
rear = NULL;
cout << \"enter size of queue: \";
cin >> sz;
}
// ~queue()
// {
//
// }
// in the enterqueue, create a new node, assign value and add it to queue
// if F = R = NULL the ( ) else ( )
void enterqueue(int yr, char el)
{
//inserting into queue if queue not overflow
if(i >= sz)
{
cout << \"queue over flow\";
return;
}
if (front == NULL )
{
front = new st;
front->next = NULL;
front->year = yr;
rear = front;
chr[i] = el;
i++;
}
else
{
st *temp = new st;
temp->year = yr;
rear->next = temp;
rear = temp;
chr[i] = el;
i++;
}
}
char deletequeue()
{
//delete front data from queue if data in the queue
char tr;
int j = 0;
st *temp;
temp = front;
if( front == NULL)
{
cout << \"queue under flow\";
return \'a\';
}
if(front == rear)
{
front = rear = NULL;
return \'a\';
}
front = front->next;
delete temp;
tr = chr[0];
i--;
for(j = 0; j < i; j++)
{
chr[j] = chr[j+1];
}
chr[j] = \'\\0\';
return tr;
}
bool isempty()
{
//it shows whether queue is empty
if( front == NULL)
{
return true;
}
else
return false;
}
bool isfull()
{
//it shows whether queue is full
if(i == sz)
return true;
else
return false;
}
void display()
{
//it displays queue data with characters of queue if invalid choice is entered
st *tmp;
int j = 0;
cout << \"\ queue is:\ character \\tyear\";
for(tmp = front; tmp != rear; tmp = tmp->next, j++)
{
cout << \"\ \" << chr[j] << \" \" << tmp->year;
}
cout << \"\ \"<year;
}
private:
st *front, *rear;
};
int main()
{
//main method to call queue functions by choice
queue s1;
int ch,yr;
char el;
bool s;
while(1)
{
cout << \"\ 1.ENTERQUEUE\ 2.DELETEQUEUE\ 3.ISEMPTY\ 4.ISFULL\ 5.EXIT\ enter
your choice: \";
cin >> ch;
switch(ch)
{
case 1:
if(!s1.isfull())
{
cout << \"\ enter a year : \";
cin >> yr;
cout << \"\ enter a character : \";
cin >> el;
s1.enterqueue(yr, el);
}
else
cout << \"insertion not possible\";
break;
case 2:
s = s1.isempty();
if(!s)
{
el = s1.deletequeue();
cout << \"\ The character of queue \" << el <<\" is deleted\" << endl;
}
else
cout << \"\ queue is empty.deletion not possible\"<< endl;
break;
case 3:
s = s1.isempty();
if(s)
cout << \"\ queue is empty\" << endl;
else
cout << \"\ queue is not empty\" << endl;
break;
case 4:
s=s1.isfull();
if(s)
cout << \"\ queue is full\" << endl;
else
cout << \"\ queue is not full\" << endl;
break;
case 5:
exit(0);
default : cout << \"invalid choice\" << endl;
if(!s1.isempty())
s1.display();
}
}
return (0);
}
Solution
#include
#include
using namespace std;
struct st
{
int year;
st *next;
};
class queue
{
public:
//int sz,i=0;//modified //code modified here// everything is fine with the code except this...
queue is implemented correctly
//you can .
Count Red Nodes. Write a program that computes the percentage of red.pdf
Count Red Nodes. Write a program that computes the percentage of red nodes in a given red-
black BST. Test your program by running at least 100 trials of the experiment of inserting N
random keys into an initially empty tree, for N = 10^4, 10^5, 10^6, and formulate an hypothesis.
*It is experiment from textbook Algorithms, 3.3.42
Does anyone know the answer?
Solution
/*
* Java Program to Implement Red Black Tree
*/
import java.util.Scanner;
/* Class Node */
class RedBlackNode
{
RedBlackNode left, right;
int element;
int color;
/* Constructor */
public RedBlackNode(int theElement)
{
this( theElement, null, null );
}
/* Constructor */
public RedBlackNode(int theElement, RedBlackNode lt, RedBlackNode rt)
{
left = lt;
right = rt;
element = theElement;
color = 1;
}
}
/* Class RBTree */
class RBTree
{
private RedBlackNode current;
private RedBlackNode parent;
private RedBlackNode grand;
private RedBlackNode great;
private RedBlackNode header;
private static RedBlackNode nullNode;
/* static initializer for nullNode */
static
{
nullNode = new RedBlackNode(0);
nullNode.left = nullNode;
nullNode.right = nullNode;
}
/* Black - 1 RED - 0 */
static final int BLACK = 1;
static final int RED = 0;
/* Constructor */
public RBTree(int negInf)
{
header = new RedBlackNode(negInf);
header.left = nullNode;
header.right = nullNode;
}
/* Function to check if tree is empty */
public boolean isEmpty()
{
return header.right == nullNode;
}
/* Make the tree logically empty */
public void makeEmpty()
{
header.right = nullNode;
}
/* Function to insert item */
public void insert(int item )
{
current = parent = grand = header;
nullNode.element = item;
while (current.element != item)
{
great = grand;
grand = parent;
parent = current;
current = item < current.element ? current.left : current.right;
// Check if two red children and fix if so
if (current.left.color == RED && current.right.color == RED)
handleReorient( item );
}
// Insertion fails if already present
if (current != nullNode)
return;
current = new RedBlackNode(item, nullNode, nullNode);
// Attach to parent
if (item < parent.element)
parent.left = current;
else
parent.right = current;
handleReorient( item );
}
private void handleReorient(int item)
{
// Do the color flip
current.color = RED;
current.left.color = BLACK;
current.right.color = BLACK;
if (parent.color == RED)
{
// Have to rotate
grand.color = RED;
if (item < grand.element != item < parent.element)
parent = rotate( item, grand ); // Start dbl rotate
current = rotate(item, great );
current.color = BLACK;
}
// Make root black
header.right.color = BLACK;
}
private RedBlackNode rotate(int item, RedBlackNode parent)
{
if(item < parent.element)
return parent.left = item < parent.left.element ? rotateWithLeftChild(parent.left) :
rotateWithRightChild(parent.left) ;
else
return parent.right = item < parent.right.element ? rotateWithLeftChild(parent.right) :
rotateWithRightChild(parent.right);
}
/* Rotate binary tree node with left child */
private RedBlackNode rot.
apple 2008 historically what were Apples major competitive advanta.pdf
apple 2008
historically what were Apples major competitive advantages and why did they fail to build on
these advantages to lead the industry?
Solution
Apples major competitive advantages:
Apple II and Macintosh were the machines that made Apple. They both had to following
features.
Brand loyalty among important segments
Reasons for why did Apple fail to build on these advantages to lead the industry:
Apple’s missteps:
Entry of Rivals:.
A partial results from a PERT analysis for a project with 50 activit.pdf
A partial results from a PERT analysis for a project with 50 activities shows that path A-K-Q
Solution
There are 50 activities shows the path A-K-Q.
With the help of PERT (Program Evaluation and Review Technique), and CPM (Critical Path
Method), the project manager can
1. Plan the project ahead of time and foresee possible sources of troubles and delays in
completion.
2. Schedule the project activities at the appropriate times to conform with proper job sequence so
that the project is completed as soon as possible.
3. Coordinate and control the projects activities so as to stay on schedule in completing the
project.
Thus PERT incorporates uncertainties in activity times in its analysis. It determines the
probabilities of completing various stages of the project by specified deadlines. It also calculates
the expected time to complete the project. An important and extremely useful by product of
PERT analysis is its identification of various “bottlenecks” in a project.
Application of PERT and CPM
A partial list of applications of PERT and CPM techniques is
1. Construction projects (e.g., buildings, highways, houses, and bridges)
2. Preparation of bids and proposals for large projects.
3. Maintenance planning of oil refineries, ship repairs, and other large operations.
4. Development of new weapons systems and new manufactured products.
5. Manufacture and assembly of large items such as airplanes, ships, and computers. 6. Simple
projects.
Six steps :
1. Define the projects and all of its significant activities or tasks.
2. Develop the relationships among the activities. Decide which activities must precede and
which must follow others.
3. Draw the network connecting all of the activities.
4. Assign time and/or cost estimates to each activity.
5. Compute the longest time path through the network; this is called the critical path.
6. Use the network to help plan, schedule, monitor and control the project..
Recommended
Give background information concerning EMT and include what is known.pdf
Give background information concerning EMT and include what is known about EMT in normal
development as well as in neoplasms. What are strategies that could be used to overcome drug
resistance in cancer cells by targeting EMT (minimum four paragraphs).
Solution
Answer:
Cancer stem cell (CSC) hypothesis: This hypothesis elucidating that the CSC are the
subpopulation and able to exert higher resistance associated with explicit capability to self-renew
finally to differentiate into different offspring of cancer cells to generate a tumor. This
hypothesis is based on the cellular mechanisms exclusively developed inside the CSC to exert
resistance such pathways are mainly through Wnt signaling, Notch signaling. There are other
targets to regulate proliferation and differentiation through ABC transporter system and receptor
tyrosine kinases. These targets are the future stem cells anti-cancer therapies in regulating the
proliferation & differentiation of cancer cells as CSCs have a potential relationship with
epithelial-mesenchymal transition (EMT).
