Implement the Queue ADT using array – based approach. Using C++ prog.pdf
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Implement the Queue ADT using array – based approach. Using C++ programming Language #include \"QueueArray.h\" template QueueArray::QueueArray(int maxNumber) { } template QueueArray::QueueArray(const QueueArray& other) { } template QueueArray& QueueArray::operator=(const QueueArray& other) { } template QueueArray::~QueueArray() { } template void QueueArray::enqueue(const DataType& newDataItem) throw (logic_error) { } template DataType QueueArray::dequeue() throw (logic_error) { DataType temp; return temp; } template void QueueArray::clear() { } template bool QueueArray::isEmpty() const { return false; } template bool QueueArray::isFull() const { return false; } template void QueueArray::putFront(const DataType& newDataItem) throw (logic_error) { } template DataType QueueArray::getRear() throw (logic_error) { DataType temp; return temp; } template int QueueArray::getLength() const { return -1; } //-------------------------------------------------------------------- template void QueueArray::showStructure() const // Array implementation. Outputs the data items in a queue. If the // queue is empty, outputs \"Empty queue\". This operation is intended // for testing and debugging purposes only. { int j; // Loop counter if ( front == -1 ) cout << \"Empty queue\" << endl; else { cout << \"Front = \" << front << \" Back = \" << back << endl; for ( j = 0 ; j < maxSize ; j++ ) cout << j << \"\\t\"; cout << endl; if ( back >= front ) for ( j = 0 ; j < maxSize ; j++ ) if ( ( j >= front ) && ( j <= back ) ) cout << dataItems[j] << \"\\t\"; else cout << \" \\t\"; else for ( j = 0 ; j < maxSize ; j++ ) if ( ( j >= front ) || ( j <= back ) ) cout << dataItems[j] << \"\\t\"; else cout << \" \\t\"; cout << endl; } } QueueArray.h ___----------------------------------------------------------------------------- #ifndef QUEUEARRAY_H #define QUEUEARRAY_H #include #include using namespace std; #include \"Queue.h\" template class QueueArray : public Queue { public: QueueArray(int maxNumber = Queue::MAX_QUEUE_SIZE); QueueArray(const QueueArray& other); QueueArray& operator=(const QueueArray& other); ~QueueArray(); void enqueue(const DataType& newDataItem) throw (logic_error); DataType dequeue() throw (logic_error); void clear(); bool isEmpty() const; bool isFull() const; void putFront(const DataType& newDataItem) throw (logic_error); DataType getRear() throw (logic_error); int getLength() const; void showStructure() const; private: int maxSize; int front; int back; DataType* dataItems; }; #endif Solution QueueArray.cpp #include \"QueueArray.h\" template QueueArray::QueueArray(int maxNumber) { maxSize = maxNumber; this->dataItems = new DataType[maxSize]; front = -1; back = -1; } template QueueArray::QueueArray(const QueueArray& other) { dataItems = new DataType[maxSize]; for (int i = 0; i <= top; i++) { dataItems[i] = other.dataItems[i]; } } template QueueArray& QueueArray::operator=(const QueueArray& other) { if (maxSize < other.maxSize) { delete[] dataItems; dataItem.
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Given below is the code for the question. Since the test files (ment.pdf
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I wrote a small test for the implementation. Output is shown at the end. Please don\'t forget to
rate the answer if it helped. Thank you.
sequence2.cpp
#include \"sequence2.h\"
using namespace CISP430_A2;
sequence::sequence(size_type entry )
{
capacity = entry;
used = 0;
current_index = 0;
data = new value_type[capacity];
}
// COPY CONSTRUCTOR
sequence::sequence(const sequence& entry)
{
data = NULL;
*this = entry;
}
// Library facilities used: cstdlib
// MODIFICATION MEMBER FUNCTIONS
void sequence::start( )
{
current_index = 0;
}
void sequence::advance( )
{
if(is_item())
current_index++;
}
void sequence::insert(const value_type& entry)
{
if(size() == capacity) //check if resizing is needed
{
resize(capacity * 1.1); //increaase by 10%
}
if(is_item() && current_index > 0)
current_index--;
else
current_index = 0;
for(size_type i = size(); i > current_index; i--)
data[i] = data[i-1];
data[current_index] = entry;
used++;
}
void sequence::attach(const value_type& entry)
{
if(size() == capacity) //check if resizing is needed
{
resize(capacity * 1.1); //increaase by 10%
}
if(!is_item())
current_index = used;
else
current_index++;
//make room for new entry by pushing elements after current to right
for(size_type i = size(); i > current_index ; i--)
data[i] = data[i-1];
data[current_index] = entry;
used++;
}
void sequence::remove_current( )
{
if(is_item())
{
for(size_type i = current_index + 1; i < size(); i++)
data[i-1] = data[i];
used--;
}
}
void sequence::resize(size_type new_capacity )
{
if(new_capacity > capacity)
{
value_type *temp = new value_type[new_capacity];
for(int i = 0; i < used; i++)
temp[i] = data[i];
delete []data;
data = temp;
}
}
void sequence::operator =(const sequence &entry)
{
if(data != NULL) delete []data;
capacity = entry.capacity;
used = entry.used;
current_index = entry.current_index;
data = new value_type[capacity];
for(int i = 0; i < used; i++)
data[i] = entry.data[i];
}
// CONSTANT MEMBER FUNCTIONS
sequence::size_type sequence::size( ) const
{
return used ;
}
bool sequence::is_item( ) const
{
return size() != 0 && current_index < used;
}
sequence::value_type sequence::current( ) const
{
return data[current_index];
}
//Destructor
sequence::~sequence()
{
delete []data;
}
output
attach 1 2 3
[1 2 3 ]
advance after print... so should not have current item
is_item() = 0
insert 4 5 6, should appear in front of seq as 6 5 4
print from current
[6 5 4 1 2 3 ]
print from current
[7 5 4 1 2 3 ]
Solution
Given below is the code for the question. Since the test files (mentioned in question) are missing,
I wrote a small test for the implementation. Output is shown at the end. Please don\'t forget to
rate the answer if it helped. Thank you.
sequence2.cpp
#include \"sequence2.h\"
using namespace CISP430_A2;
sequence::sequence(size_type entry )
{
capacity = entry;
used = 0;
current_index = 0;
data = new value_type[capacity];
}
// COPY CONSTR.
