how can we use empricism to figure out if a fact is trueSolutio.pdf
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how can we use empricism to figure out if a fact is true? Solution Empiricism is nothing but Knowledge by observation. Nothing is accepted without touching it, Experiencing it or sensing the things directly. So, empiricism helps a person to accept that fact by working on it or experimenting on it. Empiricism demands logic to believe any particular fact to accept. By using these many ways one can believe the fact is true from Empiricism..
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in C++ , Design a linked list class named IntegerList to hold a seri.pdf
in C++ , Design a linked list class named IntegerList to hold a series of integers. The class must
have public member functions for appending , inserting, deleting and displaying nodes. The
append function must append a node to the end of the list. The insert function must insert node in
ascending order. The delete function must delete the node with the given value. The display
function should display the list from head to tail. The IntegerList class should also have a
destructor that destroys the list and release memory. The list node must be declared as a structure
using keyword struct. This structure is a private member of IntegerList. You should also store the
head of the list as a private member.
Write a test program to test the implementation your IntegerList class works properly.
Solution
Here is the code for the above scenario:
#include
using namespace std;
struct ListNode
{
float value;
ListNode *next;
};
ListNode *head;
class LinkedList
{
public:
int insertNode(float num);
int appendNode(float num);
void deleteNode(float num);
void destroyList();
void displayList();
LinkedList(void) {head = NULL;}
~LinkedList(void) {destroyList();}
};
int LinkedList::appendNode(float num)
{
ListNode *newNode, *nodePtr = head;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head == NULL)
{
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while(nodePtr->next != NULL)
nodePtr = nodePtr->next;
nodePtr->next = newNode;
}
return 0;
}
int LinkedList::insertNode(float num)
{
struct ListNode *newNode, *nodePtr = head, *prevNodePtr = NULL;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head==NULL) {
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while((nodePtr != NULL) && (nodePtr->value < num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(prevNodePtr==NULL)
newNode->next = head;
else
newNode->next = nodePtr; prevNodePtr->next = newNode;
}
return 0;
}
void LinkedList::deleteNode(float num)
{
struct ListNode *nodePtr = head, *prevNodePtr = NULL;
if(head==NULL) {
cout << \"The list was empty!\ \";
return;
}
if(head->value == num) {
head = nodePtr->next;
delete [] nodePtr;
}
else
{
while((nodePtr!= NULL)&&(nodePtr->value != num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(nodePtr==NULL)
cout << \"The value \" << num << \" is not in this list!\ \";
else {
prevNodePtr->next = nodePtr->next;
delete [] nodePtr;
}
}
}
void LinkedList::destroyList()
{
struct ListNode *nodePtr = head, *nextNode = nodePtr;
if(head==NULL) {
cout << \"The list is empty!\ \";
return;
}
while (nodePtr != NULL) {
nextNode = nodePtr->next;
delete [] nodePtr;
nodePtr = nextNode;
}
}
void LinkedList.
In addition to the fossil record, what other kinds of evidence reinfo.pdf
In addition to the fossil record, what other kinds of evidence reinforce the evolutionary of life?
You may want to use your textbook as a reference.
Solution
In addition to fossil record, application of Molecular Systematics is very much helpful for easy
identification of fossilized organism. Molecular systematics involves comparisons of nucleic
acids or other molecules to presume the degree of relatedness which can disclose phylogenetic
relationships which usually cannot be determined by non-molecular methods like comparative
anatomy. For example, molecular systematics helps us to uncover evolutionary relationships
between groups and allow us to reconstruct phylogenic trees among groups of present-day
organisms for which the fossil record is poor or lacking entirely. For example, the DNA that
encodes for ribosomal RNA (rRNA) changes comparatively slowly. Therefore, comparisons of
DNA sequences in these genes are valuable for investigating relationships between taxa that
diverged hundreds of millions of years ago. Other molecular methods useful for determining
phylogeny from fossil record include Nucleic acid sequencing, Genomic fingerprinting (RFLP)
and Amino acid sequencing..
How much 0.521 M NaOH will be needed to raise the pH of 0.161 L of 4.pdf
How much 0.521 M NaOH will be needed to raise the pH of 0.161 L of 4.51 M ascorbic acid
(H2C6H6O6) to a pH of 11.60?
Solution
pH of acidic buffer = pka2 + log(salt/acid)
No of mol of ascorbic acid = 0.161*4.51 = 0.726 mol
No of mol of salt = x mol
pka2 = 11.6, pH = 11.6
11.6 = 11.6 + log(x/(0.726-x))
x = 0.363
NO of mol of NaOH required = 2x = 2*0.363 = 0.726 mol
volume of NaOH must take = 0.726/0.521 = 1.4 L
answer: 1.4 L.
I am writing a program that places bolcks on a grid the the sape of .pdf
I am writing a program that places bolcks on a grid the the sape of different letters. It can also
move the blocks and checks if they overlap. I have most of the test cases figured out except for
four. Can someone help me out?
Thanks in advance.
output:
input:
output:
input:
output:
I CAN ONLY MODIFY ONE FILE SHAPE.CPP
#ifndef SHAPE_CPP
#define SHAPE_CPP
#include
#include
#include \"Shape.h\"
using namespace std;
// Constructor for O
O::O(int posx, int posy)
{
x = new int[1];
y = new int[1];
x[0] = posx;
y[0] = posy;
}
char O::name() const
{
return \'O\';
}
int O::size() const
{
return 1;
}
// Constructor for I
I::I(int posx, int posy)
{
x = new int[2];
y = new int[2];
x[0] = x[1] = posx;
y[0] = posy;
y[1] = posy+1;
}
char I::name() const
{
return \'I\';
}
int I::size() const
{
return 2;
}
// Constructor for L
L::L(int posx, int posy)
{
x = new int[3];
y = new int[3];
x[0] = x[2] = posx;
y[0] = y[1] = posy;
x[1] = posx+1;
y[2] = posy+1;
}
char L::name() const
{
return \'L\';
}
int L::size() const
{
return 3;
}
// Constructor for S
S::S(int posx, int posy)
{
x = new int[4];
y = new int[4];
x[0] = posx;
x[1] = posx+1;
x[2] = posx+1;
x[3] = posx+2;
y[0] = y[1] = posy;
y[2] = y[3] = posy+1;
}
char S::name() const
{
return \'S\';
}
int S::size() const
{
return 4;
}
/*
Constructor for X
the constructor initialises the cell co-ordinates
X is spread across 5 cells
*
* * *
*
The Numbering of cells is done from bottom-top and left to right
(x[0], y[0]) is the position of bottom most cell
(x[1], y[1]) is the position of left most cell of the 2nd row from bottom
(x[2], y[2]) is the position of the middle cell in the 2nd row from bottom
....
*/
X::X(int posx, int posy)
{
x = new int[5];
y = new int[5];
x[0] = x[2] = x[4] =posx;
x[1] = posx-1;
x[3] = posx+1;
y[0] = posy;
y[1] = y[2] = y[3] = posy+1;
y[4] = posy+2;
}
/*
This function returns the name of type of object i.e it is a X, or L, or U ...
*/
char X::name() const
{
return \'X\';
}
int X::size() const
{
return 5;
}
// Constructor for U
U::U(int posx, int posy)
{
x = new int[7];
y = new int[7];
x[0] = x[3] = x[5] = posx;
x[1] = posx+1;
x[2] = x[4] = x[6] = posx+2;
y[0] = y[1] = y[2] = posy;
y[3] = y[4] = posy+1;
y[5] = y[6] = posy+2;
}
char U::name() const
{
return \'U\';
}
int U::size() const
{
return 7;
}
void Shape::print() const
{
//get the size of the object i.e no of cells across which this is spread
int sz = size();
// get the type of object i.e O, L, X...
char n = name();
cout< > st;
int sz1 = t.size();
int sz2 = size();
for(int i = 0 ; i < sz1 ; i++)
{
st.insert(make_pair(t.x[i], t.y[i]));
}
for(int i = 0 ; i < sz2 ; i++)
{
if(st.find(make_pair(x[i], y[i]))!= st.end())
return true;
}
return false;
}
/*
This function creates a new object of type ch
*/
Shape * Shape::makeShape(char ch, int posx, int posy)
{
if(ch == \'O\')
return new O(posx, posy);
else if(ch == \'I\')
return new I(posx, posy);
else if(ch == \'L\')
return new L(posx, posy);
else if(ch == \'S\')
return new S(posx, posy);
else if.
How do teams bring value to an organizationHow are high performin.pdf
How do teams bring value to an organization?
How are high performing teams built?
When a team is working well together and the project is proceeding as planned what can happen
to derail even the high performing teams?
Solution
Teams are of important value to any organization as they determine not only the success of an
organization but also the level with which the organization grow.
1. Teams bring about exchange of ideas which benefit the organization in terms of innovation
as well as different ways of approaching difficulties.
2. Teams use the skill sets that come varied personalities to form a comprehensive part of the
work force.
3. Teams help an organization to develop good communication and commitment values a
factor that strengthens an organization both internally and externally.
4. Teams help organizations realize goals set through combination of effort.
High performing teams are build through choosing of a visionary, enthusiastic and committed
team leader. The leader then sets to choose a group of equally visionary, skilled and hard
working individuals who are united and willing to work towards a common goal a process that
entails forging of trust between the leader and the team. The team is then put through various
projects in which it is expected to complete with utmost success a process which defines and
makes the team better with team to a level of high performance.
1. Calamities such as sickness or death or a team member may derail the high performance.
2. Lack of motivation especially by the team leader may derail the high performance.
3. Laziness may be also stand as a factor especially when the teams assume that the set
objectives have been achieved
4. Conflict amongst the team members or the team leader and members may derail the teams\'
performance..
Gene RegulationFunction of transcription factor domains, i.e. DNA.pdf
Gene Regulation:
Function of transcription factor domains, i.e. DNA binding vs. activation domain and co-
activators. Explain how they affect gene regulation in DNA.