Metastasis has many steps. The main cancer cell morphological features that support invasion,
intravasation, and extravasation are epithelial-mesenchymal transition via cell -adhesion
molecules. Invasion is mainly due to presence of cell adhesion molecules such as integrins on the
cancer cells finally promote the movement of cancer cells to another region. This process is
going to connect adjacent cells to tumor cells finally undergo epithelial-mesenchymal transition
(EMT) result in formation of \"continuously dividing neoplasms\" with higher drug resistence to
cure compared to normal cells
Strategies that could be used to overcome drug resistance in cancer cells by targeting EMT:
\"A combination therapy of neutraceuticals & chemotherapeutic agents\" are most promising
strategies to overcome drug resistance in cancer cells by targeting EMTs, cancer stem cells
(CSC) &
Modulation of microRNAs do promote effective strategy to overcome drug resistance in cancer
cells
Oligonucletodie deliver: Another strategy could be used to overcome drug resistance in cancer
cells by targeting EMT. In this method, synthesized microRNAs are going to make in the form of
\"nanoparticle or microsomal\" formulation finally used to target microRNAs in cancer cells
result in \"complete destruction of pre-mRNA\" finally there will no translation mechanism to
synthesize \"cell survival proteins\" to proliferate.
Explain how the current economic recession differs from the depressio.pdf
Explain how the current economic recession differs from the depression in the 1930\'s.
Solution
The Great depression was worldwide. It was due to the collapse of the international financial
system It also resulted from the mutual adoption by many countries ( including USA) of high-
tariff policies, which were intended to keep out foreign goods in order to protect domestic
producers. The policies were called \"beggar thy neighbour\" strategies since they attempted to
\"export\" unemployment by improving one country\'s trade position and hence demand for its
goods at the expense of its trading partners And , of course, if each country keeps out foreign
goods, the volume of world trade declines, providing a contractionary influence on the world
economy
Almost every country suffered a deep recession in the 1930s, but some countries did better than
USA Sweden began an expansionary policy in the early 1930s and reduced its unemployment
relatively quicker Britain\'s economy suffered high unemployment but in 1931 it went off the
gold standard and the ensuing devaluation of the pound sterling set the stage for some
improvement. Germany grew rapidly after Hitler came to power and expanded government
spending. China escaped the recession until after 1931, because it had a floating exchange rate.
In 1938, real GNP in US rose above its 1929 level for the first time in the decade, but it was not
until 1942 , after the US formally entered world war II , that the unemployment rate finally fell
below 5%
The experience of the US in the early 1980s- the worst recession since the Great Depression,
casts doubt on the optimistism of recovery. During the 1980s, the US experienced the largest
sustained budget deficits in its peacetime history. Even so cutting spending or raising taxes was
not politically popular. Gradually, in the 1990s, the deficit began to be brought under control,
and toward the end of the decade the budget swung into surplus.This was due to raising the
prices and government spending.
By,
Nishant Bhatt.
Does yeast uses CopI and CopII formation for vesicle budding to grow.pdf
Does yeast uses CopI and CopII formation for vesicle budding to grow asexually?Is CopII starts
from the er then goes out the cell and CopI works for the molecules comes in the cell and goes to
nucleus?Does copII carries proteins away to make the budding?Please explain in details.
Solution
Yes, yeasts reproduce asexually by an asymmetric division process known as budding. Initially,
yeast forms a small protuberance which then grows to a full size and forms a bud. Then the
nucleus splits and migrates into the daughter cell. After the movement of nucleus, the entire bud
detaches from the parent cell by forming a constriction at the base. This process will repeat until
a chain of bud cells are formed..
DNA molecules consist of chemically linked sequences of the bases ad.pdf
DNA molecules consist of chemically linked sequences of the bases adenine, guanine, cytosine
and thymine, denoted A, G, C and T. A sequence of three bases is called a codon. A base may
appear more than once in a codon.
1) Represent all possible base sequences using a tree diagram.
2) How many different codons are there?
3) The bases A and G are purines, while C and T are pyrimidines. How many codons are there
whose first and third bases are purines and whose second base is a pyrimidine?
4) How many codons consist of three different bases?
Solution.
Data StructuresPLEASE USING THIS C++ PROGRAM BELOW, I NEED HEL.pdf
Data Structures
PLEASE USING THIS C++ PROGRAM BELOW, I NEED HELP ON IMPLEMENTING
BOTH QUEUE AND STACK TO PRINT \"I able\"
REVERSE/BACKWARDS(PALINDROME)
#include
#include
using namespace std;
struct st
{
int year;
st *next;
};
class queue
{
public:
int sz,i = 0;
char chr[100];
queue()
{
//constructor for initializing front and rear to NULL
front = NULL;
rear = NULL;
cout << \"enter size of queue: \";
cin >> sz;
}
// ~queue()
// {
//
// }
// in the enterqueue, create a new node, assign value and add it to queue
// if F = R = NULL the ( ) else ( )
void enterqueue(int yr, char el)
{
//inserting into queue if queue not overflow
if(i >= sz)
{
cout << \"queue over flow\";
return;
}
if (front == NULL )
{
front = new st;
front->next = NULL;
front->year = yr;
rear = front;
chr[i] = el;
i++;
}
else
{
st *temp = new st;
temp->year = yr;
rear->next = temp;
rear = temp;
chr[i] = el;
i++;
}
}
char deletequeue()
{
//delete front data from queue if data in the queue
char tr;
int j = 0;
st *temp;
temp = front;
if( front == NULL)
{
cout << \"queue under flow\";
return \'a\';
}
if(front == rear)
{
front = rear = NULL;
return \'a\';
}
front = front->next;
delete temp;
tr = chr[0];
i--;
for(j = 0; j < i; j++)
{
chr[j] = chr[j+1];
}
chr[j] = \'\\0\';
return tr;
}
bool isempty()
{
//it shows whether queue is empty
if( front == NULL)
{
return true;
}
else
return false;
}
bool isfull()
{
//it shows whether queue is full
if(i == sz)
return true;
else
return false;
}
void display()
{
//it displays queue data with characters of queue if invalid choice is entered
st *tmp;
int j = 0;
cout << \"\ queue is:\ character \\tyear\";
for(tmp = front; tmp != rear; tmp = tmp->next, j++)
{
cout << \"\ \" << chr[j] << \" \" << tmp->year;
}
cout << \"\ \"<year;
}
private:
st *front, *rear;
};
int main()
{
//main method to call queue functions by choice
queue s1;
int ch,yr;
char el;
bool s;
while(1)
{
cout << \"\ 1.ENTERQUEUE\ 2.DELETEQUEUE\ 3.ISEMPTY\ 4.ISFULL\ 5.EXIT\ enter
your choice: \";
cin >> ch;
switch(ch)
{
case 1:
if(!s1.isfull())
{
cout << \"\ enter a year : \";
cin >> yr;
cout << \"\ enter a character : \";
cin >> el;
s1.enterqueue(yr, el);
}
else
cout << \"insertion not possible\";
break;
case 2:
s = s1.isempty();
if(!s)
{
el = s1.deletequeue();
cout << \"\ The character of queue \" << el <<\" is deleted\" << endl;
}
else
cout << \"\ queue is empty.deletion not possible\"<< endl;
break;
case 3:
s = s1.isempty();
if(s)
cout << \"\ queue is empty\" << endl;
else
cout << \"\ queue is not empty\" << endl;
break;
case 4:
s=s1.isfull();
if(s)
cout << \"\ queue is full\" << endl;
else
cout << \"\ queue is not full\" << endl;
break;
case 5:
exit(0);
default : cout << \"invalid choice\" << endl;
if(!s1.isempty())
s1.display();
}
}
return (0);
}
Solution
#include
#include
using namespace std;
struct st
{
int year;
st *next;
};
class queue
{
public:
//int sz,i=0;//modified //code modified here// everything is fine with the code except this...
queue is implemented correctly
//you can .
Count Red Nodes. Write a program that computes the percentage of red.pdf
Count Red Nodes. Write a program that computes the percentage of red nodes in a given red-
black BST. Test your program by running at least 100 trials of the experiment of inserting N
random keys into an initially empty tree, for N = 10^4, 10^5, 10^6, and formulate an hypothesis.
*It is experiment from textbook Algorithms, 3.3.42
Does anyone know the answer?