Given below is the code for the question. Since the test files (ment.pdf
Given below is the code for the question. Since the test files (mentioned in question) are missing,
I wrote a small test for the implementation. Output is shown at the end. Please don\'t forget to
rate the answer if it helped. Thank you.
sequence2.cpp
#include \"sequence2.h\"
using namespace CISP430_A2;
sequence::sequence(size_type entry )
{
capacity = entry;
used = 0;
current_index = 0;
data = new value_type[capacity];
}
// COPY CONSTRUCTOR
sequence::sequence(const sequence& entry)
{
data = NULL;
*this = entry;
}
// Library facilities used: cstdlib
// MODIFICATION MEMBER FUNCTIONS
void sequence::start( )
{
current_index = 0;
}
void sequence::advance( )
{
if(is_item())
current_index++;
}
void sequence::insert(const value_type& entry)
{
if(size() == capacity) //check if resizing is needed
{
resize(capacity * 1.1); //increaase by 10%
}
if(is_item() && current_index > 0)
current_index--;
else
current_index = 0;
for(size_type i = size(); i > current_index; i--)
data[i] = data[i-1];
data[current_index] = entry;
used++;
}
void sequence::attach(const value_type& entry)
{
if(size() == capacity) //check if resizing is needed
{
resize(capacity * 1.1); //increaase by 10%
}
if(!is_item())
current_index = used;
else
current_index++;
//make room for new entry by pushing elements after current to right
for(size_type i = size(); i > current_index ; i--)
data[i] = data[i-1];
data[current_index] = entry;
used++;
}
void sequence::remove_current( )
{
if(is_item())
{
for(size_type i = current_index + 1; i < size(); i++)
data[i-1] = data[i];
used--;
}
}
void sequence::resize(size_type new_capacity )
{
if(new_capacity > capacity)
{
value_type *temp = new value_type[new_capacity];
for(int i = 0; i < used; i++)
temp[i] = data[i];
delete []data;
data = temp;
}
}
void sequence::operator =(const sequence &entry)
{
if(data != NULL) delete []data;
capacity = entry.capacity;
used = entry.used;
current_index = entry.current_index;
data = new value_type[capacity];
for(int i = 0; i < used; i++)
data[i] = entry.data[i];
}
// CONSTANT MEMBER FUNCTIONS
sequence::size_type sequence::size( ) const
{
return used ;
}
bool sequence::is_item( ) const
{
return size() != 0 && current_index < used;
}
sequence::value_type sequence::current( ) const
{
return data[current_index];
}
//Destructor
sequence::~sequence()
{
delete []data;
}
output
attach 1 2 3
[1 2 3 ]
advance after print... so should not have current item
is_item() = 0
insert 4 5 6, should appear in front of seq as 6 5 4
print from current
[6 5 4 1 2 3 ]
print from current
[7 5 4 1 2 3 ]
Solution
Given below is the code for the question. Since the test files (mentioned in question) are missing,
I wrote a small test for the implementation. Output is shown at the end. Please don\'t forget to
rate the answer if it helped. Thank you.
sequence2.cpp
#include \"sequence2.h\"
using namespace CISP430_A2;
sequence::sequence(size_type entry )
{
capacity = entry;
used = 0;
current_index = 0;
data = new value_type[capacity];
}
// COPY CONSTR.
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head = new Node;
head->next = NULL;
length = 0;
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/* Write your code here */
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Find the below answers as the points:
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Technically speaking an application cannot be used in wide variety of purposes if the application
is not connected to any other application by the IP address and will have to innovate the other
medium of connection to and fro from the server.
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will have the capacity for communicating to the local devices.
We can do this by achieving upgrading the windows name resolution services
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I fond this way of communication as helpful..
You are burning the latest song you bought on ITunes to a disk. The .pdf
You are burning the latest song you bought on ITunes to a disk. The song is 3 minutes, 23
seconds long and was sampled as a stereo signal (two sides, left and right) at 44.1 kHz using 16-
bit encoding. How much space on the disk does this one track take up? If your CD is
640MBytes, how many songs of this size can you fit on one disk (assume 1 byte is 8-bits in
length)? Remember, you don’t want fractional songs! How many songs could be saved on this
same CD if the sampling rate was 128kHz using 32 bit encoding?
Solution
A)
This stream of audio frames, as a whole, is then subjected to CIRC encoding, which segments
and rearranges the data and expands it with parity bits in a way that allows occasional read errors
to be detected and corrected. CIRC encoding also interleaves the audio frames throughout the
disc over several consecutive frames so that the information will be more resistant to burst errors.
Therefore, a physical frame on the disc will actually contain information from multiple logical
audio frames. This process adds 64 bits of error correction data to each frame. After this, 8 bits
of subcode or subchannel data are added to each of these encoded frames, which is used for
control and addressing when playing the CD.
CIRC encoding plus the subcode byte generate 33-bytes long frames, called \"channel-data\"
frames. These frames are then modulated through eight-to-fourteen modulation (EFM), where
each 8-bit word is replaced with a corresponding 14-bit word designed to reduce the number of
transitions between 0 and 1. This reduces the density of physical pits on the disc and provides an
additional degree of error tolerance. Three \"merging\" bits are added before each 14-bit word for
disambiguation and synchronization. In total there are 33 × (14 + 3) = 561 bits. A 27-bit word (a
24-bit pattern plus 3 merging bits) is added to the beginning of each frame to assist with
synchronization, so the reading device can locate frames easily. With this, a frame ends up
containing 588 bits of \"channel data\" (which are decoded to only 192 bits music).
The frames of channel data are finally written to disc physically in the form of pits and lands,
with each pit or land representing a series of zeroes, and with the transition points—the edge of
each pit—representing 1. A Red Book-compatible CD-R has pit-and-land-shaped spots on a
layer of organic dye instead of actual pits and lands; a laser creates the spots by altering the
reflective properties of the dye.