Solution
Transcription is the process of synthesis of RNA from DNA template a- a fist step in gene
expression.The enzyme RNA polymerase copies a particular segment of DNA into RNa and it
represents the primary level at which gene expression is regulated in prokaryotes and eukaryotic
systems .the mechanism is similar in both prokayotes and eukaryotes .
In eukaryotic systems distinct factors known as the transcription initiation factors are required
and have a promoter region upstream from the gene or an enhancer region either up or
downstream from the gene and with specific motifs that are recognized by various types of
transcription factors .
The transcription factors bind and attract other transcripton factors to create a complex thus
facilitating the binding of RNA polymerase ii and hence the process of transcription. These
transcription factors are classified into genreral transcription factors which represent the basic
transcription machinery and gene specific transcription factor.
The transcrption factors are specific proteins that are required for RNA polymerase II to initiate
transcrption and contain one or more DNA binding domains (DBD) which attach to specific
DNA sequence adjacent to the gene that they regulate.
DNA binding domains is an independently folded domain which contains one motif that can
recognize a specific DNA sequence or general affinity to DNA. There are several different type
of DNA binding domains like the Helix turn Helix domain, Zinc finger domain ,Dimerization
domain , Leucine zipper etc.
Helix turn Helix domain: is characteristic of DNA binding proteins, composed of two -helices
joined by a short strand of amino acids and is found commonly in several proteins that regulate
gene expression. helix-turn-helix proteins include the homeodomain proteins, which play critical
roles in the regulation of gene expression during embryonic development
Zinc finger domain :They contain repeats of cysteine and histidine residues that bind zinc ions
and fold into looped structures that bind DNA first identified in the polymerase III transcription
factor TFIIA but are also found in transcription factors that regulate polymerse II promoters.
,Dimerization domain :,These domains bring about dimerization of two polypeptide chains as
seen in helix turn helkix domians , leucine zipper domain.
The transcrptional activator domains or the TAD are the regions of transcription factors which in
conjunction with DNA binding domain can activate transcrption from promoter by contacting
transcriptional machinery either directly or through other protein coactivators.
The transcriptional repressor domain TRDS are the regions of the transcription factors which in
conjuation with the DNA binding domain can repress transcirption from a promoter by
contacting transcrptional machinery eit.
For each step of DNA replication, predict the outcome ifconcentra.pdf
For each step of DNA replication, predict the outcome if:
concentration of nitrogenous bases is low
primase is inhibited
DNA polymerase were inhibited
DNA ligase were inhibited
Helicase were inhibited
AND
Identify cellular process(es) that require DNA replication
THANKS
Solution
concentration of nitrogenous bases is low- DNA molecule is made up of nitrogenous bases
adenine, guanine, cytosine, and thymine. These nitrogenous bases form the template and the
sequence of DNA in replication. They specify the order of these bases in the complementary
template strand being formed.
primase is inhibited- Enzyme primase is of key importance in DNA replication because it is the
only enzyme that can initiate DNA synthesis strand without an initial PNA or CNA primer.
Hence primase is essential as it catalysis the synthesis of a primer complementary to ssDNA
template aiding in the initiation of DNA replication to take place.
DNA polymerase were inhibited- DNA polymerase is an essential enzyme involved in the repair
of any damage to DNA molecule and it also plays a vital role in DNA replication; it connects the
Okazaki fragments (complementary fragements to the lagging template strand ) by deleting RNA
primers and replacing it with DNA.
DNA ligase were inhibited- Ligase enzyme is useful in DNA repair (single-strand or double-
strand breaks of DNA) and in replication by joining the double DNA strands through
phosphodiester bond formation (3\'OH end and 5\'P end). DNA double helicular structure is thus
formed through ligation of both strands on replication.
Helicase were inhibited- Helicases help in the seperation of nucleic acid strands required during
the cellular processes like DNA replication, transcription, repair. This seperation is done by
Helicase enzyme by moving through a nucleic acid phosphodiester backbone that exists in DNA,
RNA, or DNA-RNA hybrid.
cellular process that require DNA replication- In mitosis DNA replication is essential (S stage of
interphase) which gets distributed to the resulting daughter cells at the end of cytokinesis. This
ensures two similar daughter cells having common genetic material..
Explain THREE unique characteristics of fungi.SolutionThree un.pdf
Explain THREE unique characteristics of fungi.
Solution
Three unique characteristics of fungi are-
1)Fungi are filamentous and not truly multicellular. They are composed of unique thread-like
filamentous structures called hyphae. In higher fungi, these hyphae anastomose to form a
network called mycelium.The cell wall of fungi is composed of chitin or cellulose. Fungi can
build mutual relationships with other organisms in order to get nutritional benefits from such
association. For example: The symbiotic relationship between funfi and algae is called lichen.
2) They need preformed food, optimum temperature, pH and moisture to grow properly. In order
to avoid dessication, they also require some amount of water to grow. They are resistant to
osmotic pressure and hence, can grow in relatively high salt and sugar concentrations.
3)Fungi are heterotrophic and they lack plastids.They collect nutrients from the environment,
from dead or living organisms. Fungi are capable of metabolising complex carbohydrates (like
lignin)..
Recommended
in C++ , Design a linked list class named IntegerList to hold a seri.pdf
in C++ , Design a linked list class named IntegerList to hold a series of integers. The class must
have public member functions for appending , inserting, deleting and displaying nodes. The
append function must append a node to the end of the list. The insert function must insert node in
ascending order. The delete function must delete the node with the given value. The display
function should display the list from head to tail. The IntegerList class should also have a
destructor that destroys the list and release memory. The list node must be declared as a structure
using keyword struct. This structure is a private member of IntegerList. You should also store the
head of the list as a private member.
Write a test program to test the implementation your IntegerList class works properly.
Solution
Here is the code for the above scenario:
#include
using namespace std;
struct ListNode
{
float value;
ListNode *next;
};
ListNode *head;
class LinkedList
{
public:
int insertNode(float num);
int appendNode(float num);
void deleteNode(float num);
void destroyList();
void displayList();
LinkedList(void) {head = NULL;}
~LinkedList(void) {destroyList();}
};
int LinkedList::appendNode(float num)
{
ListNode *newNode, *nodePtr = head;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head == NULL)
{
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while(nodePtr->next != NULL)
nodePtr = nodePtr->next;
nodePtr->next = newNode;
}
return 0;
}
int LinkedList::insertNode(float num)
{
struct ListNode *newNode, *nodePtr = head, *prevNodePtr = NULL;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head==NULL) {
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while((nodePtr != NULL) && (nodePtr->value < num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(prevNodePtr==NULL)
newNode->next = head;
else
newNode->next = nodePtr; prevNodePtr->next = newNode;
}
return 0;
}
void LinkedList::deleteNode(float num)
{
struct ListNode *nodePtr = head, *prevNodePtr = NULL;
if(head==NULL) {
cout << \"The list was empty!\ \";
return;
}
if(head->value == num) {
head = nodePtr->next;
delete [] nodePtr;
}
else
{
while((nodePtr!= NULL)&&(nodePtr->value != num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(nodePtr==NULL)
cout << \"The value \" << num << \" is not in this list!\ \";
else {
prevNodePtr->next = nodePtr->next;
delete [] nodePtr;
}
}
}
void LinkedList::destroyList()
{
struct ListNode *nodePtr = head, *nextNode = nodePtr;
if(head==NULL) {
cout << \"The list is empty!\ \";
return;
}
while (nodePtr != NULL) {
nextNode = nodePtr->next;
delete [] nodePtr;
nodePtr = nextNode;
}
}
void LinkedList.
In addition to the fossil record, what other kinds of evidence reinfo.pdf
In addition to the fossil record, what other kinds of evidence reinforce the evolutionary of life?
You may want to use your textbook as a reference.
Solution
In addition to fossil record, application of Molecular Systematics is very much helpful for easy
identification of fossilized organism. Molecular systematics involves comparisons of nucleic
acids or other molecules to presume the degree of relatedness which can disclose phylogenetic
relationships which usually cannot be determined by non-molecular methods like comparative
anatomy. For example, molecular systematics helps us to uncover evolutionary relationships
between groups and allow us to reconstruct phylogenic trees among groups of present-day
organisms for which the fossil record is poor or lacking entirely. For example, the DNA that
encodes for ribosomal RNA (rRNA) changes comparatively slowly. Therefore, comparisons of
DNA sequences in these genes are valuable for investigating relationships between taxa that
diverged hundreds of millions of years ago. Other molecular methods useful for determining
phylogeny from fossil record include Nucleic acid sequencing, Genomic fingerprinting (RFLP)
and Amino acid sequencing..
How much 0.521 M NaOH will be needed to raise the pH of 0.161 L of 4.pdf
How much 0.521 M NaOH will be needed to raise the pH of 0.161 L of 4.51 M ascorbic acid
(H2C6H6O6) to a pH of 11.60?
Solution
pH of acidic buffer = pka2 + log(salt/acid)
No of mol of ascorbic acid = 0.161*4.51 = 0.726 mol
No of mol of salt = x mol
pka2 = 11.6, pH = 11.6
11.6 = 11.6 + log(x/(0.726-x))
x = 0.363
NO of mol of NaOH required = 2x = 2*0.363 = 0.726 mol
volume of NaOH must take = 0.726/0.521 = 1.4 L
answer: 1.4 L.
I am writing a program that places bolcks on a grid the the sape of .pdf
I am writing a program that places bolcks on a grid the the sape of different letters. It can also
move the blocks and checks if they overlap. I have most of the test cases figured out except for
four. Can someone help me out?