Solution
/*
* Java Program to Implement Red Black Tree
*/
import java.util.Scanner;
/* Class Node */
class RedBlackNode
{
RedBlackNode left, right;
int element;
int color;
/* Constructor */
public RedBlackNode(int theElement)
{
this( theElement, null, null );
}
/* Constructor */
public RedBlackNode(int theElement, RedBlackNode lt, RedBlackNode rt)
{
left = lt;
right = rt;
element = theElement;
color = 1;
}
}
/* Class RBTree */
class RBTree
{
private RedBlackNode current;
private RedBlackNode parent;
private RedBlackNode grand;
private RedBlackNode great;
private RedBlackNode header;
private static RedBlackNode nullNode;
/* static initializer for nullNode */
static
{
nullNode = new RedBlackNode(0);
nullNode.left = nullNode;
nullNode.right = nullNode;
}
/* Black - 1 RED - 0 */
static final int BLACK = 1;
static final int RED = 0;
/* Constructor */
public RBTree(int negInf)
{
header = new RedBlackNode(negInf);
header.left = nullNode;
header.right = nullNode;
}
/* Function to check if tree is empty */
public boolean isEmpty()
{
return header.right == nullNode;
}
/* Make the tree logically empty */
public void makeEmpty()
{
header.right = nullNode;
}
/* Function to insert item */
public void insert(int item )
{
current = parent = grand = header;
nullNode.element = item;
while (current.element != item)
{
great = grand;
grand = parent;
parent = current;
current = item < current.element ? current.left : current.right;
// Check if two red children and fix if so
if (current.left.color == RED && current.right.color == RED)
handleReorient( item );
}
// Insertion fails if already present
if (current != nullNode)
return;
current = new RedBlackNode(item, nullNode, nullNode);
// Attach to parent
if (item < parent.element)
parent.left = current;
else
parent.right = current;
handleReorient( item );
}
private void handleReorient(int item)
{
// Do the color flip
current.color = RED;
current.left.color = BLACK;
current.right.color = BLACK;
if (parent.color == RED)
{
// Have to rotate
grand.color = RED;
if (item < grand.element != item < parent.element)
parent = rotate( item, grand ); // Start dbl rotate
current = rotate(item, great );
current.color = BLACK;
}
// Make root black
header.right.color = BLACK;
}
private RedBlackNode rotate(int item, RedBlackNode parent)
{
if(item < parent.element)
return parent.left = item < parent.left.element ? rotateWithLeftChild(parent.left) :
rotateWithRightChild(parent.left) ;
else
return parent.right = item < parent.right.element ? rotateWithLeftChild(parent.right) :
rotateWithRightChild(parent.right);
}
/* Rotate binary tree node with left child */
private RedBlackNode rot.
apple 2008 historically what were Apples major competitive advanta.pdf
apple 2008
historically what were Apples major competitive advantages and why did they fail to build on
these advantages to lead the industry?
Solution
Apples major competitive advantages:
Apple II and Macintosh were the machines that made Apple. They both had to following
features.
Brand loyalty among important segments
Reasons for why did Apple fail to build on these advantages to lead the industry:
Apple’s missteps:
Entry of Rivals:.
A partial results from a PERT analysis for a project with 50 activit.pdf
A partial results from a PERT analysis for a project with 50 activities shows that path A-K-Q
Solution
There are 50 activities shows the path A-K-Q.
With the help of PERT (Program Evaluation and Review Technique), and CPM (Critical Path
Method), the project manager can
1. Plan the project ahead of time and foresee possible sources of troubles and delays in
completion.
2. Schedule the project activities at the appropriate times to conform with proper job sequence so
that the project is completed as soon as possible.
3. Coordinate and control the projects activities so as to stay on schedule in completing the
project.
Thus PERT incorporates uncertainties in activity times in its analysis. It determines the
probabilities of completing various stages of the project by specified deadlines. It also calculates
the expected time to complete the project. An important and extremely useful by product of
PERT analysis is its identification of various “bottlenecks” in a project.
Application of PERT and CPM
A partial list of applications of PERT and CPM techniques is
1. Construction projects (e.g., buildings, highways, houses, and bridges)
2. Preparation of bids and proposals for large projects.
3. Maintenance planning of oil refineries, ship repairs, and other large operations.
4. Development of new weapons systems and new manufactured products.
5. Manufacture and assembly of large items such as airplanes, ships, and computers. 6. Simple
projects.
Six steps :
1. Define the projects and all of its significant activities or tasks.
2. Develop the relationships among the activities. Decide which activities must precede and
which must follow others.
3. Draw the network connecting all of the activities.
4. Assign time and/or cost estimates to each activity.
5. Compute the longest time path through the network; this is called the critical path.
6. Use the network to help plan, schedule, monitor and control the project..
A graph is symmetric with respect to the vertical line corresponding.pdf
A graph is symmetric with respect to the vertical line corresponding to t= pi/2 and r > 0. What
does this say about the algebraic symmetry of the function?
A. r(t)=r(pi-t)
B. r(t)=r(-t)
C. r(t)=r(pi/2-t)
D. r(t)=-r(t)
Solution
Symmetrical wrt to vertical line: t=pi/2 and r>0
which above the horizontal axisi.e. 0--- pi rad
So, it means r(t) = r(180 -t) = r(pi -t)
Option A.
X-Linked Recessive Write three rules to keep in mind when counseling .pdf
X-Linked Recessive Write three rules to keep in mind when counseling someone about the
likelihood of inheriting an X-linked recessive condition: About parents\' genotype. About
parents\' phenotype. About the probability of the offspring showing the trait. If instead of having
Hemophilia C (autosomal recessive), Lisa Keller\'s brother had Hemophilia A (X-linked
recessive), what would be the probability that she was a carrier? 25% 50% 67% 100% More
information is required to answer this question.
Solution
About genotype of parents- If the mother is a carrier i.e. carrying the recessive allele on one of
her X chromosome and the father do not have the recessive allele in his genotype, then none of
the daughters will suffer from the disease (since they will inherit only one recessive allele from
the mother) but sons will have a 50% probability of suffering from the disease as they will
inherit the recessive allele from the mother and the Y chromosome from the father, although
there is a 50% probability of inheriting the wild type allele from the mother.
Now if the mother is suffering from the disease i.e. has the recessive allele on both the X
chromosome while the father is healthy having no allele for the disease, the daughters will
always be a carrier but never suffer from the disease as they eill inherit the healthy wild type
allele from the father as one X chromosome comes from the father and the other X chromosome
from the mother. The sons will however all suffer from the disease as they inherit the Y
chromosome from the father and the X chromosome having the recessive allele causing the
disease will be inherited from the mother.
Now suppose the mother is a carrier i.e. carries the recessive allele only on one X chromosome
but the father suffers from the disease making his genotype XrY (Xr denoting recessive allele
carrying X chromosome), then, 50%of their son will duffer from the disease while 50% of the
daughters will also have a chance of suffering from the disease.
If both the parents do not have a single recessive X linked disease gene on any of their X
chromosome then none of the children will suffer from the disease.
About phenotype of parents- If the mother is a carrier i.e. have the diseased allele on only one X
Chromosome, then the phenotype of mother will be normal. But the father can never be a carrier
for a X linked recessive disease as males have only one X chromosome. If the diseased allele in
present on that X chromosome in males it will be expressed as there is no dominant healthy wild
type allele to dominate the diseased gene. Thus if the father has the recessive diseased allele on
X chromosome, the father\'s phenotype will be expresding the disease while if the father do not
have the allele he will be healthy. On the other hand for the mother to express the disease the
recessive allele has to be present on bith her X chromosome.
Probability of offsprings showing the trait- If mother is a carrier and father is normal, then 5.
Write a Java Class to Implement a Generic Linked ListYour list mus.pdf
Write a Java Class to Implement a Generic Linked List
Your list must be implemented as a singly-linked list of generic nodes, where each Node object
has two instance variables: an object of the “type variable” class, and a pointer to the next node
on the list.
Your class will contain separate methods to handle each of the operations read from a data file
(see II., below)
Your class will also override toString() to return the objects on the list in the order in which they
occur.
Write a Test Class for Your LinkedList Class
Your main method will read list operation instructions from a data file, until end-of-file and call
the appropriate LinkedList method to execute each one.
After each operation is executed, print out the operation and the updated list.
The data file to be used is on the class web site and the operations are:
APPEND X - Append object X to the end of the list
ADD N X - Insert object X as the new Nth element in the list, increasing the size of the list by 1
E.g. Suppose the list is:
head -> 1 -> 2 -> 3 -> 4->null
After ADD 3 7 it would be:
head -> 1 -> 2 -> 7 -> 3 -> 4->null
DELETE N – Remove the Nth object from the list
SWAP M N - Interchange the positions of the Mth and Nth
objects on the list
For credit, the two nodes must actually \"trade places\" in the list, and not merely swap their data
values
REVERSE - Reverse the order of the objects on the list
This must be done by reversing the order of the nodes themselves, rather than by swapping the
data stored
To get credit for your reverse() method, it must use either one of these 2 algorithms:
For each node on the list except the current “head”
node, delete the node and insert it as the new head
Use your swap() method
6. CLEAR – Clear the list (make it empty)
No credit will be given for programs that use any additional data structures – either from the Java
API or programmer defined -, other than your own LinkedList class
txt file:
Solution
//Java Program to Implement Singly Linked List
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
/* Class Node */
class Nodes
{
protected Object data;
protected Nodes next;
/* Constructor */
public Nodes()
{
next = null;
data = null;
}
/* Constructor */
public Nodes(Object d,Nodes n)
{
data = d;
next = n;
}
/* Function to set link to next Node */
public void setLink(Nodes n)
{
next = n;
}
/* Function to set data to current Node */
public void setData(Object d)
{
data = d;
}
/* Function to get link to next node */
public Nodes getLink()
{
return next;
}
/* Function to get data from current Node */
public Object getData()
{
return data;
}
}
/* Class linkedList */
class linkList
{
protected Nodes start;
protected Nodes end ;
public int size ;
/* Constructor */
public linkList()
{
start = null;
end = null;
size = 0;
}
/* Function to check if list is empty */
public boolean isEmpty()
{
return start == null;
}
/* Function to get size of list */
public int getSize()
{
return .