The audio data stream in an audio CD is continuous, but has three parts. The main portion, which
is further divided into playable audio tracks, is the program area. This section is preceded by a
lead-in track and followed by a lead-out track. The lead-in and lead-out tracks encode only silent
audio, but all three sections contain subcode data streams.
The lead-in\'s subcode contains repeated copies of the disc\'s Table Of Contents (TOC), which
provides an index of the start positions of the track.
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This document describes two implementations of a queue data structure in C++. A queue uses the First In First Out (FIFO) principle to add and remove elements. Elements can be added to the rear of the queue using the Enque operation and removed from the front using the Deque operation. The first implementation uses an array to store queue elements, while the second uses linked nodes to dynamically allocate memory for elements as they are added. Both implementations provide functions for Enqueuing, Dequeuing, checking if the queue is empty, and displaying the queue elements.
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Given below is the code for the question. Since the test files (ment.pdf
Given below is the code for the question. Since the test files (mentioned in question) are missing,
I wrote a small test for the implementation. Output is shown at the end. Please don\'t forget to
rate the answer if it helped. Thank you.
sequence2.cpp
#include \"sequence2.h\"
using namespace CISP430_A2;
sequence::sequence(size_type entry )
{
capacity = entry;
used = 0;
current_index = 0;
data = new value_type[capacity];
}
// COPY CONSTRUCTOR
sequence::sequence(const sequence& entry)
{
data = NULL;
*this = entry;
}
// Library facilities used: cstdlib
// MODIFICATION MEMBER FUNCTIONS
void sequence::start( )
{
current_index = 0;
}
void sequence::advance( )
{
if(is_item())
current_index++;
}
void sequence::insert(const value_type& entry)
{
if(size() == capacity) //check if resizing is needed
{
resize(capacity * 1.1); //increaase by 10%
}
if(is_item() && current_index > 0)
current_index--;
else
current_index = 0;
for(size_type i = size(); i > current_index; i--)
data[i] = data[i-1];
data[current_index] = entry;
used++;
}
void sequence::attach(const value_type& entry)
{
if(size() == capacity) //check if resizing is needed
{
resize(capacity * 1.1); //increaase by 10%
}
if(!is_item())
current_index = used;
else
current_index++;
//make room for new entry by pushing elements after current to right
for(size_type i = size(); i > current_index ; i--)
data[i] = data[i-1];
data[current_index] = entry;
used++;
}
void sequence::remove_current( )
{
if(is_item())
{
for(size_type i = current_index + 1; i < size(); i++)
data[i-1] = data[i];
used--;
}
}
void sequence::resize(size_type new_capacity )
{
if(new_capacity > capacity)
{
value_type *temp = new value_type[new_capacity];
for(int i = 0; i < used; i++)
temp[i] = data[i];
delete []data;
data = temp;
}
}
void sequence::operator =(const sequence &entry)
{
if(data != NULL) delete []data;
capacity = entry.capacity;
used = entry.used;
current_index = entry.current_index;
data = new value_type[capacity];
for(int i = 0; i < used; i++)
data[i] = entry.data[i];
}
// CONSTANT MEMBER FUNCTIONS
sequence::size_type sequence::size( ) const
{
return used ;
}
bool sequence::is_item( ) const
{
return size() != 0 && current_index < used;
}
sequence::value_type sequence::current( ) const
{
return data[current_index];
}
//Destructor
sequence::~sequence()
{
delete []data;
}
output
attach 1 2 3
[1 2 3 ]
advance after print... so should not have current item
is_item() = 0
insert 4 5 6, should appear in front of seq as 6 5 4
print from current
[6 5 4 1 2 3 ]
print from current
[7 5 4 1 2 3 ]
Solution
Given below is the code for the question. Since the test files (mentioned in question) are missing,
I wrote a small test for the implementation. Output is shown at the end. Please don\'t forget to
rate the answer if it helped. Thank you.
sequence2.cpp
#include \"sequence2.h\"
using namespace CISP430_A2;
sequence::sequence(size_type entry )
{
capacity = entry;
used = 0;
current_index = 0;
data = new value_type[capacity];
}
// COPY CONSTR.
Given below is the code for the question. Since the test files (ment.pdf
Given below is the code for the question. Since the test files (mentioned in question) are missing,
I wrote a small test for the implementation. Output is shown at the end. Please don\'t forget to
rate the answer if it helped. Thank you.
sequence2.cpp
#include \"sequence2.h\"
using namespace CISP430_A2;
sequence::sequence(size_type entry )
{
capacity = entry;
used = 0;
current_index = 0;
data = new value_type[capacity];
}
// COPY CONSTRUCTOR
sequence::sequence(const sequence& entry)
{
data = NULL;
*this = entry;
}
// Library facilities used: cstdlib
// MODIFICATION MEMBER FUNCTIONS
void sequence::start( )
{
current_index = 0;
}
void sequence::advance( )
{
if(is_item())
current_index++;
}
void sequence::insert(const value_type& entry)
{
if(size() == capacity) //check if resizing is needed
{
resize(capacity * 1.1); //increaase by 10%
}
if(is_item() && current_index > 0)
current_index--;
else
current_index = 0;
for(size_type i = size(); i > current_index; i--)
data[i] = data[i-1];
data[current_index] = entry;
used++;
}
void sequence::attach(const value_type& entry)
{
if(size() == capacity) //check if resizing is needed
{
resize(capacity * 1.1); //increaase by 10%
}
if(!is_item())
current_index = used;
else
current_index++;
//make room for new entry by pushing elements after current to right
for(size_type i = size(); i > current_index ; i--)
data[i] = data[i-1];
data[current_index] = entry;
used++;
}
void sequence::remove_current( )
{
if(is_item())
{
for(size_type i = current_index + 1; i < size(); i++)
data[i-1] = data[i];
used--;
}
}
void sequence::resize(size_type new_capacity )
{
if(new_capacity > capacity)
{
value_type *temp = new value_type[new_capacity];
for(int i = 0; i < used; i++)
temp[i] = data[i];
delete []data;
data = temp;
}
}
void sequence::operator =(const sequence &entry)
{
if(data != NULL) delete []data;
capacity = entry.capacity;
used = entry.used;
current_index = entry.current_index;
data = new value_type[capacity];
for(int i = 0; i < used; i++)
data[i] = entry.data[i];
}
// CONSTANT MEMBER FUNCTIONS
sequence::size_type sequence::size( ) const
{
return used ;
}
bool sequence::is_item( ) const
{
return size() != 0 && current_index < used;
}
sequence::value_type sequence::current( ) const
{
return data[current_index];
}
//Destructor
sequence::~sequence()
{
delete []data;
}
output
attach 1 2 3
[1 2 3 ]
advance after print... so should not have current item
is_item() = 0
insert 4 5 6, should appear in front of seq as 6 5 4
print from current
[6 5 4 1 2 3 ]
print from current
[7 5 4 1 2 3 ]
Solution
Given below is the code for the question. Since the test files (mentioned in question) are missing,
I wrote a small test for the implementation. Output is shown at the end. Please don\'t forget to
rate the answer if it helped. Thank you.