Thanks in advance.
output:
input:
output:
input:
output:
I CAN ONLY MODIFY ONE FILE SHAPE.CPP
#ifndef SHAPE_CPP
#define SHAPE_CPP
#include
#include
#include \"Shape.h\"
using namespace std;
// Constructor for O
O::O(int posx, int posy)
{
x = new int[1];
y = new int[1];
x[0] = posx;
y[0] = posy;
}
char O::name() const
{
return \'O\';
}
int O::size() const
{
return 1;
}
// Constructor for I
I::I(int posx, int posy)
{
x = new int[2];
y = new int[2];
x[0] = x[1] = posx;
y[0] = posy;
y[1] = posy+1;
}
char I::name() const
{
return \'I\';
}
int I::size() const
{
return 2;
}
// Constructor for L
L::L(int posx, int posy)
{
x = new int[3];
y = new int[3];
x[0] = x[2] = posx;
y[0] = y[1] = posy;
x[1] = posx+1;
y[2] = posy+1;
}
char L::name() const
{
return \'L\';
}
int L::size() const
{
return 3;
}
// Constructor for S
S::S(int posx, int posy)
{
x = new int[4];
y = new int[4];
x[0] = posx;
x[1] = posx+1;
x[2] = posx+1;
x[3] = posx+2;
y[0] = y[1] = posy;
y[2] = y[3] = posy+1;
}
char S::name() const
{
return \'S\';
}
int S::size() const
{
return 4;
}
/*
Constructor for X
the constructor initialises the cell co-ordinates
X is spread across 5 cells
*
* * *
*
The Numbering of cells is done from bottom-top and left to right
(x[0], y[0]) is the position of bottom most cell
(x[1], y[1]) is the position of left most cell of the 2nd row from bottom
(x[2], y[2]) is the position of the middle cell in the 2nd row from bottom
....
*/
X::X(int posx, int posy)
{
x = new int[5];
y = new int[5];
x[0] = x[2] = x[4] =posx;
x[1] = posx-1;
x[3] = posx+1;
y[0] = posy;
y[1] = y[2] = y[3] = posy+1;
y[4] = posy+2;
}
/*
This function returns the name of type of object i.e it is a X, or L, or U ...
*/
char X::name() const
{
return \'X\';
}
int X::size() const
{
return 5;
}
// Constructor for U
U::U(int posx, int posy)
{
x = new int[7];
y = new int[7];
x[0] = x[3] = x[5] = posx;
x[1] = posx+1;
x[2] = x[4] = x[6] = posx+2;
y[0] = y[1] = y[2] = posy;
y[3] = y[4] = posy+1;
y[5] = y[6] = posy+2;
}
char U::name() const
{
return \'U\';
}
int U::size() const
{
return 7;
}
void Shape::print() const
{
//get the size of the object i.e no of cells across which this is spread
int sz = size();
// get the type of object i.e O, L, X...
char n = name();
cout< > st;
int sz1 = t.size();
int sz2 = size();
for(int i = 0 ; i < sz1 ; i++)
{
st.insert(make_pair(t.x[i], t.y[i]));
}
for(int i = 0 ; i < sz2 ; i++)
{
if(st.find(make_pair(x[i], y[i]))!= st.end())
return true;
}
return false;
}
/*
This function creates a new object of type ch
*/
Shape * Shape::makeShape(char ch, int posx, int posy)
{
if(ch == \'O\')
return new O(posx, posy);
else if(ch == \'I\')
return new I(posx, posy);
else if(ch == \'L\')
return new L(posx, posy);
else if(ch == \'S\')
return new S(posx, posy);
else if.
How do teams bring value to an organizationHow are high performin.pdf
How do teams bring value to an organization?
How are high performing teams built?
When a team is working well together and the project is proceeding as planned what can happen
to derail even the high performing teams?
Solution
Teams are of important value to any organization as they determine not only the success of an
organization but also the level with which the organization grow.
1. Teams bring about exchange of ideas which benefit the organization in terms of innovation
as well as different ways of approaching difficulties.
2. Teams use the skill sets that come varied personalities to form a comprehensive part of the
work force.
3. Teams help an organization to develop good communication and commitment values a
factor that strengthens an organization both internally and externally.
4. Teams help organizations realize goals set through combination of effort.
High performing teams are build through choosing of a visionary, enthusiastic and committed
team leader. The leader then sets to choose a group of equally visionary, skilled and hard
working individuals who are united and willing to work towards a common goal a process that
entails forging of trust between the leader and the team. The team is then put through various
projects in which it is expected to complete with utmost success a process which defines and
makes the team better with team to a level of high performance.
1. Calamities such as sickness or death or a team member may derail the high performance.
2. Lack of motivation especially by the team leader may derail the high performance.
3. Laziness may be also stand as a factor especially when the teams assume that the set
objectives have been achieved
4. Conflict amongst the team members or the team leader and members may derail the teams\'
performance..
Gene RegulationFunction of transcription factor domains, i.e. DNA.pdf
Gene Regulation:
Function of transcription factor domains, i.e. DNA binding vs. activation domain and co-
activators. Explain how they affect gene regulation in DNA.
Solution
Transcription is the process of synthesis of RNA from DNA template a- a fist step in gene
expression.The enzyme RNA polymerase copies a particular segment of DNA into RNa and it
represents the primary level at which gene expression is regulated in prokaryotes and eukaryotic
systems .the mechanism is similar in both prokayotes and eukaryotes .
In eukaryotic systems distinct factors known as the transcription initiation factors are required
and have a promoter region upstream from the gene or an enhancer region either up or
downstream from the gene and with specific motifs that are recognized by various types of
transcription factors .
The transcription factors bind and attract other transcripton factors to create a complex thus
facilitating the binding of RNA polymerase ii and hence the process of transcription. These
transcription factors are classified into genreral transcription factors which represent the basic
transcription machinery and gene specific transcription factor.
The transcrption factors are specific proteins that are required for RNA polymerase II to initiate
transcrption and contain one or more DNA binding domains (DBD) which attach to specific
DNA sequence adjacent to the gene that they regulate.
DNA binding domains is an independently folded domain which contains one motif that can
recognize a specific DNA sequence or general affinity to DNA. There are several different type
of DNA binding domains like the Helix turn Helix domain, Zinc finger domain ,Dimerization
domain , Leucine zipper etc.
Helix turn Helix domain: is characteristic of DNA binding proteins, composed of two -helices
joined by a short strand of amino acids and is found commonly in several proteins that regulate
gene expression. helix-turn-helix proteins include the homeodomain proteins, which play critical
roles in the regulation of gene expression during embryonic development
Zinc finger domain :They contain repeats of cysteine and histidine residues that bind zinc ions
and fold into looped structures that bind DNA first identified in the polymerase III transcription
factor TFIIA but are also found in transcription factors that regulate polymerse II promoters.
,Dimerization domain :,These domains bring about dimerization of two polypeptide chains as
seen in helix turn helkix domians , leucine zipper domain.
The transcrptional activator domains or the TAD are the regions of transcription factors which in
conjunction with DNA binding domain can activate transcrption from promoter by contacting
transcriptional machinery either directly or through other protein coactivators.
The transcriptional repressor domain TRDS are the regions of the transcription factors which in
conjuation with the DNA binding domain can repress transcirption from a promoter by
contacting transcrptional machinery eit.
For each step of DNA replication, predict the outcome ifconcentra.pdf
For each step of DNA replication, predict the outcome if:
concentration of nitrogenous bases is low
primase is inhibited
DNA polymerase were inhibited
DNA ligase were inhibited
Helicase were inhibited
AND
Identify cellular process(es) that require DNA replication
THANKS
Solution
concentration of nitrogenous bases is low- DNA molecule is made up of nitrogenous bases
adenine, guanine, cytosine, and thymine. These nitrogenous bases form the template and the
sequence of DNA in replication. They specify the order of these bases in the complementary
template strand being formed.
primase is inhibited- Enzyme primase is of key importance in DNA replication because it is the
only enzyme that can initiate DNA synthesis strand without an initial PNA or CNA primer.
Hence primase is essential as it catalysis the synthesis of a primer complementary to ssDNA
template aiding in the initiation of DNA replication to take place.
DNA polymerase were inhibited- DNA polymerase is an essential enzyme involved in the repair
of any damage to DNA molecule and it also plays a vital role in DNA replication; it connects the
Okazaki fragments (complementary fragements to the lagging template strand ) by deleting RNA
primers and replacing it with DNA.
DNA ligase were inhibited- Ligase enzyme is useful in DNA repair (single-strand or double-
strand breaks of DNA) and in replication by joining the double DNA strands through
phosphodiester bond formation (3\'OH end and 5\'P end). DNA double helicular structure is thus
formed through ligation of both strands on replication.
Helicase were inhibited- Helicases help in the seperation of nucleic acid strands required during
the cellular processes like DNA replication, transcription, repair. This seperation is done by
Helicase enzyme by moving through a nucleic acid phosphodiester backbone that exists in DNA,
RNA, or DNA-RNA hybrid.
cellular process that require DNA replication- In mitosis DNA replication is essential (S stage of
interphase) which gets distributed to the resulting daughter cells at the end of cytokinesis. This
ensures two similar daughter cells having common genetic material..
Explain THREE unique characteristics of fungi.SolutionThree un.pdf
Explain THREE unique characteristics of fungi.
Solution
Three unique characteristics of fungi are-
1)Fungi are filamentous and not truly multicellular. They are composed of unique thread-like
filamentous structures called hyphae. In higher fungi, these hyphae anastomose to form a
network called mycelium.The cell wall of fungi is composed of chitin or cellulose. Fungi can
build mutual relationships with other organisms in order to get nutritional benefits from such
association. For example: The symbiotic relationship between funfi and algae is called lichen.
2) They need preformed food, optimum temperature, pH and moisture to grow properly. In order
to avoid dessication, they also require some amount of water to grow. They are resistant to
osmotic pressure and hence, can grow in relatively high salt and sugar concentrations.
3)Fungi are heterotrophic and they lack plastids.They collect nutrients from the environment,
from dead or living organisms. Fungi are capable of metabolising complex carbohydrates (like
lignin)..