Write a class (BasketballTeam) encapsulating the concept of a tea.pdf
Write a class (BasketballTeam) encapsulating the concept of a team of basketball players. This
method should have a single instance variable which is an array of basketball player objects. In
addition to that class, you will need to design and a code a Playerclass to encapsulate the concept
of a basketball player, assuming a basketball player has the following attributes: a name, a
position, number of shots taken, and number of shots made.
In your Player class you should write the following methods:
A constructor taking a String for the name, a String for the position, an int for the number of
shots taken, and an int for the number of shots made.
Accessor, mutator, toString, and equals methods.
A method returning the individual shooting percentage. Avg = Shots Made / Shots Taken. Hint:
This method should return a double!!!
In your BasketballTeam class you should write the following methods:
A constructor taking an array of Player objects as its only parameter and assigning that array to
the array data member of the class, its only instance variable.
Accessor, mutator, toString, and equals methods.
A method checking if all positions are different, returning true if they are, false if they are not.
A method returning the shooting percentage of the team. Team Shooting Percentage = Total
Shots Made / Total Shots Taken Hint: This should return a double!!! Also, this is not the average
of the the averages.
A method checking that we have a center (that is, the name of the position) on the team. If we
do not have any return false; otherwise, it returns true;
A method returning void that sorts the array of Player objects in ascending order using the
number of shots taken as the sorting key.
A method returning the array of Player objects sorted in ascending alphabetical order by name.
Hint: Use compareTo method of the String class.
A method returning the name of the player with the most shots made.
A method returning the name of the player with the highest shooting average.
In your BasketballTeamClient class, when you test all your methods, you can hard-code five
basketball Player objects. This client should call all 12 methods including the constructor for full
credit.
When writing the Player class you should write a PlayerClient class that tests all its methods
before integrating into BasketballTeam class.
Solution
public class Player {
//Attributes
private String name;
private String position;
private int shotsTaken;
private int shotsMade;
/**
* Constructor
* @param name
* @param position
* @param shotsTaken
* @param shotsMade
*/
public Player(String name, String position, int shotsTaken, int shotsMade) {
this.name = name;
this.position = position;
this.shotsTaken = shotsTaken;
this.shotsMade = shotsMade;
}
/**
* @return the name
*/
public String getName() {
return name;
}
/**
* @param name the name to set
*/
public void setName(String name) {
this.name = name;
}
/**
* @return the position
*/
public String getPosition() {
return position;
}
/**
* @p.
Which of the following would you expect to happen in a plant that doe.pdf
Which of the following would you expect to happen in a plant that does not have enough water
(multiple answers are correct) Without water, the plant cannot make the sugar it needs to grow
the stem begins to dry out growth would be stunted because it is difficult to move nutrients
around Without water, the plant would become limp.
Solution
Answer: All the choices given are correct! If Plant that does not have enough water;
(a) Without water, plants cannot make the sugar it needs to grow.
Plants make their own food by the process of photosynthesis.The amount of water, sunlight and
carbon dioxide directly influences the process of food production for plant growth. Thus the
process of photosynthesis slows down and even stops when water is absent, which causes
internal food supplies to diminish.
(b) the stem begins to dry out
Scarcity of water leads to loss of turgor pressure. Stomata on the stem surface close as a result of
less water (for water conservation) and as a result stem dries out.
(c) the growth would be stunted because it is difficult to move nutrients around.
As water moves above the plant system it carries vital minerals and nutrients to different areas of
the plant. The stomas on the leaf and stem surface release waste products (oxygen) into the
environment and takes CO2 in. This process of transpiration maintains the turgor pressure and
also cools the plant in sunlight. In scarcity of water, transpiration process slows and, the plant
show stunded growth due to lack of nutrients supply.
(d) Without water the plant would become limp
Turgor pressure holds the plant cells and keeps them erect. This pressure is lost without water
and plants begin to collapse (wilting!) or give a limp appearance..
What is the value of including personal information about yourself a.pdf
What is the value of including personal information about yourself and your lifestyle when
researchers are looking at your DNA? How does study different than many other similar sets of
data gathered from GWAS and SNP Haplotypes?What do you think about the scope and purpose
of this type of genetics data study.
Solution
Answer to including personal information and lifestyle lies in phenotype differences. Phenotype
is influenced by genotype and environment. Hence although scientists and researchers look for
DNA or genetic material of an individual but also take a look on the environmental conditions or
life style which influences the phenotype of an organism.
GWAS (Genome wide association study) is study of genetic variants which are associated with a
trait like any major disease. By using SNPs (Short nuclear polymorphisms) short interspersed
regions can be identified to locate the traits of interest. The environmental factors like sunlight
can differ the response of an individual to it. One individual can be sensitiove to it but the other
could be diseased on exposure to sunlight, thus having a knowledge about individual\'s lifestyle
helps in tracking previous information.
The scope of GWAS has wider implications. Disease dignosis using genetics data helps in
developing drugs relating to it. It can be used to identify genes from different organisms.
Ancestral hierarchy can be established..
What is a voluntary response sampleSolutionVoluntary response.pdf
What is a voluntary response sample?
Solution
Voluntary response sampling is heavily biased because it focuses on volunteered survey answers
rather than a random sampling. What makes this biased is because the only people who volunteer
are those who have a particular interest in the topic of the survey. They only include people who
choose volunteer, whereas a random sample would need to include people whether or not they
choose to volunteer. Often, voluntary response samples oversample people who have strong
opinions and undersample people who don\'t care much about the topic of the survey. Thus
inferences from a voluntary response sample are not as trustworthy as conclusions based on a
random sample of the entire population under consideration..
What adaptive benefit do these abilities give wolbachia In oth.pdf
What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating
the host reproduction help its survival survival as parasite?
What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating
the host reproduction help its survival survival as parasite?
What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating
the host reproduction help its survival survival as parasite?
Solution
Manipulating the host reproduction help its survival as parasite.The improvement of complex
lifestyles cycles can growth parasite health. Some parasites have, via host manipulation, further
improved their reproductive success. Host manipulation in a few parasites, where they
effectively can change the behavior or morphology of their intermediate host in a way that
increases the parasite\'s trophic transmission
It is a method of parasite transmission wherein the parasite manipulates the behavior of its host
to growth transmission success to better trophic levels.
Think about autumn (Fall season) in places like New England (Pennsyl.pdf
Think about autumn (Fall season) in places like New England (Pennsylvania, New York,
Vermont, New Hampshire), where the leaves change color with the seasons. During the fall,
which pigments are the first to be broken down in the leaves? How does that explain the fall
colors we see? Why does the green color disappears during autumn (Fall season)? Please fully
and clearly explain why the green color disappeares in trees located in New England during the
autumn.
Solution
A number of factors play crucial role in changing the color of leaves in winter. Some are listed
below:
Similar is the reason with the New England trees..
The orange is which type of fruit Simple - Legume Simple - Drupe .pdf
The orange is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry Multiple
Aggregate The peas is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry
Multiple Aggregate The pistachio is which type of fruit? Simple - Legume Simple - Drupe
Simple - Berry Multiple Aggregate
Solution
Please find the answers below and reasons in parentheses:
Answer 22: Chocie 3 (Botanically, orange fruit is considered as a multiple aggregate of berry and
the juicy part or pulp is considered as succulent testa of the ovary. The peel of orange is
considered as ovarian wall.)
Answer 23: Choice 1 (A pea pod is considered as a simple legume by botanical nature. It can
open across both walls which eliminates the seeds inside the pod)
Answer 24: Choice2 (Pistachios belong to family of cashew or family Anacardiaceae
botanically. The fruits of this family are simple drupes in nature. These fruits contain the ovarian
wall as the hard covering of the fruit and the ovary itself modifies into edible seed).
The _________ is the protective chamber that houses the ovule and Lat.pdf
The _________ is the protective chamber that houses the ovule and Later matures to become the
fruit. A. Petal B. Ovary C. Carpel D. Stigma E. Sepals. What adaptation enhances bird\'s
flying ability? A. hollow bones B. sternum. C. air sacs D. homeothermy E. all of the above.
What plant characteristics are adaptations to land? A. internal fertilizations B. vascular tissues.
C. seeds with waterproof coats. D. waxy leaves. E. all of the above. Which of which is
characteristic of alternation of generation in land plants? A. haploid sporophytes make haploid
spores. B. gametophytes produce spores that develop into gametes C. sporophytes and
gametophytes are typically similar in appearance. D. meiosis in sporophytes produces haploid
spores. E. either the gametophyte or the sporophyte is unicellular. The haploid stage in a plant
life in a plant life cycle is called A. embryo B. zygote C. gametophyte D. sporophyte E.
conspicuous Animals in which the blastopo
Solution
31. ovary is the protective chamber that houses the ovule and later matures the fruit
32. tha hollow bones, air sacs , the sternum and homeothermy helps the birds to fly. the hollow
bones helps in reducing the body weight which helps in flying. the air sacs are responsible for
extracting more oxygen which hepls them to frly high. the sternum bones are present in the
wings which are strong and flexible for the movement of the wings
33. all of the above are the plants characteristics for the plant to adapt land
34.