sequence2.cpp
#include \"sequence2.h\"
using namespace CISP430_A2;
sequence::sequence(size_type entry )
{
capacity = entry;
used = 0;
current_index = 0;
data = new value_type[capacity];
}
// COPY CONSTR.
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Please teach me how to fix the errors and where should be modified. .pdf
Please teach me how to fix the errors and where should be modified. Thank you!
"Program generated too much output. Output restricted to 50000 characters. Check program for
any unterminated loops generating output."
LinkedList.cpp
LinkedList::LinkedList() {
head = new Node;
head->next = NULL;
length = 0;
}
void LinkedList::insertNode(Sales dataIn) {
/* Write your code here */
Node *newNode;
Node *pCur;
Node *pPre;
newNode = new Node;
newNode->data = dataIn;
pPre = head;
pCur = head->next;
while (pCur != NULL && pCur->data.getName() < dataIn.getName()) {
pPre = pCur;
pCur = pCur->next;
}
pPre->next = newNode;
newNode->next = pCur;
length++;
}
bool LinkedList::deleteNode(string target){
Node *pCur;
Node *pPre;
bool deleted = false;
pPre = head;
pCur = head->next;
while (pCur != NULL && pCur->data.getName() < target) {
pPre = pCur;
pCur = pCur->next;
}
if (pCur != NULL && pCur->data.getName() == target) {
pPre->next = pCur->next;
delete pCur;
deleted = true;
length--;
}
return deleted;
}
void LinkedList::displayList() const{
Node *pCur;
pCur = head->next;
while (pCur) {
cout << pCur->data;
pCur = pCur->next;
}
cout << endl;
}
void LinkedList::displayList(int year) const {
Node *pCur;
pCur = head->next;
int cnt = 0;
while (pCur) {
if (pCur->data.getYear() == year){
cout << pCur->data;
cnt++;
}
pCur = pCur->next;
}
if (cnt == 0){
cout << "N/A" << endl;
}
}
double LinkedList::average() const {
double total = 0.0;
double average;
Node *pCur;
pCur = head;
while (pCur != NULL) {
total += pCur->data.getAmount();
pCur = pCur->next;
}
average = total / static_cast(length);
return average;
}
bool LinkedList::searchList(string find, Sales &dataOut) const {
bool found = false;
Node *pCur;
pCur = head->next;
while (pCur != NULL && pCur->data.getName() < find) {
pCur = pCur->next;
}
if (pCur != NULL && pCur->data.getName() == find) {
dataOut = pCur->data;
found = true;
}
return found;
}
LinkedList::~LinkedList() {
Node *pCur;
Node *pNext;
pCur = head->next;
while(pCur != NULL) {
pNext = pCur->next;
delete pCur;
pCur = pNext;
}
delete head;
}
main.cpp
void buildList(const string &filename, LinkedList &list);
void deleteManager(LinkedList &list);
void searchManager(const LinkedList &list);
void displayManager(const LinkedList &list);
int main() {
string inputFileName;
LinkedList list;
cout << "Enter file name: ";
getline(cin, inputFileName);
buildList(inputFileName, list);
displayManager(list);
searchManager(list);
deleteManager(list);
displayManager(list);
return 0;
}
void buildList(const string &filename, LinkedList &list) {
ifstream inFile(filename);
cout <<"Reading data from \"" << filename << "\"\n";
if(!inFile) {
cout << "Error opening the input file: \""<< filename << "\"" << endl;
exit(EXIT_FAILURE);
}
string id;
int year;
string name;
int amount;
string line;
while (getline(inFile, line) ) {
stringstream temp(line);
temp >> id >> year;
temp.ignore();
getline(temp, name, ';');
temp >> amount;
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Given below is the code for the question. Since the test files (ment.pdf
Given below is the code for the question. Since the test files (ment.pdf
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Your company is preparing to migrate from IPv4 to IPv6, and you are .pdf
Your company is preparing to migrate from IPv4 to IPv6, and you are responsible for
determining the steps that take the company up to deploying a transition technology. Your
company uses Windows 7 and Windows 10 with the lastest service packs for client computers as
well as Windows Server 2012 and Windows Server 2016 for your server infrastructure. You
have determined that there are three critical steps that must occur before deplying a transition
technology: 1. Adopt applications that are independent of IP version. 2. Upgrade name resolution
services. Upgrade client computers. Briefly define the activitied involved in each of these steps.
Solution
Find the below answers as the points:
1.Adopt applications that are independent of IP version
Technically speaking an application cannot be used in wide variety of purposes if the application
is not connected to any other application by the IP address and will have to innovate the other
medium of connection to and fro from the server.
Adopting an app that is independent of IP version can be an independent of protocol specific and
will have the capacity for communicating to the local devices.
We can do this by achieving upgrading the windows name resolution services
2.Upgrade name resolution services.