Explain the basic relationship between humans and microorganisms.pdf
Explain the basic relationship between humans and microorganisms
Solution
Answer=
The basic relationship between humans & microorganisms are as follows
The surfaces of human body inside & outside ,for example skin ,mouth & intestines are covered
in millions of individual microbes
The microbes that do not harm in fact they helps to protect us becoming infected with harmful
microbes they are known as normal body flora
There are 3 types of symbiotic relationships between humans & bacteria
1] Commensalism = Is a relationship that is beneficial to the bacteria ,but does not help or
harmthe host
most are reside on epithelial surfaces that come in contact with the external environment
commonly found on skin ,respiratory tract & gastrointestinal tract
They aquire nutrients & place to live & grow from their host
some are pathogenic & cause disease may provide benefit to host
2] Mutuliastic relationship-=
Both bacteria & host are beneficial
for example there are several kinds of bacteria that live on skin & inside the mouth ,nose,throat
& intestines of humans
These bacteria received a place to live & feed while keeping other harmful microbes from taking
up residence
bacteria in digestive system assist in nutrient metabolism ,vitamin production & waste
processing
3] Parasitic relatioship= is one in which the bacteria benefit while the host is harmed
pathogenic parasites ,which cause disease,do so by resisting the hosts defences & growing at the
expense of host
These bacteria produces poisonous substances called endotoxins & exotoxins are responsible for
the symptoms to occure with an illness
Disease causing bacteria are responsible for a number of diseases including meningitis
,pneumonia .tuberculosis& several types of food borne diseases.
discuss the rationale for the global harmonization of financial repo.pdf
discuss the rationale for the global harmonization of financial reporting and disclosure
requirements
Solution
The focus and rationale of global harmonization of financial reporting and disclosure
requirements is for achieving a level of compatibility for the statements. There is less
complication for the companies and individuals to present the financial statements and its
interpretations. The differences between financial reporting standards are erased with
international harmonization of financial reporting.
When there is achievement of international harmonization in finial reporting then it benefits
countries and improves the access effectively for the financial market. The confidence and
investor\'s K owned he is increased. According to Wiley (2000) even though the rationale behind
harmonization is smooth preparation and presentation of financial statements; there may be
differences based on the factors like taxation and political aspects.
It focuses on creation of similar set of procedures with the help of boundary establishment and
how they can differ globally.
It also benefits accounting standards on national and international levels.
The disclosure requirements include the information related to nature and the extent of risks
which arises from financial instruments. The importance of financial instutments for the purpose
of finial position and performance of entity. It includes financial assets, held to maturity
investments, loans and receivables, sale assets etc..
Describe the need to multitask in BBC (behavior-based control) syste.pdf
Describe the need to multitask in BBC (behavior-based control) systems?
Solution
Behavior-based control employs a set of distributed, in-teracting modules, called behaviors that
collectively achieve the desired system-level behavior. To an ex-ternal observer, behaviors are
patterns of the robot’s activity emerging from interactions between the robot and its
environment. To a programmer, behaviors are control modules that cluster sets of constraints in
order to achieve and maintain a goal. Each behavior receives inputs from sensors and/or other
behaviors in the system, and provides outputs to the robot’s actuators or to other behaviors. Thus,
a behavior-based controller is a structured network of interacting behaviors, with no centralized
world representation or focus of control. In-stead, individual behaviors and networks of
behaviors maintain any state information and models.
The basic principles of behavior-based control can be summarized briefly as follows:
• Behaviors are implemented as control laws (some-times similar to those used in control
theory), either in software or hardware, as a processing element or as a procedure.
• Each behavior can take inputs from the robot’s sen-sors (e.g., proximity sensors, range
detectors, contact sensors, camera) and/or from other modules in the
system, and send outputs to the robot’s effectors (e.g., wheels, grippers, arm, speech) and/or to
other modules.
• Many different behaviors may independently re- ceive input from the same sensors and output
action commands to the same actuators.
• Behaviors are encoded to be relatively simple, and are added to the system incrementally.
• Behaviors (or subsets thereof) are executed con- currently, not sequentially, in order to exploit
parallelism and speed of computation, as well as the interaction dynamics among behaviors and
between behaviors and the environment.
The ability to improve performance over time and to reason about the world, in the context of a
chang-ing and dynamic environment, is an important area of research in situated robotics. Unlike
in classical ma-chine learning, where the goal is typically to optimize performance over a long
period of time, in situated learning the aim is to adapt relatively quickly, toward
attaining efficiency in the light of uncertainty. Models from biology are often considered, given
its proper- ties of learning directly from environmental feedback. Variations and adaptations of
machine learning, and in particular reinforcement learning, have been effectively applied to
behavior-based robots, which have demon- strated learning to walk [38.
, navigate and create topological maps, di-vide tasks, behave socially , and even identify
opponents and score goals in robot soc-
cer. Methods from artificial life, evolutionary computation/genetic algorithms, fuzzy logic,
vision and learning, multi-agent systems, and many other research areas continue to be actively
explored and applied to behavior-based robots as their role in ani.
A spinner with 7 equally sized slices is shown below. The dial is spu.pdf
A spinner with 7 equally sized slices is shown below. The dial is spun and stops on a slice at
random. What are the odds in favor of landing on a black slice?
Solution
4 slice grey
1 black slice
2 white slice
odds are ratio of favorable outcomes to non favorable outcomes
Odds in favour of landing on a black slice = 1:6.
ble below to calculate the Gross Domestic Product uning the ral Gro.pdf
ble below to calculate the Gross Domestic Product uning the ral Gross Domestic Product? Fully
explain your anower. (4 mak) or rea overnment expenditures on goods and service Wages,
salaries and supplementary labour income 2,000 reciation nvestment 400 sumption expenditure
et exports direct taxes 150
Solution
Gross domestic product be expenditure method= consumption expenditure+ government
expenditure+ investment+ net export
=2200+500+400+(-50)
=3050
This is the nominal GDP as well as real GDP because base year price and current year price is
not given.
Hence real GDP and nominal GDP will be same..
1.The limbic systemA. matures before the prefrontal cortexB. is.pdf
1.The limbic system:
A. matures before the prefrontal cortex
B. is another name for the prefrontal cortex
C. is responsible for emotional regulation and impulse control
D. matures before the amygdala
Solution
The limbic system is responsible for emotional regulationand impules control.
Limbic system is that portion of the brain which deals with the emotions, arousal, learning aw
well as formation of memories. It has a complicated structure that lies on the both side of the
thalamus just beneath the cerebrum. It includes amygdala, hippocampus, thalamus,
hypothalamus, basal ganglia and cingulate gyrus.
Amygdala also known as the emotional centre of the brain. Hippocampus helps in the formation
of new memories from our past experinces. While thalamus and hypothalamus are governs the
changes occuring to the emotional activities. Basal ganglia involved in motor control and habbit
learning. And cingulate gyrus manage smell and sight and regulate our behaviour..
Why is leadership a critical factor in implementing change and qua.pdf
Why is leadership a critical factor in implementing change and quality improvement initiatives?
Why is leadership a critical factor in implementing change and quality improvement initiatives?
Solution
Leadership is required for inspiring and motivating followers to work towards common
objectives. The leader employs enthusiasm and charisma to motivate the employees to think out
of the box.Leader lays emphasis on the values, morals, ideas and needs of the organization and
the followers.
Leadership is critical for transformation and change in the firm in a turbulent and challenging
environment. By concentrating on innovation and creation rather than just daily planning and
execution, the leader ensures that the changes are implemented with minimum resistance.
Below is how the leaders can implement change and quality improvement initiatives while
overcoming resistance-
1. Communicating the objective of the change and how it would positively affect the
productivity of the employees and the organization
2. Developing training and mentoring programs to train the employees to adopt to the new
process or system
3. Removing fears from the minds of employees through openhouse discussions etc.
4. Explaining the employees how the change would help in the development of the individuals..
Why are there bubbles in beer and champagneSolution Bubbles of .pdf
Why are there bubbles in beer and champagne?
Solution
Bubbles of gas, typically carbon dioxide are produced from beer and champagne , rising to the
surface. Because . yeasts ferment sugars and form carbon dioxide gas during fermentation. In the
bottle, where the pressure is high, the carbon dioxide disappears. However, when you open the
bottle and lower the pressure, the carbon dioxide escapes in the form of bubbles. In Champagne
and other sparkling wines, the bubbles pop, while in beer, the bubbles remain and form the
beer\'shead..
Which of the following is true of a traditional master budget a. It.pdf
Which of the following is true of a traditional master budget? a. It uses the zero-base budgeting
approach. b. It is department oriented. c. It recognizes interdependencies. d. It is dynamic in
nature.
Solution
Correct Answer is b. it is department Oriented
One of the critisism of Traditional Master Budget is that it is Department oriented and does not
recognise independencies among departments, further it is static not dynamic in nature..
Which ethnic groups were a part of the former YugoslaviaCroats, S.pdf
Which ethnic groups were a part of the former Yugoslavia?
Croats, Serbs, Bosnian Muslims, Macedonians
Janjaweed
Israelis, Palestinians, Arabs
Hutu & Tutsis
Croats, Serbs, Bosnian Muslims, Macedonians
Janjaweed
Israelis, Palestinians, Arabs
Hutu & Tutsis
Solution
The ethnic group can be defined as a social category based on common culture or nationality.
In other words, it can be said that an ethnic group is a category of people who recognize with
each other based on similarities for example as a common ancestry, language, society, culture or
nation.
First Yugoslavia contains people of the Kingdom of Serbs, Croats and Slovenes (1918–29).
Hence it can be said that the ethnic groups which were a part of the former Yugoslavia were
Croats, Serbs, Bosnian Muslims, Macedonians.
Hence option first is the correct answer.
Option first; Croats, Serbs, Bosnian Muslims, Macedonians.
What would the components of the development be in an early learning.pdf
What would the components of the development be in an early learning(1st – 3rd) website
environment be?
Solution
For a website,major components include Front end,images,contents navigation,page structure.
For early learning website.As it for the children,content along with effective graphic design is
most preferable.