35. the haploid stage of plant life in plant life cycle is known as gametophyte.
36.in protostomes the blastopore becomes the mouth.
Suppose X follows a normal distribution with mean=1 and variance=4. .pdf
Suppose X follows a normal distribution with mean=1 and variance=4. Let\'s randomly select 4
numbers from X. What is the distribution of the mean of these four numbers? What is the
probability that the mean is between -1 and 1?
Solution
The distribution of mean is also normal distribution with mean 1 and variance 1
Probability that the mean is between -1 and 1 is 0.47725.
Question is about the Enron Scandal in 2001... There is no case stud.pdf
Question is about the Enron Scandal in 2001... There is no case study... it is just based off of the
case in general. What impacts, if any, do emerging technologies have on the ethics of this case?
If not applicable to this case, discuss possible overall impacts on ethical situations in general.
Please provide sources
Solution
Enron was a company that thrived use of latest technologies and had a code of ethics that
prohibited managers and executives from being involved in another business entity that did
business with their own company. But, the codes of ethics that was voluntary and was set aside
by the top management and the board of directors. The legal structure permitted top management
to enter the arrangements, that constituted an interest conflict and while the fiduciary duty states
the managers and executives hato d to behave and act in the best interest of the shareholders and
the company. However there was discretion allowed to exercise their own business judgment
related what was in the best company\'s interest. A unethical and illegal activities path was
followed. Managerst was succumbing to greed and dishonesty by secretly exercising stock
options and constructed fake falsely financial report which enabled to hide billions of dollars of
debt. Using this approach Enron management created a noncompliance that existed different than
GAAP or SEC reporting standards and abandoned the basic accounting standards of integrity.
The main failure of the Enron accounting in terms of ethicas are as follows:
-- Poor Management Accountability: Weak management was virtually not accountable to anyone
as long as the company showed dramatic stock increases that were justified by the growth in the
earnings.
--Accounting Rules: The accounting activities became more complex and rule-oriented.
Accounting permitted practitioners to take obscure pronouncements, like dealing with Special
Purpose Entities that were primarily designed for transactions of leasing, and apply the
pronouncement to certain entities for which these accounting was never intended.
-- Weak Corporate Governance: Although the board seemed to be independent, but majority of
the board members had close ties to company managers through philanthropic organizations.
--Enthusiastic Financial Analyst Community: The analysts that were riding the bubble of the
financial dot-com economy made a conclusion that they did not have proper tools for the
valuation of the emerging companies.
--Biased Banking and Investment Banking: Several big financial institutions were willing
participants in the process as they were promoted with large underwriting fees for other Enron
work.
-- Lack of an Independent External Auditor: During the time of Enron, there existed mainly five
large external auditing firms who were referred as professional service firms with their business
in diverse line. All such firms had big consulting practices. The internal audit work for Enron
was performed by Arthur Andersen. Moreov.
Security Essay This is a 1 -2 page essay on an issue you found inte.pdf
Security Essay: This is a 1 -2 page essay on an issue you found interesting while reading
Cuckoos Egg. Examples include the continuing conflict between the FBI and Apple or Kevin
Mitnick or Stuxnet or how terrorists use the Internet or how the US government decision to
attack ISIS digitally. These are master’s level essays not college descriptive essays. You must
find and analyze interesting and challenging issue. Prior to submitting the paper, you must
submit a five source annotated bibliography with two paragraphs about each source. The first is a
summary of the article. The second is your reflection on the article – what it makes you think
about. Each paragraph contains a minimum of five sentences. Submit to me on blackboard. The
document should include the paper and the annotated bibliography
Solution
Carter and Barker (2010) describe listing as a twofold bookish discipline—the organized listing
of books (enumerative bibliography) and therefore the systematic, description of books as
physical objects (descriptive bibliography). These 2 distinct ideas and practices have separate
rationales and serve differing functions. Innovators and originators within the field embrace W.
W. Greg, Fredson Bowers, Duke of Edinburgh Elizabeth Gaskell, G. Thomas Tanselle.
Bowers (1949) refers to enumerative listing as a procedure that identifies books in “specific
collections or libraries,” in an exceedingly specific discipline, by Associate in Nursing author,
printer, or amount of production (3). He refers to descriptive listing because the systematic
description of a book as a fabric or physical object. Analytical listing, the cornerstone of
descriptive listing, investigates the printing and every one physical options of a book that yield
proof establishing a book\'s history and transmission (Feather 10). it\'s the preliminary part of
listing description and provides the vocabulary, principles and techniques of research that
descriptive bibliographers apply and on that they base their descriptive follow.
Descriptive bibliographers follow specific conventions and associated classification in their
description. Titles and title pages ar transcribed in an exceedingly quasi-facsimile vogue and
illustration. Illustration, typeface, binding, paper, and every one physical parts associated with
distinctive a book follow conventional conventions, as Bower\'s established in his foundational
piece, The Principles of listing Description. The thought expressed during this book expands
substantively on W. W. Greg\'s groundbreaking theory that argued for the adoption of formal
listing principles (Greg 29). basically, analytical listing is bothered with objective, physical
analysis and history of a book whereas descriptive listing employs all information that analytical
listing furnishes then codifies it with a read to distinctive the perfect copy or style of a book that
almost all nearly represents the printer’s initial conception and intention in printing.
In addition .
1. Which of the following brain structures are visible on the inferi.pdf
1. Which of the following brain structures are visible on the inferior surface of the brain. (select
all that apply)
A.brainstem
B.cerebellum
C.frontal lobe
D.parietal lobe
E.temporal lobe
2. Match the decription listed with the corresponding structure
A. corpus callosum
B. hypothalamus
C. mammillary bodies
D. pineal body (gland)
1. controls hormone secretion from the pituitary gland
2. fiber tract that connects the left cerebral hemisphere to the right cerebral hemisphere
3. involved in suckling reflex and chewing
4. primary relay center for sensory information coming into the brain
5. secretes the hormone melatonin from its precursor molecule serotonin
Solution
1.all of the above mentioned structures except the parietal lobe are visible on the inferior surface
of the brain. Parietal lobe will be visible only from the lateral view.
2. A-2
B- 1
C-4
D-5.
11. The cloning vectors used for human gene therapy are mainly based.pdf
11. The cloning vectors used for human gene therapy are mainly based on _________.
A. bacterial viruses (i.e. phages)
b.animal viruses
c.plasmids
d. BACs
e.YACs
Solution
Answers : The correct answer is B (Animal viruses)
Supplementary information :The introduction of DNA into the cells of live animals in order to
treat disease. Viral gene-delivery vectors are favoured for therapeutic applications because of
their efficiency. Commonly used viral vectors used for gene therapy includes, adenovirus,
herpes simplex and adeno-associated virus..
Present and explain all the elements in Freud theory.SolutionI.pdf
Present and explain all the elements in Freud theory.
Solution
In the year 1905, Freud developed a model of the mind, which he used to describe the features of
the mind’s structure and function. He stated that on the surface of the mind is consciousness,
which consists of those thoughts that are the focus of our attention now. The preconscious
consists of all which can be retrieved from memory. The third and most significant region is the
unconscious. Here lie the processes that are the real cause of most behaviour.
In 1923, Freud developed a structural model of mind. Mind has entities like id, ego and
superego. Freud assumed the id operated at an unconscious level according to the pleasure
principle. Eros, or life instinct, helps the individual to survive; it directs life-sustaining activities
such as respiration, eating. The ego develops from the id during infancy. The ego\'s goal is to
satisfy the demands of the id in a safe a socially acceptable way. The superego develops during
early childhood (when the child identifies with the same sex parent) and is responsible for
ensuring moral standards are followed.
Freud\'s theory is very good at explaining behavior. His theory of cognitive psychology has
identified unconscious processes, such as procedural memory and automatic processing..
please answes these multiple choices above on.pdf
please answes these multiple choices above one by one thanks
Solution
1.
E.Execution always begins in the Main() method for console applications.
2
D. System.Windows.Forms
3.
E. all of the above
4
E. framework
5.
E. handler
6
C. window
7.
C.Title
8
D. A Form inherits the members of the Form class.
9
C.Constructor
10
E. all of the above
11
A.Event
13.
B. textbox1 = new TextBox();
14
D. buttonl.Name = \"A button\";
15
C. this.Controls.AddRange(this.buttonlClickEvent);
16
C.ToUpper
17
E. none of the above
18
E. all of the above
19
E. collapsible segment of code that must be used for Windows applications
20.
D. Create separate classes for each..
Metabolism using an electron transport system is classified based on .pdf
Metabolism using an electron transport system is classified based on the nature of the initial
electron donors and terminal electron acceptors. Lithotrophy s defined as the use of A. organic
molecules as the initial electron donors. B. organic molecules as the final electron acceptors. C.
inorganic molecules as the initial electron donors. D. inorganic molecules as the final electron
acceptors. E. Two of the above are correct. Which of the following forms of energy production
does NOT involve the formation of a proton gradient to synthesize ATP? a. Fermentation b.