Making the upgrade of NMS will help you intranslating numerical Internet Protocol addresses
that are used for TCP/IP communications to computer names, which are easier for users to
remember.
With these, the computer names are kept to the IP Address of the systems of the source and
destination hosts.
I fond this way of communication as helpful..
You are burning the latest song you bought on ITunes to a disk. The .pdf
You are burning the latest song you bought on ITunes to a disk. The song is 3 minutes, 23
seconds long and was sampled as a stereo signal (two sides, left and right) at 44.1 kHz using 16-
bit encoding. How much space on the disk does this one track take up? If your CD is
640MBytes, how many songs of this size can you fit on one disk (assume 1 byte is 8-bits in
length)? Remember, you don’t want fractional songs! How many songs could be saved on this
same CD if the sampling rate was 128kHz using 32 bit encoding?
Solution
A)
This stream of audio frames, as a whole, is then subjected to CIRC encoding, which segments
and rearranges the data and expands it with parity bits in a way that allows occasional read errors
to be detected and corrected. CIRC encoding also interleaves the audio frames throughout the
disc over several consecutive frames so that the information will be more resistant to burst errors.
Therefore, a physical frame on the disc will actually contain information from multiple logical
audio frames. This process adds 64 bits of error correction data to each frame. After this, 8 bits
of subcode or subchannel data are added to each of these encoded frames, which is used for
control and addressing when playing the CD.
CIRC encoding plus the subcode byte generate 33-bytes long frames, called \"channel-data\"
frames. These frames are then modulated through eight-to-fourteen modulation (EFM), where
each 8-bit word is replaced with a corresponding 14-bit word designed to reduce the number of
transitions between 0 and 1. This reduces the density of physical pits on the disc and provides an
additional degree of error tolerance. Three \"merging\" bits are added before each 14-bit word for
disambiguation and synchronization. In total there are 33 × (14 + 3) = 561 bits. A 27-bit word (a
24-bit pattern plus 3 merging bits) is added to the beginning of each frame to assist with
synchronization, so the reading device can locate frames easily. With this, a frame ends up
containing 588 bits of \"channel data\" (which are decoded to only 192 bits music).
The frames of channel data are finally written to disc physically in the form of pits and lands,
with each pit or land representing a series of zeroes, and with the transition points—the edge of
each pit—representing 1. A Red Book-compatible CD-R has pit-and-land-shaped spots on a
layer of organic dye instead of actual pits and lands; a laser creates the spots by altering the
reflective properties of the dye.
The audio data stream in an audio CD is continuous, but has three parts. The main portion, which
is further divided into playable audio tracks, is the program area. This section is preceded by a
lead-in track and followed by a lead-out track. The lead-in and lead-out tracks encode only silent
audio, but all three sections contain subcode data streams.
The lead-in\'s subcode contains repeated copies of the disc\'s Table Of Contents (TOC), which
provides an index of the start positions of the track.
CASE 3-12 Information System Project Steering Committee The Informat.pdf
CASE 3-12 Information System Project Steering Committee The Information System Steering
Committee in conjunction with administration has decided to implement a computerized
physician order entry (CPOE) system. A separate CPOE committee is being created to control
the implementation of the CPOE with the Information System Steering Committee providing
guidance. As project manager, you have been given the responsibility for selecting the
appropriate individuals—those with a vested interest in the CPOE—for inclusion on this
committee.
1. Identify the departments that should be represented on the CPOE committee.
2. Identify whom the CPOE committee should report to.
3. Recommend the frequency you would recommend that the CPOE committee report to the
Information System Steering Committee.
4. Identify the responsibilities the CPOE committee should have.
Solution
1. The departments that must be represented in the committee are Physicians and clinicians, IT
staff, staff from clinical and other departments, training personnel, Governing authorities,
government bodies and regulatory agencies.
2.CPOE committee should report to the project head overseeing the implementation of the
project. This person should be the head ofInformation System Project Steering Committee, and a
seasoned healthcare practitioner with commercial acumen and exposure to such projects.
3. The frequency of report would depend upon the project duration. For a project duration of six
months, it should be monthly for first three months, fortnightly for next two months and weekly
thereafter.
4.Responsibilities of CPOE committee.
(a) Identifying the issues with the physicians and clinicians as well as those with the IT experts
and act as a bridge between them.
(b) Identify the stakeholders and communicate with them to include them in decision making.
(c) Ensure that compliance with the regulations are met.
(d) Track the progress of the project with the schedule.
(e) Continuous evaluation of the outcomes through trials during the project and take corrective
actions.
(f) Final evaluation and mapping it with project objectives. Record learnings..
Can you date the latest financial crisis in the United States or in .pdf
Can you date the latest financial crisis in the United States or in Europe?Are there reasons to
think that these crises might have been related? Why?
Solution
The last financial crisis both in the United States as well as in UK is the financial crisis of 2008.
This is a global financial crisis and the crisises are related to one another where the factors
contributing are same including the asset liability mismatch, mortgages of banks etc. Sometimes,
it is also said one financial crisis is a warning to the next crisis on the whole..
B1. State the components of organization. Give good examples to justi.pdf
B1. State the components of organization. Give good examples to justify your answer.
Solution
Components of organisation are
1: Purpose: This element is all about a proper mission and objectives of an organisation. ( eg: To
become a market leader in electrnoc industry). Inorder to achieve the purpose organisation has to
set their objectives clearly.
2: Organisation structure: The structure is about the heirarch of organisation ( like formal,
informal, matrix , open etc). Some organisation are having very tight organisation structure and
some will be very flexible and communication flow is based on the organisation structure.
3: Division of labour: This division comprises of activities to be done and responbilities to be
fulfiled based on the roles given. ( like shop floor staffs should do the plant activities, logisitcs
teams should provide the facilties and transportation of required materials etc)
4: Co ordinated effort: Mutual effort and colloborative system is required to produce the desired
result in any organisation. It helps in sustainability of organisation..