Graphic design:
Many of the elements described, such as the logo, the navigation menus, the layout, images, etc.,
fall under the general category of graphic design. But graphic design is more than these. It is the
look and feel the website will have as a result of proper use and integration of all these elements.
Content management is considered.
The website is not recommended to be a static as it should be more creative and changes is
necessary.
Content management system:
This is the ability to update your website without having to directly edit the html. Simple
systems create areas on a web page that can be easily changed on a regular basis.
Include multiedia and databases.
Multi-media.
It is for than just text and images. Photo-tours, video-clips, sound-clips all can add to the
experience if they are well matched to the type of site and profile of the target audience..
What is true of all fungi They area. eukaryotic, heterotrophic pl.pdf
What is true of all fungi? They are
a. eukaryotic, heterotrophic plants.
b. eukaryotic, parasitic plants.
c. saprobic/decomposing plants.
d. eukaryotic heterotrophs with cell walls.
e. saprobic heterotrophs.
Solution
Answer: a. eukaryotic, heterotrophic plants.
Fungi are non-chlorophyllous(absence of chloroplasts) plants. They are eukaryotes due to the
presence of true nucleus. Fungi live as heterotrophic, parasitic, saprophytic, etc..
What is the possible reason that secondaries arise from parents that.pdf
What is the possible reason that secondaries arise from parents that have unpaired chromosomes
but not from parents that are normal diploids? Map Genetics: A Conceptual Approach 6th
Edition Pierce MHE/Freeman presented by Sapling Learning Progeny of triploid tomato plants
often contain parts of an extra chromosome, in addition to the normal complement of 24
chromosomes. Mutants with a part of an extra chromosome are referred to as secondaries. James
and Margaret Lesley observed that secondaries arise from triploid (3n), trisomic (3n 1), and
double trisomic (3n 1 1) parents, but never from diploids (2n). (J. W. Lesley and M. M. Lesley.
1929. Genetics 14:321-336) What is a possible reason that secondaries arise from parents that
have unpaired chromosomes but not from parents that are normal diploids? Triploid and trisomic
conditions in tomato plants are unnatural, and this genomic imbalance leads to higher levels of
mitotic recombination. The trivalent chromosomes would still be able to associate through
chiasma prior to segregation during meiosis. Due to some unknown mechanism, triploid and
trisom c tomato plants O occasionally over-replicate part of a chromosome, which leads to the
generation of mutant tomato plants. Triploid and trisomic plants are vulnerable to infection by
the tomato mosaic O virus, so they acquire extra chromosomal material by the integration of
viral DNA, whereas diploid plants are not infected by the virus.
Solution
Ans: B
2n+1 condition is called trisomic in which an organism contains two complete genome + 1 extra
chromosome. During the process of meiosis 1 chromosomal page separate so that 1 members of
each state is distributed to one daughter nucleus and other members goes to other daughter
nucleus but rarely one pair of chromosomes fail to separate and moves to the pole giving half of
the daughter cells extra chromosome and remaining half lost one chromosome. As a result n - 1
and n+ 1 chromosomes appear. When the gamut with n+ 1 chromosome fuses with normal
gamut the resulting zygote will have 2n + 1 chromosomes producing trisomics organism.
It was found that trisomies or white study nature and studied in plants like tomatoes tobacco
wheat datura and so on..
What is a trigonometric identity How and why do we use trigonometri.pdf
What is a trigonometric identity? How and why do we use trigonometric identities in
mathematics? How can you verify an identity? Give some examples.
Solution
Trigonometric identities are equalities that involve trigonometric functions and are true for every
single value of the occurring variables where both sides of the equality are defined
Sometimes it is possible to write some complex trignmetric functions in really simple form fr
example instead f writing Sinx/Cosx we can write that as Tanx so as we can see here it is easy to
identify Tanx rather than first finding Sinx and then finding Cosx and then divide them to get our
answer
Here is an example below:
we need to prove that left side is equal to right side
cot x/csc x=cos x
So let us simplify left side,
(cos x/sin x)/(1/sin x)
applying keep change flip
(cos x/sin x)*sin x
now we can cancel out sin x
so we are left with cos x
which is equal to right side of the equation
hence proved..
This is for $7.70 per share... not $7.20Thank you.. The Holtz .pdf
This is for $7.70 per share... not $7.20
Thank you.. The Holtz Corporation acquired 80 percent of the 100,000 outstanding voting shares
of Devine, Inc., for $7.70 per share on January 1, 2014. The remaining 20 percent of Devine\'s
shares also traded actively at $7.70 per share before and after Holtz\'s acquisition. An appraisal
made on that date determined that all book values appropriately reflected the fair values of
Devine\'s underlying accounts except that a building with a 5-year life was undervalued by
remaining life had a $75,000 fair value. At the acquisition date, Devine reported common stock
of $100,000 and a retained earnings balance of $290,000 $81,000 and a fully amortized
trademark with an estimated 10-year Following are the separate financial statements for the year
ending December 31, 2015 HoltzDevine Corporation Inc. $ (760,000) $(365,250) Sales Cost of
goods sold Operating expenses Dividend income 234,000 133,000 319,000 107,250 (16,000) Net
income $ (223,000) $(125,000) Retained eamings, 1/1/15 Net income (above) Dividends
declared $ (709,000) $(360,000) (223,000) (125,000) 90,00020,000 Retained earnings, 12/31/15
S (842,000) $(465,000) Current assets Investment in Devine, Inc Buildings and equipment (net)
Trademarks s 166,000 $ 199,000 616,000 905,000389,000 160,000226,000 Total assets
$1,847,000 $ 814,000 s (685,000 $249.000) Liabilities Common stock Retained eamings,
12/31/15 (above) (320,000) (100,000) (842,000) (465,000) Total liabilities and equities
$(1,847,000) $(814,000) At year-end, there were no intra-entity receivables or payables
Solution
SOLUTION:
a) Consolidation Worksheet:
HOLTZ CORPORATION AND DEVINE INC
Consolidated Worksheet for the Year Ending December 31, 2015
Consolidated Entries
Holtz Corporation
Devine Inc.
Debit
Credit
NCI
Consolidated
Sales
-760000
-365250
-1125250
Cost of Goods Sold
234000
133000
367000
Operating Expense
319000
107250
23700
449950
Dividend Income
-16000
16000
0
Net Income
-223000
-125000
Consolidated Net Income
-308300
NCI in Consolidated Net Income
-20260
20260
Holtz\'s Interest in Consolidated Income
-288040
Retained Earnings on 1/1
-709000
-360000
360000
37040
-746040
Net Income (Consolidated Above)
-223000
-125000
-288040
Dividend Declared
90000
20000
16000
4000
70000
Retained earnings on 12/31
-842000
-465000
-964080
Current Assets
166000
199000
365000
Investment in Devine
616000
37040
653040
0
Building and Equipment
905000
389000
64800
16200
1342600
Trademarks
160000
226000
67500
7500
446000
Goodwill
0
0
224000
224000
Total Assets
1847000
814000
2377600
Liabilities
-685000
-249000
-934000
Common Stock
-320000
-100000
100000
-320000
Retained earnings (Consolidated Above)
-842000
-465000
-964080
NCI in Devine on 1/1
78000
71260
-149260
NCI in Devine on 12/31
-165520
-165520
Total Liabilities and Equities
-1847000
-814000
-2383600
Working Notes:
1. Calculation of Goodwill or Gain on Bargain purchase
Total
Holtz
NCI
Fair Value of Consideration
770000
616000
154000
Less: Book Val.
This project is the first projects you will be working on this quart.pdf
This project is the first projects you will be working on this quarter. There are a number of
objectives to this assignment. First, to start gaining experience in developing programs in C++.
Second, in case you haven’t already you are going to work in pairs to get the sense of
collaborative development of software. Third, because you are allowed to use any references you
find online, this assignment will help you get a sense for just how many sources for
programming in C++ are available via the web. Project specification: In this project you are
going to implement a word search engine using C++. The search engine would help users to
count number of occurrence of a word among given set of text files. Input specification: The
program takes as an input a set of files, containing of multiple words. Words can appear multiple
times in a single file or in different files. The input files should be stored in a dedicated directory
(e.g. /cs/class/cs24/project1/input) and loaded in memory upon program execution. Program
functionality: Upon execution, the program should load all the words from the input files in the
memory. The program should run until the user explicitly specifies “exit”. The following
functionality should be supported: 1. Count how many times a user specified word appears in the
files. 3. Display the name of input files and corresponding number of occurrences of the
specified word. Upon execution of you program, you should specify as input parameter the path
to a directory containing the input files. Below you can find the format specification:
$./wordsearch Examples for execution format: $./wordsearch /cs/class/cs24/project1/input Enter
word: cat If executed as specified above, the program should return the name of files that contain
word “cat” and the number of occurrences of the word in each file Implementation requirements:
First, you need to implement a class for an abstract data type, in which you are going to store
files in memory. This step is very specific, depending on the functionality of the program you are
developing. For the current implementation you are required to use linked lists. A basic layout of
the data structure you would need to implement is illustrated on the figure below: You need to
declare a class Word that would store a word and a pointer to the beginning of a linked list and
another class File that would store a file-name and number of times that the word occurred in this
file. The process of “loading files in the memory” consists of (i) creating an object of type Word
for each new word that occurs in the set of input files, (ii) appending this object to a linked list
(e.g. the green linked list from the picture), (iii) creating an object File for each occurrence of the
word in a file and (iv) updating the corresponding (blue) linked list with the object File. Once
you have the files loaded in such structure, the searching would be as easy as finding the queried
word in the “green” linked list and tracing.
The total energy for the hydrogen molecule is -1.1789737 Hartrees. w.pdf
The total energy for the hydrogen molecule is -1.1789737 Hartrees. what is the binding energy (
in KJ/mol ) of the molecule ?
Solution
Eb =Et-E0
whr Et = total energy of H2 molecule which is -32.08ev
and E0 = ground state energy of hydrogen molecuel which is -1.17ev
hence Eb= -32.08+1.17
= -30.31ev
= -4.84 *10^-21 kJ
as 1ev = 1.6* 106-22kj
and 1 hartree = 27.2107 ev.