Oxidative respiration c. Phototrophy d. All forms at energy production require a proton motive
force. Which is false among A-C regarding glycolysis? A. It initially requires energy input in
the form of ATP hydrolysis B. It yields ATP and NADH C. It forms ATP via substrate level
phosphorylation D. none of A-C are false; all are true
Solution
Answer:
1). C. Inorganic molecules as the initial electron donrs
Lithotrophy is the use of an inorganic compound as a source of energy. Most lithotrophic
bacteria are aerobic respirers that produce energy in the same manner as all aerobic respiring
organisms: they remove electrons from a substrate and put them through an electron transport
system that will produce ATP by electron transport phosphorylation. Lithotrophs just happen to
get those electrons from an inorganic, rather than an organic compound.
Lithotrophic sulfur oxidizers include both Bacteria (e.g. Thiobacillus) and Archaea (e.g.
Sulfolobus). Sulfur oxidizers oxidize H2S (sulfide) or S (elemental sulfur) as a source of energy.
Similarly, the purple and green sulfur bacteria oxidize H2S or S as an electron donor for
photosynthesis, and use the electrons for CO2 fixation
2). D. All forms of energy production require a proton motive force
3). D. none of A-C are false; all are true.
National Insurance Company and Owen put their agency agreement into .pdf
National Insurance Company and Owen put their agency agreement into a written document that
describes the rights and duties of both parties. Owen, as the agent, has
apparent authority.
express authority.
implied authority.
equal authority.
Fred is legally incompetent. Fred can be
none of the other answers.
an agent only.
an agent or a principal.
a principal only.A.
apparent authority.B.
express authority.C.
implied authority.D.
equal authority.
Solution
Answer - B Express Authority.
There is a contract between two companies and Owen is an agent that means Owen is expressly
told by other company what to do and it will result in Express Authority.
Answer - B An Agent only.
As he is not legally Incompetent so he can\'t be principal so he must be Agent only..
The geography of Taylor Swift - some ideas
Geographical themes connected with Taylor Swift's ERAS tour - coming to the UK in June 2024
Home assignment II on Spectroscopy 2024 Answers.pdf
Answers to Home assignment on UV-Visible spectroscopy: Calculation of wavelength of UV-Visible absorption
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A graph is symmetric with respect to the vertical line corresponding.pdf
A graph is symmetric with respect to the vertical line corresponding to t= pi/2 and r > 0. What
does this say about the algebraic symmetry of the function?
A. r(t)=r(pi-t)
B. r(t)=r(-t)
C. r(t)=r(pi/2-t)
D. r(t)=-r(t)
Solution
Symmetrical wrt to vertical line: t=pi/2 and r>0
which above the horizontal axisi.e. 0--- pi rad
So, it means r(t) = r(180 -t) = r(pi -t)
Option A.
X-Linked Recessive Write three rules to keep in mind when counseling .pdf
X-Linked Recessive Write three rules to keep in mind when counseling someone about the
likelihood of inheriting an X-linked recessive condition: About parents\' genotype. About
parents\' phenotype. About the probability of the offspring showing the trait. If instead of having
Hemophilia C (autosomal recessive), Lisa Keller\'s brother had Hemophilia A (X-linked
recessive), what would be the probability that she was a carrier? 25% 50% 67% 100% More
information is required to answer this question.
Solution
About genotype of parents- If the mother is a carrier i.e. carrying the recessive allele on one of
her X chromosome and the father do not have the recessive allele in his genotype, then none of
the daughters will suffer from the disease (since they will inherit only one recessive allele from
the mother) but sons will have a 50% probability of suffering from the disease as they will
inherit the recessive allele from the mother and the Y chromosome from the father, although
there is a 50% probability of inheriting the wild type allele from the mother.
Now if the mother is suffering from the disease i.e. has the recessive allele on both the X
chromosome while the father is healthy having no allele for the disease, the daughters will
always be a carrier but never suffer from the disease as they eill inherit the healthy wild type
allele from the father as one X chromosome comes from the father and the other X chromosome
from the mother. The sons will however all suffer from the disease as they inherit the Y
chromosome from the father and the X chromosome having the recessive allele causing the
disease will be inherited from the mother.
Now suppose the mother is a carrier i.e. carries the recessive allele only on one X chromosome
but the father suffers from the disease making his genotype XrY (Xr denoting recessive allele
carrying X chromosome), then, 50%of their son will duffer from the disease while 50% of the
daughters will also have a chance of suffering from the disease.
If both the parents do not have a single recessive X linked disease gene on any of their X
chromosome then none of the children will suffer from the disease.
About phenotype of parents- If the mother is a carrier i.e. have the diseased allele on only one X
Chromosome, then the phenotype of mother will be normal. But the father can never be a carrier
for a X linked recessive disease as males have only one X chromosome. If the diseased allele in
present on that X chromosome in males it will be expressed as there is no dominant healthy wild
type allele to dominate the diseased gene. Thus if the father has the recessive diseased allele on
X chromosome, the father\'s phenotype will be expresding the disease while if the father do not
have the allele he will be healthy. On the other hand for the mother to express the disease the
recessive allele has to be present on bith her X chromosome.
Probability of offsprings showing the trait- If mother is a carrier and father is normal, then 5.
Write a Java Class to Implement a Generic Linked ListYour list mus.pdf
Write a Java Class to Implement a Generic Linked List
Your list must be implemented as a singly-linked list of generic nodes, where each Node object
has two instance variables: an object of the “type variable” class, and a pointer to the next node
on the list.
Your class will contain separate methods to handle each of the operations read from a data file
(see II., below)
Your class will also override toString() to return the objects on the list in the order in which they
occur.
Write a Test Class for Your LinkedList Class
Your main method will read list operation instructions from a data file, until end-of-file and call
the appropriate LinkedList method to execute each one.
After each operation is executed, print out the operation and the updated list.
The data file to be used is on the class web site and the operations are:
APPEND X - Append object X to the end of the list
ADD N X - Insert object X as the new Nth element in the list, increasing the size of the list by 1
E.g. Suppose the list is:
head -> 1 -> 2 -> 3 -> 4->null
After ADD 3 7 it would be:
head -> 1 -> 2 -> 7 -> 3 -> 4->null
DELETE N – Remove the Nth object from the list
SWAP M N - Interchange the positions of the Mth and Nth
objects on the list
For credit, the two nodes must actually \"trade places\" in the list, and not merely swap their data
values
REVERSE - Reverse the order of the objects on the list
This must be done by reversing the order of the nodes themselves, rather than by swapping the
data stored
To get credit for your reverse() method, it must use either one of these 2 algorithms:
For each node on the list except the current “head”
node, delete the node and insert it as the new head
Use your swap() method
6. CLEAR – Clear the list (make it empty)
No credit will be given for programs that use any additional data structures – either from the Java
API or programmer defined -, other than your own LinkedList class
txt file:
Solution
//Java Program to Implement Singly Linked List
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
/* Class Node */
class Nodes
{
protected Object data;
protected Nodes next;
/* Constructor */
public Nodes()
{
next = null;
data = null;
}
/* Constructor */
public Nodes(Object d,Nodes n)
{
data = d;
next = n;
}
/* Function to set link to next Node */
public void setLink(Nodes n)
{
next = n;
}
/* Function to set data to current Node */
public void setData(Object d)
{
data = d;
}
/* Function to get link to next node */
public Nodes getLink()
{
return next;
}
/* Function to get data from current Node */
public Object getData()
{
return data;
}
}
/* Class linkedList */
class linkList
{
protected Nodes start;
protected Nodes end ;
public int size ;
/* Constructor */
public linkList()
{
start = null;
end = null;
size = 0;
}
/* Function to check if list is empty */
public boolean isEmpty()
{
return start == null;
}
/* Function to get size of list */
public int getSize()
{
return .
Write a class (BasketballTeam) encapsulating the concept of a tea.pdf
Write a class (BasketballTeam) encapsulating the concept of a team of basketball players. This
method should have a single instance variable which is an array of basketball player objects. In
addition to that class, you will need to design and a code a Playerclass to encapsulate the concept
of a basketball player, assuming a basketball player has the following attributes: a name, a
position, number of shots taken, and number of shots made.
In your Player class you should write the following methods:
A constructor taking a String for the name, a String for the position, an int for the number of
shots taken, and an int for the number of shots made.
Accessor, mutator, toString, and equals methods.
A method returning the individual shooting percentage. Avg = Shots Made / Shots Taken. Hint:
This method should return a double!!!
In your BasketballTeam class you should write the following methods:
A constructor taking an array of Player objects as its only parameter and assigning that array to
the array data member of the class, its only instance variable.
Accessor, mutator, toString, and equals methods.
A method checking if all positions are different, returning true if they are, false if they are not.
A method returning the shooting percentage of the team. Team Shooting Percentage = Total
Shots Made / Total Shots Taken Hint: This should return a double!!! Also, this is not the average
of the the averages.
A method checking that we have a center (that is, the name of the position) on the team. If we
do not have any return false; otherwise, it returns true;
A method returning void that sorts the array of Player objects in ascending order using the
number of shots taken as the sorting key.
A method returning the array of Player objects sorted in ascending alphabetical order by name.
Hint: Use compareTo method of the String class.
A method returning the name of the player with the most shots made.
A method returning the name of the player with the highest shooting average.