Assignment of SOS operating systemThe file lmemman.c has one incom.pdf
Assignment of SOS operating system
The file lmemman.c has one incomplete function with TODO blocks. Finish these blocks to
complete the assignment.
bool init_logical_memory(PCB *p, uint32_t code_size);
This function initializes a process’ logical address space. This involves allocating frames for the
process code, stacks, page directory and page tables, and then setting them up. The PCB is
passed in the argument p, and the number of bytes in the process’ code and data is in the
argument code_size. The function should return TRUE if all goes well; otherwise FALSE. Any
frames allocated in the function must be freed if FALSE is returned.
This should be able to work under SOS operating system.
Here is the lmenman.c file:
////////////////////////////////////////////////////////
// The NAIVE Logical Memory Manager
//
// Divides memory into 4KB pages and allocates from them
#include \"kernel_only.h\"
// kernel page directory will be placed at frame 257
PDE *k_page_directory = (PDE *)(0xC0101000);
// page table entries for the 768th page directory entry will be placed
// at frame 258; 768th entry corresponds to virtual address range
// 3GB to 3GB+4MB-1 (0xC0000000 to 0xC03FFFFF)
PTE *pages_768 = (PTE *)(0xC0102000);
/*** Initialize logical memory for a process ***/
// Allocates physical memory and sets up page tables;
// we need to allocate memory to hold the program code and
// data, the stack, the page directory, and the required
// page tables
// called by runprogram.c; this function does not load the
// program from disk to memory (done in scheduler.c)
bool init_logical_memory(PCB *p, uint32_t code_size) {
// TODO: see background material on what this function should
// do. High-level objectives are:
// 1) calculate the number of frames necessary for program code,
// user-mode stack and kernel-mode stack
// 2) allocate frames for above
// 3) determine beginning physical address of program code,
// user-mode stack and kernel-mode stack
// 4) calculate the number of frames necessary for page directory
// and page tables
// 5) allocate frames for above
// 6) set up page directory and page tables
// 7) set mem struct in PCB: start_code, end_code, start_stack,
// start_brk, brk, and page_directory
// Return value: TRUE if everything goes well; FALSE if allocation
// of frames failed (you should dealloc any frames that may
// have already been allocated before returning)
// TODO: uncomment following line when you start working in
// this function
return FALSE;
}
/*** Initialize kernel\'s page directory and table ***/
void init_kernel_pages(void) {
uint32_t i;
// set up kernel page directory (users cannot touch this)
for (i=0; i<1024; i++) k_page_directory[i] = 0;
k_page_directory[768] = ((uint32_t)pages_768-KERNEL_BASE) | PDE_PRESENT |
PDE_READ_WRITE;
// map virtual (0xC0000000--0xC03FFFFF) to physical (0--0x3FFFFF)
for (i=0; i<1024; i++)
pages_768[i] = (i*4096) | PTE_PRESENT | PTE_READ_WRITE | PTE_GLOBAL;
// load page directory
load_CR3((uint32_t)k_pa.
Any kind of help would gladly be appreciated. (C-programming)Probl.pdf
Any kind of help would gladly be appreciated. (C-programming)
Problem:
Charity Runs are a popular way for organizations to raise awareness and get fit at the same time.
However, these events can be tough to organize. The Princess Half Marathon Weekend, for
example, has as many as 40,000 runners raising money and awareness for the Children’s Miracle
Network Hospitals. People can run both 5ks and 10ks and there’s even a children’s race.
After nearly a semester of C programming, you\'ve decided that you\'d like to donate your skills
to create a program that organizes the typical activities for charity run weekend. In particular,
your program will help manage the following:
1) Individual Registration
2) Team Registration
3) Running Events
4) Donation Totals
Your program will log the number of teams and individual who are signed up for the different
races, process the racing events to see who has the fastest times, and track the total amount of
money raised by teams and individuals for charity.
Header Specification
To facilitate grading, include in your header comment which portions of the assignment you
have completed. You must complete all portions in order to earn full credit, but partial credit is
available for completing some of the steps. The primary steps are as follows:
(1) Processing Individual Registrations
(2) Processing Team Registrations
(3) Processing the running events
(4) Calculating total donations
If your comment is accurate, meaning that you pass the appropriate tests cases corresponding to
your choice, you\'ll earn 10 points. If your comment is roughly accurate, meaning that you
sincerely attempted the items you listed, and most of them work minus a tiny bug, then you\'ll
get 5 of these points. If your comment isn\'t accurate to a reasonable degree, you\'ll get 0 of these
points.
The reason this is here is because it\'s very important to communicate to others accurately what
you\'ve accomplished and what is left to accomplish. This sort of honesty and ability to appraise
your own work is a critical skill in a job.
Program Details
Individual RegistrationDetails
There are a two registration windows: early registration and regular registration. Prices for
registering early and regularly will be given. Each individual will have a unique number and
must be processed separately to record their name, age, running event, and donations raised. The
maximum number of runners for our program will be 10000.
Team RegistrationDetails
Teams may be created and registered to sign up several participators at once. Teams may only
sign up during early registration, have between 5 and 50 members, and must pay the team
registration fee for every member on the team. Teams should be recorded with their name and
number of members. Each member of the team still needs to be recorded with their name, age,
running event, and donations raised. We can organize at most 200 teams.
Running EventsDetails
There will be three running events: a 5k race, a 10k race, .
Which of the following solutions will turn red litmus blue pOH 1.pdf
Which of the following solutions will turn red litmus blue?
pOH 11.50,
2.3 × 10−4 M [OH−],
pH 5.50 or
4.1 × 10−5 M [H3O+]
Solution
 2.3 × 10−4 M [OH−],Â
tip: pH = 14 - pOH = 14 -(-log(2.3x 10-4) = 14-3.64 = 10.36 > pH = 7. The solution is basic and
thus has turn red litmus blue. all the other cases, pH < 7 and thus they are acidic..
What serves as the most reliable source of information about the .pdf
What serves as the most reliable source of information about the world in general?
What serves as the most reliable source of information about the world in general?
Solution
Answer: Things that one learns from others
Own observations may be wrong; own interest may not be adequate unless this is able to produce
information; therefore, learning from others could be a good source because this is a general
tendency – we used to learn if someone teaches us..