The Queue is a special memory structure with accesses limited to PUS.pdf
The Queue is a special memory structure with accesses limited to PUSH and POP operations
which result in a firstinfirstout (FIFO) access. The Stack is a related memory structure which
also has two operations PUSH and POP, but unlike the Queue, the POP operation results in the
data value entered using the most recent PUSH operation.
This results in a lastinfirstout (LIFO) access. Implementing a queue treats the memory as a
circle. How would you modify this design to implement a Stack? How are the Stack Empty and
Stack conditions checked?
Solution
we can Implement stack with using 2 Queues.
In this the first method we ensure that the new entry of element is always at top of queue1 , so
that the pop operation just dequeue from queue 1. We here are using the other queue \"queue2\"
to make sure that the new element is always at the top of queue1.
Below is the operation of the implemented stack:
PUSH :
step 1 : Enqueue element to Queue2.
step 2 : Now gradually dequeueing each element from Queue1 and enqueue to Queue2.
step 3 : Interchanging the name of the Queues with each other. So that we can dont allow one
more movement of all elements from Queue2 Queue1.
POP :
step 1 : Dequeue a element from Queue1 and return it.
STACK_EMPTY :
step 1 : we can check that by cheking the state of Queue1 , if it is empty than the stack is
empty..
The liver has a portal vein as well as a hepatic vein. It also has u.pdf
The liver has a portal vein as well as a hepatic vein. It also has unique exchange blood vessels
similar to capillaries, called “sinusoids.” How do these unique structures determine the function
of the organ?
Solution
Hepatic veins:
Hepatic veins are arisen in the liver lobule\'s central vein. Valves are absent in Hepatic veins.
These vessels can transports deoxygenated blood of liver and blood which has been filtered by
the liver (this is blood from the pancreas, colon, small intestine, and stomach) to the inferior vena
cava.
Hepatic portal vein:
The hepatic portal vein is one of the most important vein that originates from a networks in the
stomach, pancreas, intestines and spleen. this vessel can carry blood from these organs through
the hepatic portal vein to the liver. In the liver this blood enters into a capillary-like hepatic
sinusoids called as hepatic portal system. Later this blood passes through the portal veins of the
liver, blood is carried through a series of merging vessels into the hepatic veins. These empty
into the inferior vena cava.
Sinusoids:
These are capillary like vessels. In the digestive system, they\'re found near the liver and blood is
conveyed via the sinusoids to the inferior vena cava by the hepatic veins.
Sinusoid pathways, or sinusoidal pathways:
This pathway describes the hepatic portal system. From the organs (stomach, intestines,
pancreas, and spleen) and blood moves through the hepatic portal vein to the capillary like
hepatic sinusoids, which together are called the hepatic portal system. From the sinusoid
pathways, blood is transported by the hepatic veins back to the inferior vena cava..
The effects of the Vietnam War and the outcomes of the warThe ef.pdf
The effects of the Vietnam War and the outcomes of the war
The effects of the Vietnam War and the outcomes of the war
Solution
The Effect of the Vietnam War on the American People Essay. The Vietnam War was marked by
brutality, death, protests, and psychological tolls. No war caused such great division among the
American people like the Vietnam War. The war was extremely costly, and it left long-term
effects on people all over the world.
The Vietnam War was the first major war American’s had suffered defeat. The Vietnam war
was a war of confusion, competition and biasness. The outcome of the war was far greater than
an upset American nation, but a severe breakdown of the Vietnamese culture, economy,
environment and government. It also had a tremendous impact on American society even up to
present day. It was unclear from the beginning of the war if the American’s should even be
involved. It was a war between Northern and Southern Vietnam but the U.S saw it as an indirect
way to challenge the USSR’s sphere of influence in Southern Asia and to prevent the domino
effect and the further spread of communism. The Vietnam War completely changed the way the
United State
The War not only affected those in Vietnam, but those still at home in America. Americans had
split into two groups, known as the hawks and the doves. The hawks were people who supported
the war and wanted the country to be victorious in its fight with Communism. The doves were
the people that disagreed with the war. These were traditionally students, and people of the
younger generation. Many veterans when they would return home, would join in these mass
protests. One popular protest, which took place at Kent State University, resulted in 9 wounded
people with 4 dead. The police at Kent State took action against a mass protest by opening fire.
This is the evil that is created through war. Without the Vietnam war, there is no protest at Kent
State. Without the Vietnam war, the college kid’s lives are spared. Without the Vietnam War,
thousands of lives could have been saved. Neil Young said it best in his song Ohio. Ohio
discussed the brutality of war but mainly the police brutality against the kids at Kent State.
Vietnam was a tragic stain in American history. Vietnam was an entirely new type of war for the
United States. It still remains morally and historically problematic in today’s society.
Embracing GenAI - A Strategic Imperative
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
More Related Content
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Explain the basic relationship between humans and microorganisms.pdf
Explain the basic relationship between humans and microorganisms
Solution
Answer=
The basic relationship between humans & microorganisms are as follows
The surfaces of human body inside & outside ,for example skin ,mouth & intestines are covered
in millions of individual microbes
The microbes that do not harm in fact they helps to protect us becoming infected with harmful
microbes they are known as normal body flora
There are 3 types of symbiotic relationships between humans & bacteria
1] Commensalism = Is a relationship that is beneficial to the bacteria ,but does not help or
harmthe host
most are reside on epithelial surfaces that come in contact with the external environment
commonly found on skin ,respiratory tract & gastrointestinal tract
They aquire nutrients & place to live & grow from their host
some are pathogenic & cause disease may provide benefit to host
2] Mutuliastic relationship-=
Both bacteria & host are beneficial
for example there are several kinds of bacteria that live on skin & inside the mouth ,nose,throat
& intestines of humans
These bacteria received a place to live & feed while keeping other harmful microbes from taking
up residence
bacteria in digestive system assist in nutrient metabolism ,vitamin production & waste
processing
3] Parasitic relatioship= is one in which the bacteria benefit while the host is harmed
pathogenic parasites ,which cause disease,do so by resisting the hosts defences & growing at the
expense of host
These bacteria produces poisonous substances called endotoxins & exotoxins are responsible for
the symptoms to occure with an illness
Disease causing bacteria are responsible for a number of diseases including meningitis
,pneumonia .tuberculosis& several types of food borne diseases.
discuss the rationale for the global harmonization of financial repo.pdf
discuss the rationale for the global harmonization of financial reporting and disclosure
requirements
Solution
The focus and rationale of global harmonization of financial reporting and disclosure
requirements is for achieving a level of compatibility for the statements. There is less
complication for the companies and individuals to present the financial statements and its
interpretations. The differences between financial reporting standards are erased with
international harmonization of financial reporting.
When there is achievement of international harmonization in finial reporting then it benefits
countries and improves the access effectively for the financial market. The confidence and
investor\'s K owned he is increased. According to Wiley (2000) even though the rationale behind
harmonization is smooth preparation and presentation of financial statements; there may be
differences based on the factors like taxation and political aspects.
It focuses on creation of similar set of procedures with the help of boundary establishment and
how they can differ globally.
It also benefits accounting standards on national and international levels.
The disclosure requirements include the information related to nature and the extent of risks
which arises from financial instruments. The importance of financial instutments for the purpose
of finial position and performance of entity. It includes financial assets, held to maturity
investments, loans and receivables, sale assets etc..
Describe the need to multitask in BBC (behavior-based control) syste.pdf
Describe the need to multitask in BBC (behavior-based control) systems?
Solution
Behavior-based control employs a set of distributed, in-teracting modules, called behaviors that
collectively achieve the desired system-level behavior. To an ex-ternal observer, behaviors are
patterns of the robot’s activity emerging from interactions between the robot and its
environment. To a programmer, behaviors are control modules that cluster sets of constraints in
order to achieve and maintain a goal. Each behavior receives inputs from sensors and/or other
behaviors in the system, and provides outputs to the robot’s actuators or to other behaviors. Thus,
a behavior-based controller is a structured network of interacting behaviors, with no centralized
world representation or focus of control. In-stead, individual behaviors and networks of
behaviors maintain any state information and models.
The basic principles of behavior-based control can be summarized briefly as follows:
• Behaviors are implemented as control laws (some-times similar to those used in control
theory), either in software or hardware, as a processing element or as a procedure.
• Each behavior can take inputs from the robot’s sen-sors (e.g., proximity sensors, range
detectors, contact sensors, camera) and/or from other modules in the
system, and send outputs to the robot’s effectors (e.g., wheels, grippers, arm, speech) and/or to
other modules.
• Many different behaviors may independently re- ceive input from the same sensors and output
action commands to the same actuators.
• Behaviors are encoded to be relatively simple, and are added to the system incrementally.
• Behaviors (or subsets thereof) are executed con- currently, not sequentially, in order to exploit
parallelism and speed of computation, as well as the interaction dynamics among behaviors and
between behaviors and the environment.
The ability to improve performance over time and to reason about the world, in the context of a
chang-ing and dynamic environment, is an important area of research in situated robotics. Unlike
in classical ma-chine learning, where the goal is typically to optimize performance over a long
period of time, in situated learning the aim is to adapt relatively quickly, toward
attaining efficiency in the light of uncertainty. Models from biology are often considered, given
its proper- ties of learning directly from environmental feedback. Variations and adaptations of
machine learning, and in particular reinforcement learning, have been effectively applied to
behavior-based robots, which have demon- strated learning to walk [38.
, navigate and create topological maps, di-vide tasks, behave socially , and even identify
opponents and score goals in robot soc-
cer. Methods from artificial life, evolutionary computation/genetic algorithms, fuzzy logic,
vision and learning, multi-agent systems, and many other research areas continue to be actively
explored and applied to behavior-based robots as their role in ani.