In your BasketballTeamClient class, when you test all your methods, you can hard-code five
basketball Player objects. This client should call all 12 methods including the constructor for full
credit.
When writing the Player class you should write a PlayerClient class that tests all its methods
before integrating into BasketballTeam class.
Solution
public class Player {
//Attributes
private String name;
private String position;
private int shotsTaken;
private int shotsMade;
/**
* Constructor
* @param name
* @param position
* @param shotsTaken
* @param shotsMade
*/
public Player(String name, String position, int shotsTaken, int shotsMade) {
this.name = name;
this.position = position;
this.shotsTaken = shotsTaken;
this.shotsMade = shotsMade;
}
/**
* @return the name
*/
public String getName() {
return name;
}
/**
* @param name the name to set
*/
public void setName(String name) {
this.name = name;
}
/**
* @return the position
*/
public String getPosition() {
return position;
}
/**
* @p.
Which of the following would you expect to happen in a plant that doe.pdf
Which of the following would you expect to happen in a plant that does not have enough water
(multiple answers are correct) Without water, the plant cannot make the sugar it needs to grow
the stem begins to dry out growth would be stunted because it is difficult to move nutrients
around Without water, the plant would become limp.
Solution
Answer: All the choices given are correct! If Plant that does not have enough water;
(a) Without water, plants cannot make the sugar it needs to grow.
Plants make their own food by the process of photosynthesis.The amount of water, sunlight and
carbon dioxide directly influences the process of food production for plant growth. Thus the
process of photosynthesis slows down and even stops when water is absent, which causes
internal food supplies to diminish.
(b) the stem begins to dry out
Scarcity of water leads to loss of turgor pressure. Stomata on the stem surface close as a result of
less water (for water conservation) and as a result stem dries out.
(c) the growth would be stunted because it is difficult to move nutrients around.
As water moves above the plant system it carries vital minerals and nutrients to different areas of
the plant. The stomas on the leaf and stem surface release waste products (oxygen) into the
environment and takes CO2 in. This process of transpiration maintains the turgor pressure and
also cools the plant in sunlight. In scarcity of water, transpiration process slows and, the plant
show stunded growth due to lack of nutrients supply.
(d) Without water the plant would become limp
Turgor pressure holds the plant cells and keeps them erect. This pressure is lost without water
and plants begin to collapse (wilting!) or give a limp appearance..
What is the value of including personal information about yourself a.pdf
What is the value of including personal information about yourself and your lifestyle when
researchers are looking at your DNA? How does study different than many other similar sets of
data gathered from GWAS and SNP Haplotypes?What do you think about the scope and purpose
of this type of genetics data study.
Solution
Answer to including personal information and lifestyle lies in phenotype differences. Phenotype
is influenced by genotype and environment. Hence although scientists and researchers look for
DNA or genetic material of an individual but also take a look on the environmental conditions or
life style which influences the phenotype of an organism.
GWAS (Genome wide association study) is study of genetic variants which are associated with a
trait like any major disease. By using SNPs (Short nuclear polymorphisms) short interspersed
regions can be identified to locate the traits of interest. The environmental factors like sunlight
can differ the response of an individual to it. One individual can be sensitiove to it but the other
could be diseased on exposure to sunlight, thus having a knowledge about individual\'s lifestyle
helps in tracking previous information.
The scope of GWAS has wider implications. Disease dignosis using genetics data helps in
developing drugs relating to it. It can be used to identify genes from different organisms.
Ancestral hierarchy can be established..
What is a voluntary response sampleSolutionVoluntary response.pdf
What is a voluntary response sample?
Solution
Voluntary response sampling is heavily biased because it focuses on volunteered survey answers
rather than a random sampling. What makes this biased is because the only people who volunteer
are those who have a particular interest in the topic of the survey. They only include people who
choose volunteer, whereas a random sample would need to include people whether or not they
choose to volunteer. Often, voluntary response samples oversample people who have strong
opinions and undersample people who don\'t care much about the topic of the survey. Thus
inferences from a voluntary response sample are not as trustworthy as conclusions based on a
random sample of the entire population under consideration..
What adaptive benefit do these abilities give wolbachia In oth.pdf
What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating
the host reproduction help its survival survival as parasite?
What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating
the host reproduction help its survival survival as parasite?
What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating
the host reproduction help its survival survival as parasite?
Solution
Manipulating the host reproduction help its survival as parasite.The improvement of complex
lifestyles cycles can growth parasite health. Some parasites have, via host manipulation, further
improved their reproductive success. Host manipulation in a few parasites, where they
effectively can change the behavior or morphology of their intermediate host in a way that
increases the parasite\'s trophic transmission
It is a method of parasite transmission wherein the parasite manipulates the behavior of its host
to growth transmission success to better trophic levels.
Think about autumn (Fall season) in places like New England (Pennsyl.pdf
Think about autumn (Fall season) in places like New England (Pennsylvania, New York,
Vermont, New Hampshire), where the leaves change color with the seasons. During the fall,
which pigments are the first to be broken down in the leaves? How does that explain the fall
colors we see? Why does the green color disappears during autumn (Fall season)? Please fully
and clearly explain why the green color disappeares in trees located in New England during the
autumn.
Solution
A number of factors play crucial role in changing the color of leaves in winter. Some are listed
below:
Similar is the reason with the New England trees..
The orange is which type of fruit Simple - Legume Simple - Drupe .pdf
The orange is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry Multiple
Aggregate The peas is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry
Multiple Aggregate The pistachio is which type of fruit? Simple - Legume Simple - Drupe
Simple - Berry Multiple Aggregate
Solution
Please find the answers below and reasons in parentheses:
Answer 22: Chocie 3 (Botanically, orange fruit is considered as a multiple aggregate of berry and
the juicy part or pulp is considered as succulent testa of the ovary. The peel of orange is
considered as ovarian wall.)
Answer 23: Choice 1 (A pea pod is considered as a simple legume by botanical nature. It can
open across both walls which eliminates the seeds inside the pod)
Answer 24: Choice2 (Pistachios belong to family of cashew or family Anacardiaceae
botanically. The fruits of this family are simple drupes in nature. These fruits contain the ovarian
wall as the hard covering of the fruit and the ovary itself modifies into edible seed).
The _________ is the protective chamber that houses the ovule and Lat.pdf
The _________ is the protective chamber that houses the ovule and Later matures to become the
fruit. A. Petal B. Ovary C. Carpel D. Stigma E. Sepals. What adaptation enhances bird\'s
flying ability? A. hollow bones B. sternum. C. air sacs D. homeothermy E. all of the above.
What plant characteristics are adaptations to land? A. internal fertilizations B. vascular tissues.
C. seeds with waterproof coats. D. waxy leaves. E. all of the above. Which of which is
characteristic of alternation of generation in land plants? A. haploid sporophytes make haploid
spores. B. gametophytes produce spores that develop into gametes C. sporophytes and
gametophytes are typically similar in appearance. D. meiosis in sporophytes produces haploid
spores. E. either the gametophyte or the sporophyte is unicellular. The haploid stage in a plant
life in a plant life cycle is called A. embryo B. zygote C. gametophyte D. sporophyte E.
conspicuous Animals in which the blastopo
Solution
31. ovary is the protective chamber that houses the ovule and later matures the fruit
32. tha hollow bones, air sacs , the sternum and homeothermy helps the birds to fly. the hollow
bones helps in reducing the body weight which helps in flying. the air sacs are responsible for
extracting more oxygen which hepls them to frly high. the sternum bones are present in the
wings which are strong and flexible for the movement of the wings
33. all of the above are the plants characteristics for the plant to adapt land
34.
35. the haploid stage of plant life in plant life cycle is known as gametophyte.
36.in protostomes the blastopore becomes the mouth.
Suppose X follows a normal distribution with mean=1 and variance=4. .pdf
Suppose X follows a normal distribution with mean=1 and variance=4. Let\'s randomly select 4
numbers from X. What is the distribution of the mean of these four numbers? What is the
probability that the mean is between -1 and 1?
Solution
The distribution of mean is also normal distribution with mean 1 and variance 1
Probability that the mean is between -1 and 1 is 0.47725.
Question is about the Enron Scandal in 2001... There is no case stud.pdf
Question is about the Enron Scandal in 2001... There is no case study... it is just based off of the
case in general. What impacts, if any, do emerging technologies have on the ethics of this case?
If not applicable to this case, discuss possible overall impacts on ethical situations in general.
Please provide sources
Solution
Enron was a company that thrived use of latest technologies and had a code of ethics that
prohibited managers and executives from being involved in another business entity that did
business with their own company. But, the codes of ethics that was voluntary and was set aside
by the top management and the board of directors. The legal structure permitted top management
to enter the arrangements, that constituted an interest conflict and while the fiduciary duty states
the managers and executives hato d to behave and act in the best interest of the shareholders and
the company. However there was discretion allowed to exercise their own business judgment
related what was in the best company\'s interest. A unethical and illegal activities path was
followed. Managerst was succumbing to greed and dishonesty by secretly exercising stock
options and constructed fake falsely financial report which enabled to hide billions of dollars of
debt. Using this approach Enron management created a noncompliance that existed different than
GAAP or SEC reporting standards and abandoned the basic accounting standards of integrity.