What is the difference between the terms “earnings and profits” and .pdf
What is the difference between the terms “earnings and profits” and “retained earnings”?
Solution
Earnings and profits indicates the current year profit earned from the operations while retained
earnings indicate the profits of previous year adjusted with dividends and other distributions
made to stockholders from the profits earned till date. Retained earnings is part of the capital of
the busienss which increases when the company earns profits and decreases when have losses or
pay out dividend.
what are three effects of transistor scaling on computer architectur.pdf
what are three effects of transistor scaling on computer architecture development. Please also
explain these effects in details and make a comparison between old-time computer architecture
and current computer architecture
Solution
Computer architecture: is a specification detailing how a set of software and hardware
technology standards interact to form a computer system or platform. In short, computer
architecture refers to how a computer system is designed and what technologies it is compatible
with.
Let’s understand Moore\'s Law in simple words:
Every two years the Smartphone you are carrying, the computer or tablet you are using or your
TV at home gets twice as powerful/smart while the cost of that computing power dramatically
reduces over time.
Now coming to question, effects of Computer architecture:
1. Higher Level of integration enables more complex architectures.
Ex: On chip memory, super scalar processor.
2. Higher level of integration enables more application specific architectures.
Ex: A variety of microcontrollers
3. Larger logic capacity and higher performance allow more freedom in architecture trade-off.
Comp arch can focus more on what should be done rather than worrying about physical
constraints.
4. Lower cost generates a wider mater. Profitability and competition stimulates arch innovations.
Generations of Computer:
First Generation (1940-1956) Vacuum Tubes: The first computers used vacuum tubes for
circuitry and magnetic drums for memory, and were often enormous, taking up entire rooms.
They were very expensive to operate and in addition to using a great deal of electricity, the first
computers generated a lot of heat, which was often the cause of malfunctions.
Second Generation (1956-1963) Transistors
Transistors replace vacuum tubes and ushered in the second generation of computers. The
transistor was invented in 1947 but did not see widespread use in computers until the late 1950s.
The transistor was far superior to the vacuum tube, allowing computers to become smaller,
faster, cheaper, more energy-efficient and more reliable than their first-generation predecessors.
Third Generation (1964-1971) Integrated Circuits
The development of the integrated circuit was the hallmark of the third generation of computers.
Transistors were miniaturized and placed on silicon chips, called semiconductors, which
drastically increased the speed and efficiency of computers.
Fourth Generation (1971-Present) Microprocessors
The microprocessor brought the fourth generation of computers, as thousands of integrated
circuits were built onto a single silicon chip. What in the first generation filled an entire room
could now fit in the palm of the hand. The Intel 4004 chip, developed in 1971, located all the
components of the computer—from the central processing unit and memory to input/output
controls—on a single chip.
Fifth Generation (Present and Beyond) Artificial Intelligence
Fifth generation computing devices, based on artificial intellig.
What are some of the motives for employee theft What are some .pdf
What are some of the motives for employee theft?
What are some of the motives for employee theft?
Solution
ANSWER:
There can be numerous motives of executives to commit theft discussed as follows:
1) Personal needs:
The personal requirements of the executive’s may drive them to commit a theft. The reason is
that the executives have a lifestyle they need to uphold in the company and in the society. They
will not prefer a situation where they feel inferior to others and thus will use the opportunity
accorded to them to make more money
2) The pressure on the individual
It entails to the pressure on the individual on emanating from a financial situation such as debts,
as well as the shortfall regarding the revenues. Majority of the executive see such problems as
difficult to get solved via the proper methods, therefore resorting to engaging in fraudulent
activities to solve the financial challenges
3) Company requirements
The executives are required to drive the company to success or out of difficult times. Thus there
is a need to show the good performance, as a poor performance is an indicator of poor leadership
in the firm. For its avoidance few executives will inflate some of the results to satisfy the
shareholders requirement, who may have pressured them much to produce certain results in a
given time frame.
4) Comparison with peers
The feel of self-worth among the executives in the company is a major motivator of the
executives to commit fraud. It occurs generally when they compare themselves across their
industry peers as well as how much they are receiving as a reward in their firm. This comparison
generates pressure on the executives as a result they wanting to be equal with their peers
commits theft
5) Opportunity factor
The opportunity factor is natural, because executives may receive number of opportunities in the
company as they understand the internal controls and as well as the command structure of the
organization
6) Ability to rationalize crime
Ability to rationalize crime is a stage of theft triangle calls for the executive officer to make use
of their cognitive skills in order to justify their crime activities within the company. Their
explanations tend to be so convincing to the rest of the company\'s members, that their action
gets seen as just normal
7) Rational approach
Rationalization has its own significance as again human is capable of creating any kind of
justification for their theft activities, thus the executives often don’t consider themselves as
criminals. However they think that they are taking a rational approach..
Twitter is a popular social media. It allows its users to exchange tw.pdf
Twitter is a popular social media. It allows its users to exchange tweets (short messages of a
maximum of 140 Characters). Write a program that prompts the user for a tweet and then flips
the letter-case of each letter in the tweet. Apply the flipping code again to get the original
message If the user enters an empty tweet then the program shall show the message \"Empty
tweet !\". Refer to the sample input/output given below. Your program should have one c-string
(call it twt []).
Solution
#include
#include
#include
using namespace std;
void flipCase( char* s )
{
for(int i=0;s[i];i++)
if (islower(s[i])) s[i] = toupper(s[i]); //if letter is in uppercase convert to lowercase . If letter
is lowercase convert to upper case
else s[i] = tolower(s[i]);
}
int main()
{
char twt[ 100 ];
cout << \"Enter your tweet: \";
cin.getline(twt,141);
flipCase(twt);
cout<<\"\ Tweet after flipping letter-case is: \";
cout << twt << endl;
cout<<\"\ Tweet after flipping letter-case again is: \";
flipCase(twt);
cout << twt << endl;
return 0;
}
output:.