A spinner with 7 equally sized slices is shown below. The dial is spu.pdf
A spinner with 7 equally sized slices is shown below. The dial is spun and stops on a slice at
random. What are the odds in favor of landing on a black slice?
Solution
4 slice grey
1 black slice
2 white slice
odds are ratio of favorable outcomes to non favorable outcomes
Odds in favour of landing on a black slice = 1:6.
ble below to calculate the Gross Domestic Product uning the ral Gro.pdf
ble below to calculate the Gross Domestic Product uning the ral Gross Domestic Product? Fully
explain your anower. (4 mak) or rea overnment expenditures on goods and service Wages,
salaries and supplementary labour income 2,000 reciation nvestment 400 sumption expenditure
et exports direct taxes 150
Solution
Gross domestic product be expenditure method= consumption expenditure+ government
expenditure+ investment+ net export
=2200+500+400+(-50)
=3050
This is the nominal GDP as well as real GDP because base year price and current year price is
not given.
Hence real GDP and nominal GDP will be same..
1.The limbic systemA. matures before the prefrontal cortexB. is.pdf
1.The limbic system:
A. matures before the prefrontal cortex
B. is another name for the prefrontal cortex
C. is responsible for emotional regulation and impulse control
D. matures before the amygdala
Solution
The limbic system is responsible for emotional regulationand impules control.
Limbic system is that portion of the brain which deals with the emotions, arousal, learning aw
well as formation of memories. It has a complicated structure that lies on the both side of the
thalamus just beneath the cerebrum. It includes amygdala, hippocampus, thalamus,
hypothalamus, basal ganglia and cingulate gyrus.
Amygdala also known as the emotional centre of the brain. Hippocampus helps in the formation
of new memories from our past experinces. While thalamus and hypothalamus are governs the
changes occuring to the emotional activities. Basal ganglia involved in motor control and habbit
learning. And cingulate gyrus manage smell and sight and regulate our behaviour..
Why is leadership a critical factor in implementing change and qua.pdf
Why is leadership a critical factor in implementing change and quality improvement initiatives?
Why is leadership a critical factor in implementing change and quality improvement initiatives?
Solution
Leadership is required for inspiring and motivating followers to work towards common
objectives. The leader employs enthusiasm and charisma to motivate the employees to think out
of the box.Leader lays emphasis on the values, morals, ideas and needs of the organization and
the followers.
Leadership is critical for transformation and change in the firm in a turbulent and challenging
environment. By concentrating on innovation and creation rather than just daily planning and
execution, the leader ensures that the changes are implemented with minimum resistance.
Below is how the leaders can implement change and quality improvement initiatives while
overcoming resistance-
1. Communicating the objective of the change and how it would positively affect the
productivity of the employees and the organization
2. Developing training and mentoring programs to train the employees to adopt to the new
process or system
3. Removing fears from the minds of employees through openhouse discussions etc.
4. Explaining the employees how the change would help in the development of the individuals..
Why are there bubbles in beer and champagneSolution Bubbles of .pdf
Why are there bubbles in beer and champagne?
Solution
Bubbles of gas, typically carbon dioxide are produced from beer and champagne , rising to the
surface. Because . yeasts ferment sugars and form carbon dioxide gas during fermentation. In the
bottle, where the pressure is high, the carbon dioxide disappears. However, when you open the
bottle and lower the pressure, the carbon dioxide escapes in the form of bubbles. In Champagne
and other sparkling wines, the bubbles pop, while in beer, the bubbles remain and form the
beer\'shead..
Which of the following is true of a traditional master budget a. It.pdf
Which of the following is true of a traditional master budget? a. It uses the zero-base budgeting
approach. b. It is department oriented. c. It recognizes interdependencies. d. It is dynamic in
nature.
Solution
Correct Answer is b. it is department Oriented
One of the critisism of Traditional Master Budget is that it is Department oriented and does not
recognise independencies among departments, further it is static not dynamic in nature..
Which ethnic groups were a part of the former YugoslaviaCroats, S.pdf
Which ethnic groups were a part of the former Yugoslavia?
Croats, Serbs, Bosnian Muslims, Macedonians
Janjaweed
Israelis, Palestinians, Arabs
Hutu & Tutsis
Croats, Serbs, Bosnian Muslims, Macedonians
Janjaweed
Israelis, Palestinians, Arabs
Hutu & Tutsis
Solution
The ethnic group can be defined as a social category based on common culture or nationality.
In other words, it can be said that an ethnic group is a category of people who recognize with
each other based on similarities for example as a common ancestry, language, society, culture or
nation.
First Yugoslavia contains people of the Kingdom of Serbs, Croats and Slovenes (1918–29).
Hence it can be said that the ethnic groups which were a part of the former Yugoslavia were
Croats, Serbs, Bosnian Muslims, Macedonians.
Hence option first is the correct answer.
Option first; Croats, Serbs, Bosnian Muslims, Macedonians.
What would the components of the development be in an early learning.pdf
What would the components of the development be in an early learning(1st – 3rd) website
environment be?
Solution
For a website,major components include Front end,images,contents navigation,page structure.
For early learning website.As it for the children,content along with effective graphic design is
most preferable.
Graphic design:
Many of the elements described, such as the logo, the navigation menus, the layout, images, etc.,
fall under the general category of graphic design. But graphic design is more than these. It is the
look and feel the website will have as a result of proper use and integration of all these elements.
Content management is considered.
The website is not recommended to be a static as it should be more creative and changes is
necessary.
Content management system:
This is the ability to update your website without having to directly edit the html. Simple
systems create areas on a web page that can be easily changed on a regular basis.
Include multiedia and databases.
Multi-media.
It is for than just text and images. Photo-tours, video-clips, sound-clips all can add to the
experience if they are well matched to the type of site and profile of the target audience..
What is true of all fungi They area. eukaryotic, heterotrophic pl.pdf
What is true of all fungi? They are
a. eukaryotic, heterotrophic plants.
b. eukaryotic, parasitic plants.
c. saprobic/decomposing plants.
d. eukaryotic heterotrophs with cell walls.
e. saprobic heterotrophs.
Solution
Answer: a. eukaryotic, heterotrophic plants.
Fungi are non-chlorophyllous(absence of chloroplasts) plants. They are eukaryotes due to the
presence of true nucleus. Fungi live as heterotrophic, parasitic, saprophytic, etc..
What is the possible reason that secondaries arise from parents that.pdf
What is the possible reason that secondaries arise from parents that have unpaired chromosomes
but not from parents that are normal diploids? Map Genetics: A Conceptual Approach 6th
Edition Pierce MHE/Freeman presented by Sapling Learning Progeny of triploid tomato plants
often contain parts of an extra chromosome, in addition to the normal complement of 24
chromosomes. Mutants with a part of an extra chromosome are referred to as secondaries. James
and Margaret Lesley observed that secondaries arise from triploid (3n), trisomic (3n 1), and
double trisomic (3n 1 1) parents, but never from diploids (2n). (J. W. Lesley and M. M. Lesley.
1929. Genetics 14:321-336) What is a possible reason that secondaries arise from parents that
have unpaired chromosomes but not from parents that are normal diploids? Triploid and trisomic
conditions in tomato plants are unnatural, and this genomic imbalance leads to higher levels of
mitotic recombination. The trivalent chromosomes would still be able to associate through
chiasma prior to segregation during meiosis. Due to some unknown mechanism, triploid and
trisom c tomato plants O occasionally over-replicate part of a chromosome, which leads to the
generation of mutant tomato plants. Triploid and trisomic plants are vulnerable to infection by
the tomato mosaic O virus, so they acquire extra chromosomal material by the integration of
viral DNA, whereas diploid plants are not infected by the virus.
Solution
Ans: B
2n+1 condition is called trisomic in which an organism contains two complete genome + 1 extra
chromosome. During the process of meiosis 1 chromosomal page separate so that 1 members of
each state is distributed to one daughter nucleus and other members goes to other daughter
nucleus but rarely one pair of chromosomes fail to separate and moves to the pole giving half of
the daughter cells extra chromosome and remaining half lost one chromosome. As a result n - 1
and n+ 1 chromosomes appear. When the gamut with n+ 1 chromosome fuses with normal
gamut the resulting zygote will have 2n + 1 chromosomes producing trisomics organism.
It was found that trisomies or white study nature and studied in plants like tomatoes tobacco
wheat datura and so on..
What is a trigonometric identity How and why do we use trigonometri.pdf
What is a trigonometric identity? How and why do we use trigonometric identities in
mathematics? How can you verify an identity? Give some examples.
Solution
Trigonometric identities are equalities that involve trigonometric functions and are true for every
single value of the occurring variables where both sides of the equality are defined
Sometimes it is possible to write some complex trignmetric functions in really simple form fr
example instead f writing Sinx/Cosx we can write that as Tanx so as we can see here it is easy to
identify Tanx rather than first finding Sinx and then finding Cosx and then divide them to get our
answer
Here is an example below:
we need to prove that left side is equal to right side
cot x/csc x=cos x
So let us simplify left side,
(cos x/sin x)/(1/sin x)
applying keep change flip
(cos x/sin x)*sin x
now we can cancel out sin x
so we are left with cos x
which is equal to right side of the equation
hence proved..