The main failure of the Enron accounting in terms of ethicas are as follows:
-- Poor Management Accountability: Weak management was virtually not accountable to anyone
as long as the company showed dramatic stock increases that were justified by the growth in the
earnings.
--Accounting Rules: The accounting activities became more complex and rule-oriented.
Accounting permitted practitioners to take obscure pronouncements, like dealing with Special
Purpose Entities that were primarily designed for transactions of leasing, and apply the
pronouncement to certain entities for which these accounting was never intended.
-- Weak Corporate Governance: Although the board seemed to be independent, but majority of
the board members had close ties to company managers through philanthropic organizations.
--Enthusiastic Financial Analyst Community: The analysts that were riding the bubble of the
financial dot-com economy made a conclusion that they did not have proper tools for the
valuation of the emerging companies.
--Biased Banking and Investment Banking: Several big financial institutions were willing
participants in the process as they were promoted with large underwriting fees for other Enron
work.
-- Lack of an Independent External Auditor: During the time of Enron, there existed mainly five
large external auditing firms who were referred as professional service firms with their business
in diverse line. All such firms had big consulting practices. The internal audit work for Enron
was performed by Arthur Andersen. Moreov.
Security Essay This is a 1 -2 page essay on an issue you found inte.pdf
Security Essay: This is a 1 -2 page essay on an issue you found interesting while reading
Cuckoos Egg. Examples include the continuing conflict between the FBI and Apple or Kevin
Mitnick or Stuxnet or how terrorists use the Internet or how the US government decision to
attack ISIS digitally. These are master’s level essays not college descriptive essays. You must
find and analyze interesting and challenging issue. Prior to submitting the paper, you must
submit a five source annotated bibliography with two paragraphs about each source. The first is a
summary of the article. The second is your reflection on the article – what it makes you think
about. Each paragraph contains a minimum of five sentences. Submit to me on blackboard. The
document should include the paper and the annotated bibliography
Solution
Carter and Barker (2010) describe listing as a twofold bookish discipline—the organized listing
of books (enumerative bibliography) and therefore the systematic, description of books as
physical objects (descriptive bibliography). These 2 distinct ideas and practices have separate
rationales and serve differing functions. Innovators and originators within the field embrace W.
W. Greg, Fredson Bowers, Duke of Edinburgh Elizabeth Gaskell, G. Thomas Tanselle.
Bowers (1949) refers to enumerative listing as a procedure that identifies books in “specific
collections or libraries,” in an exceedingly specific discipline, by Associate in Nursing author,
printer, or amount of production (3). He refers to descriptive listing because the systematic
description of a book as a fabric or physical object. Analytical listing, the cornerstone of
descriptive listing, investigates the printing and every one physical options of a book that yield
proof establishing a book\'s history and transmission (Feather 10). it\'s the preliminary part of
listing description and provides the vocabulary, principles and techniques of research that
descriptive bibliographers apply and on that they base their descriptive follow.
Descriptive bibliographers follow specific conventions and associated classification in their
description. Titles and title pages ar transcribed in an exceedingly quasi-facsimile vogue and
illustration. Illustration, typeface, binding, paper, and every one physical parts associated with
distinctive a book follow conventional conventions, as Bower\'s established in his foundational
piece, The Principles of listing Description. The thought expressed during this book expands
substantively on W. W. Greg\'s groundbreaking theory that argued for the adoption of formal
listing principles (Greg 29). basically, analytical listing is bothered with objective, physical
analysis and history of a book whereas descriptive listing employs all information that analytical
listing furnishes then codifies it with a read to distinctive the perfect copy or style of a book that
almost all nearly represents the printer’s initial conception and intention in printing.
In addition .
1. Which of the following brain structures are visible on the inferi.pdf
1. Which of the following brain structures are visible on the inferior surface of the brain. (select
all that apply)
A.brainstem
B.cerebellum
C.frontal lobe
D.parietal lobe
E.temporal lobe
2. Match the decription listed with the corresponding structure
A. corpus callosum
B. hypothalamus
C. mammillary bodies
D. pineal body (gland)
1. controls hormone secretion from the pituitary gland
2. fiber tract that connects the left cerebral hemisphere to the right cerebral hemisphere
3. involved in suckling reflex and chewing
4. primary relay center for sensory information coming into the brain
5. secretes the hormone melatonin from its precursor molecule serotonin
Solution
1.all of the above mentioned structures except the parietal lobe are visible on the inferior surface
of the brain. Parietal lobe will be visible only from the lateral view.
2. A-2
B- 1
C-4
D-5.
11. The cloning vectors used for human gene therapy are mainly based.pdf
11. The cloning vectors used for human gene therapy are mainly based on _________.
A. bacterial viruses (i.e. phages)
b.animal viruses
c.plasmids
d. BACs
e.YACs
Solution
Answers : The correct answer is B (Animal viruses)
Supplementary information :The introduction of DNA into the cells of live animals in order to
treat disease. Viral gene-delivery vectors are favoured for therapeutic applications because of
their efficiency. Commonly used viral vectors used for gene therapy includes, adenovirus,
herpes simplex and adeno-associated virus..
Present and explain all the elements in Freud theory.SolutionI.pdf
Present and explain all the elements in Freud theory.
Solution
In the year 1905, Freud developed a model of the mind, which he used to describe the features of
the mind’s structure and function. He stated that on the surface of the mind is consciousness,
which consists of those thoughts that are the focus of our attention now. The preconscious
consists of all which can be retrieved from memory. The third and most significant region is the
unconscious. Here lie the processes that are the real cause of most behaviour.
In 1923, Freud developed a structural model of mind. Mind has entities like id, ego and
superego. Freud assumed the id operated at an unconscious level according to the pleasure
principle. Eros, or life instinct, helps the individual to survive; it directs life-sustaining activities
such as respiration, eating. The ego develops from the id during infancy. The ego\'s goal is to
satisfy the demands of the id in a safe a socially acceptable way. The superego develops during
early childhood (when the child identifies with the same sex parent) and is responsible for
ensuring moral standards are followed.
Freud\'s theory is very good at explaining behavior. His theory of cognitive psychology has
identified unconscious processes, such as procedural memory and automatic processing..
please answes these multiple choices above on.pdf
please answes these multiple choices above one by one thanks
Solution
1.
E.Execution always begins in the Main() method for console applications.
2
D. System.Windows.Forms
3.
E. all of the above
4
E. framework
5.
E. handler
6
C. window
7.
C.Title
8
D. A Form inherits the members of the Form class.
9
C.Constructor
10
E. all of the above
11
A.Event
13.
B. textbox1 = new TextBox();
14
D. buttonl.Name = \"A button\";
15
C. this.Controls.AddRange(this.buttonlClickEvent);
16
C.ToUpper
17
E. none of the above
18
E. all of the above
19
E. collapsible segment of code that must be used for Windows applications
20.
D. Create separate classes for each..
Metabolism using an electron transport system is classified based on .pdf
Metabolism using an electron transport system is classified based on the nature of the initial
electron donors and terminal electron acceptors. Lithotrophy s defined as the use of A. organic
molecules as the initial electron donors. B. organic molecules as the final electron acceptors. C.
inorganic molecules as the initial electron donors. D. inorganic molecules as the final electron
acceptors. E. Two of the above are correct. Which of the following forms of energy production
does NOT involve the formation of a proton gradient to synthesize ATP? a. Fermentation b.
Oxidative respiration c. Phototrophy d. All forms at energy production require a proton motive
force. Which is false among A-C regarding glycolysis? A. It initially requires energy input in
the form of ATP hydrolysis B. It yields ATP and NADH C. It forms ATP via substrate level
phosphorylation D. none of A-C are false; all are true
Solution
Answer:
1). C. Inorganic molecules as the initial electron donrs
Lithotrophy is the use of an inorganic compound as a source of energy. Most lithotrophic
bacteria are aerobic respirers that produce energy in the same manner as all aerobic respiring
organisms: they remove electrons from a substrate and put them through an electron transport
system that will produce ATP by electron transport phosphorylation. Lithotrophs just happen to
get those electrons from an inorganic, rather than an organic compound.
Lithotrophic sulfur oxidizers include both Bacteria (e.g. Thiobacillus) and Archaea (e.g.
Sulfolobus). Sulfur oxidizers oxidize H2S (sulfide) or S (elemental sulfur) as a source of energy.
Similarly, the purple and green sulfur bacteria oxidize H2S or S as an electron donor for
photosynthesis, and use the electrons for CO2 fixation
2). D. All forms of energy production require a proton motive force
3). D. none of A-C are false; all are true.
National Insurance Company and Owen put their agency agreement into .pdf
National Insurance Company and Owen put their agency agreement into a written document that
describes the rights and duties of both parties. Owen, as the agent, has
apparent authority.
express authority.
implied authority.
equal authority.
Fred is legally incompetent. Fred can be
none of the other answers.
an agent only.
an agent or a principal.
a principal only.A.
apparent authority.B.
express authority.C.
implied authority.D.
equal authority.
Solution
Answer - B Express Authority.
There is a contract between two companies and Owen is an agent that means Owen is expressly
told by other company what to do and it will result in Express Authority.
Answer - B An Agent only.
As he is not legally Incompetent so he can\'t be principal so he must be Agent only..
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http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
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