This is for $7.70 per share... not $7.20Thank you.. The Holtz .pdf
This is for $7.70 per share... not $7.20
Thank you.. The Holtz Corporation acquired 80 percent of the 100,000 outstanding voting shares
of Devine, Inc., for $7.70 per share on January 1, 2014. The remaining 20 percent of Devine\'s
shares also traded actively at $7.70 per share before and after Holtz\'s acquisition. An appraisal
made on that date determined that all book values appropriately reflected the fair values of
Devine\'s underlying accounts except that a building with a 5-year life was undervalued by
remaining life had a $75,000 fair value. At the acquisition date, Devine reported common stock
of $100,000 and a retained earnings balance of $290,000 $81,000 and a fully amortized
trademark with an estimated 10-year Following are the separate financial statements for the year
ending December 31, 2015 HoltzDevine Corporation Inc. $ (760,000) $(365,250) Sales Cost of
goods sold Operating expenses Dividend income 234,000 133,000 319,000 107,250 (16,000) Net
income $ (223,000) $(125,000) Retained eamings, 1/1/15 Net income (above) Dividends
declared $ (709,000) $(360,000) (223,000) (125,000) 90,00020,000 Retained earnings, 12/31/15
S (842,000) $(465,000) Current assets Investment in Devine, Inc Buildings and equipment (net)
Trademarks s 166,000 $ 199,000 616,000 905,000389,000 160,000226,000 Total assets
$1,847,000 $ 814,000 s (685,000 $249.000) Liabilities Common stock Retained eamings,
12/31/15 (above) (320,000) (100,000) (842,000) (465,000) Total liabilities and equities
$(1,847,000) $(814,000) At year-end, there were no intra-entity receivables or payables
Solution
SOLUTION:
a) Consolidation Worksheet:
HOLTZ CORPORATION AND DEVINE INC
Consolidated Worksheet for the Year Ending December 31, 2015
Consolidated Entries
Holtz Corporation
Devine Inc.
Debit
Credit
NCI
Consolidated
Sales
-760000
-365250
-1125250
Cost of Goods Sold
234000
133000
367000
Operating Expense
319000
107250
23700
449950
Dividend Income
-16000
16000
0
Net Income
-223000
-125000
Consolidated Net Income
-308300
NCI in Consolidated Net Income
-20260
20260
Holtz\'s Interest in Consolidated Income
-288040
Retained Earnings on 1/1
-709000
-360000
360000
37040
-746040
Net Income (Consolidated Above)
-223000
-125000
-288040
Dividend Declared
90000
20000
16000
4000
70000
Retained earnings on 12/31
-842000
-465000
-964080
Current Assets
166000
199000
365000
Investment in Devine
616000
37040
653040
0
Building and Equipment
905000
389000
64800
16200
1342600
Trademarks
160000
226000
67500
7500
446000
Goodwill
0
0
224000
224000
Total Assets
1847000
814000
2377600
Liabilities
-685000
-249000
-934000
Common Stock
-320000
-100000
100000
-320000
Retained earnings (Consolidated Above)
-842000
-465000
-964080
NCI in Devine on 1/1
78000
71260
-149260
NCI in Devine on 12/31
-165520
-165520
Total Liabilities and Equities
-1847000
-814000
-2383600
Working Notes:
1. Calculation of Goodwill or Gain on Bargain purchase
Total
Holtz
NCI
Fair Value of Consideration
770000
616000
154000
Less: Book Val.
This project is the first projects you will be working on this quart.pdf
This project is the first projects you will be working on this quarter. There are a number of
objectives to this assignment. First, to start gaining experience in developing programs in C++.
Second, in case you haven’t already you are going to work in pairs to get the sense of
collaborative development of software. Third, because you are allowed to use any references you
find online, this assignment will help you get a sense for just how many sources for
programming in C++ are available via the web. Project specification: In this project you are
going to implement a word search engine using C++. The search engine would help users to
count number of occurrence of a word among given set of text files. Input specification: The
program takes as an input a set of files, containing of multiple words. Words can appear multiple
times in a single file or in different files. The input files should be stored in a dedicated directory
(e.g. /cs/class/cs24/project1/input) and loaded in memory upon program execution. Program
functionality: Upon execution, the program should load all the words from the input files in the
memory. The program should run until the user explicitly specifies “exit”. The following
functionality should be supported: 1. Count how many times a user specified word appears in the
files. 3. Display the name of input files and corresponding number of occurrences of the
specified word. Upon execution of you program, you should specify as input parameter the path
to a directory containing the input files. Below you can find the format specification:
$./wordsearch Examples for execution format: $./wordsearch /cs/class/cs24/project1/input Enter
word: cat If executed as specified above, the program should return the name of files that contain
word “cat” and the number of occurrences of the word in each file Implementation requirements:
First, you need to implement a class for an abstract data type, in which you are going to store
files in memory. This step is very specific, depending on the functionality of the program you are
developing. For the current implementation you are required to use linked lists. A basic layout of
the data structure you would need to implement is illustrated on the figure below: You need to
declare a class Word that would store a word and a pointer to the beginning of a linked list and
another class File that would store a file-name and number of times that the word occurred in this
file. The process of “loading files in the memory” consists of (i) creating an object of type Word
for each new word that occurs in the set of input files, (ii) appending this object to a linked list
(e.g. the green linked list from the picture), (iii) creating an object File for each occurrence of the
word in a file and (iv) updating the corresponding (blue) linked list with the object File. Once
you have the files loaded in such structure, the searching would be as easy as finding the queried
word in the “green” linked list and tracing.
The total energy for the hydrogen molecule is -1.1789737 Hartrees. w.pdf
The total energy for the hydrogen molecule is -1.1789737 Hartrees. what is the binding energy (
in KJ/mol ) of the molecule ?
Solution
Eb =Et-E0
whr Et = total energy of H2 molecule which is -32.08ev
and E0 = ground state energy of hydrogen molecuel which is -1.17ev
hence Eb= -32.08+1.17
= -30.31ev
= -4.84 *10^-21 kJ
as 1ev = 1.6* 106-22kj
and 1 hartree = 27.2107 ev.
The Queue is a special memory structure with accesses limited to PUS.pdf
The Queue is a special memory structure with accesses limited to PUSH and POP operations
which result in a firstinfirstout (FIFO) access. The Stack is a related memory structure which
also has two operations PUSH and POP, but unlike the Queue, the POP operation results in the
data value entered using the most recent PUSH operation.
This results in a lastinfirstout (LIFO) access. Implementing a queue treats the memory as a
circle. How would you modify this design to implement a Stack? How are the Stack Empty and
Stack conditions checked?
Solution
we can Implement stack with using 2 Queues.
In this the first method we ensure that the new entry of element is always at top of queue1 , so
that the pop operation just dequeue from queue 1. We here are using the other queue \"queue2\"
to make sure that the new element is always at the top of queue1.
Below is the operation of the implemented stack:
PUSH :
step 1 : Enqueue element to Queue2.
step 2 : Now gradually dequeueing each element from Queue1 and enqueue to Queue2.
step 3 : Interchanging the name of the Queues with each other. So that we can dont allow one
more movement of all elements from Queue2 Queue1.
POP :
step 1 : Dequeue a element from Queue1 and return it.
STACK_EMPTY :
step 1 : we can check that by cheking the state of Queue1 , if it is empty than the stack is
empty..
The liver has a portal vein as well as a hepatic vein. It also has u.pdf
The liver has a portal vein as well as a hepatic vein. It also has unique exchange blood vessels
similar to capillaries, called “sinusoids.” How do these unique structures determine the function
of the organ?
Solution
Hepatic veins:
Hepatic veins are arisen in the liver lobule\'s central vein. Valves are absent in Hepatic veins.
These vessels can transports deoxygenated blood of liver and blood which has been filtered by
the liver (this is blood from the pancreas, colon, small intestine, and stomach) to the inferior vena
cava.
Hepatic portal vein:
The hepatic portal vein is one of the most important vein that originates from a networks in the
stomach, pancreas, intestines and spleen. this vessel can carry blood from these organs through
the hepatic portal vein to the liver. In the liver this blood enters into a capillary-like hepatic
sinusoids called as hepatic portal system. Later this blood passes through the portal veins of the
liver, blood is carried through a series of merging vessels into the hepatic veins. These empty
into the inferior vena cava.
Sinusoids:
These are capillary like vessels. In the digestive system, they\'re found near the liver and blood is
conveyed via the sinusoids to the inferior vena cava by the hepatic veins.
Sinusoid pathways, or sinusoidal pathways:
This pathway describes the hepatic portal system. From the organs (stomach, intestines,
pancreas, and spleen) and blood moves through the hepatic portal vein to the capillary like
hepatic sinusoids, which together are called the hepatic portal system. From the sinusoid
pathways, blood is transported by the hepatic veins back to the inferior vena cava..
The effects of the Vietnam War and the outcomes of the warThe ef.pdf
The effects of the Vietnam War and the outcomes of the war
The effects of the Vietnam War and the outcomes of the war
Solution
The Effect of the Vietnam War on the American People Essay. The Vietnam War was marked by
brutality, death, protests, and psychological tolls. No war caused such great division among the
American people like the Vietnam War. The war was extremely costly, and it left long-term
effects on people all over the world.
The Vietnam War was the first major war American’s had suffered defeat. The Vietnam war
was a war of confusion, competition and biasness. The outcome of the war was far greater than
an upset American nation, but a severe breakdown of the Vietnamese culture, economy,
environment and government. It also had a tremendous impact on American society even up to
present day. It was unclear from the beginning of the war if the American’s should even be
involved. It was a war between Northern and Southern Vietnam but the U.S saw it as an indirect
way to challenge the USSR’s sphere of influence in Southern Asia and to prevent the domino
effect and the further spread of communism. The Vietnam War completely changed the way the
United State
The War not only affected those in Vietnam, but those still at home in America. Americans had
split into two groups, known as the hawks and the doves. The hawks were people who supported
the war and wanted the country to be victorious in its fight with Communism. The doves were
the people that disagreed with the war. These were traditionally students, and people of the
younger generation. Many veterans when they would return home, would join in these mass
protests. One popular protest, which took place at Kent State University, resulted in 9 wounded
people with 4 dead. The police at Kent State took action against a mass protest by opening fire.
This is the evil that is created through war. Without the Vietnam war, there is no protest at Kent
State. Without the Vietnam war, the college kid’s lives are spared. Without the Vietnam War,
thousands of lives could have been saved. Neil Young said it best in his song Ohio. Ohio
discussed the brutality of war but mainly the police brutality against the kids at Kent State.
Vietnam was a tragic stain in American history. Vietnam was an entirely new type of war for the
United States. It still remains morally and historically problematic in today’s society.
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