A. State the domai and ranga. 1. y find the inverse and state the dom.pdf
A. State the domai and ranga. 1. y find the inverse and state the domamamd Tange of f Domain:
Rang Graph and sraph.
Solution
f(x) = 3+ 2^x
Domain : ( - inf , inf)
Range : ( 2 , inf )
Inverse : y = 3 +2^x
Plug x= y and y = x:
x = 3 + 2^y
(x -3) = 2^y
take ln on both sides: ln( x-3 ) = yln2
y = ln2( x-3)
f^-1(x) = ln2( x-3)
Range : ( - inf , inf)
Domain : ( 2 , inf ).
The Puritan faith community shaped the New England colonies in virtu.pdf
The Puritan faith community shaped the New England colonies in virtually every way during
much of the seventeenth century. Discuss the ideas and religious principles that characterized
Puritanism, and explore the significant differences between the separatists led by William
Bradford, who founded Plymouth colony, and the group of Puritians led by John Winthrop, who
founded Massachusetts Bay colony.
Solution
Plymouth Colony and the Massachusetts Bay Colony were two distinct Puritan
commonwealths.he Plymouth Colony was founded by the Pilgrims, who arrived in
Massachusetts on the Mayflower in 1620 and celebrated the first Thanksgiving, which was a
harvest festival, the following year. The colony lasted from 1620 to 1691.
Plymouth ultimately proved less successful than its neighbor to the north, the Massachusetts Bay
Colony, which was founded 10 years later in 1630 and also ceased to exist in 1691. When people
talk about Puritan influence in America, they are referring, albeit usually unwittingly, to the
Puritanism of the Massachusetts Bay Colony, not Plymouth or the Pilgrims.
Boston was the chief center of the Bay Colony. Colonial jurisdiction eventually extended to
much of present-day New England, including portions of Massachusetts, Maine, New
Hampshire, Rhode Island, and Connecticut. The Colony’s leading figure was John Winthrop.
William Bradford’s Plymouth Colony was from the start more tolerant than its neighbor.
Although witchcraft, for example, was a capital offense in Plymouth, only two charges for the
crime were ever brought. The first was dismissed before trial; the second resulted in the acquittal
of the defendant and punishment of the accuser for making false allegations.
But even Massachusetts Bay itself was a remarkably argumentative and politically contentious
society, where authority was frequently and vigorously contested. Despite his great natural
leadership gifts and authoritarian style, for example, John Winthrop was deposed more than
once.
The two colonies differed in other important ways.
The Massachusetts Bay colonists were committed to reforming the Church of England from
within rather than separating from it. The Church of England remained the official church.
However, as a practical matter, there was no ecclesiastical hierarchy to sustain a state church
bureaucracy. During the entire colonial period, for example, not a single Anglican bishop was
appointed to rule the American flock.
Thus, over time, migration from England brought about de facto separatism in Massachusetts
Bay, and the adoption of Plymouth Colony’s Congregational form of church polity, wherein each
local congregation controlled its own affairs.
There were also significant wealth and status differences between the two populations.
The Pilgrims at Plymouth were mostly farmers and artisans. The colonists of Massachusetts Bay,
in contrast, were better educated, more economically and socially successful, and brought with
them educated clergy to provide leader.
savings account d. the value of the shares is based on the amount of .pdf
savings account d. the value of the shares is based on the amount of money Drew owes his
creditors. 30. Ali and Alecia are promoters of a new u-pick farm named AA-Apples. Before
incorporating, Ali and Alecia draft a document that explains the rights of shareholders, the
structure of the business, and how many shares form a quorum. This document is named a.
articles of incorporation. b. bylaws c. articles of organization. d. a partnership agreement. 31.
Wild &Scenic; River Tours, Inc., is a corporation. Wild & Scenic has the implie power to a. sell
the majority of assets. b. replace the board of directors. c. change the bylaws. d. perform all acts
reasonably appropriate and necessary to accomplishi corporate purposes
Solution
30. The article of incorporation is the foundation document for any organization. It contains the
essential details on the incorporation of any organization. It contains details on how and why the
organization has come into existence, what are the terms and conditions relating to incorporation,
who are the people involved in the incorporation.
It also explains the very detailed view of who all the shareholders, their rights, nature, and
structure of the business and also what are the requirements on the quorum required for any
decision making.
31. Any corporation which has been formed and registered as a corporation has all the power to
perform all acts which are reasonable, appropriate and necessary to achieve/accomplish what has
been laid down in the mission statement/vision of the corporation.
Thus Option D is the correct answer..
QuestionIt was reported on June 11, 1997, by NBC Nightly News that.pdf
Question
It was reported on June 11, 1997, by NBC Nightly News that according to a study completed,
women are genetically predisposed to developing interpersonal skills such as perceiving the
attitudes and needs of others.
1. Which personality dimensions relate to emotional intelligence, such as sensitivity to the needs
of others mentioned in the case?
2. Does culture affect development of certain personality characteristics by gender? If so, how?
3. Do you think that this phenomenon has changed over time? In which countries do you think
the conclusion reported in the case is true today?
Solution
1. there are five types of personality dimensions, they are openness to experience,
conscientiousness, extraversion, agreeableness and neuroticism.
openness to experience, conscientiousness, extraversion, agreeableness, and neuroticism. and the
given situation comes under openness to experience.
2. yes, culture diefinitely influece gender to think or the way they think. because in some cultures
women is only treated as an instrment to serve the male people. in some cultures, they are treated
like as nature, so, the way they treat and grown will helps them to think, to act and to react.
3. usually in developed countries, we can observe the above said changes. the people who are
educated, and financially and economically better than other nations, they are treating women in
good manner. in developing and under developed nations, it is still a problem only..
946 LTE Labs Le.chateliers-lab.pdf Before beginning this experiment.pdf
9:46 LTE Labs Le.chateliers-lab.pdf Before beginning this experiment in the laboratory you
should be alieto answer the following questions PRE LAB QUESTIONS 1. Beiefly stabe
LeChidelier\'s Principle. 2. Consider the following equilibrium In which direction will the
equilibrium shit if a. HSO,ibs added? Why? b. Buicl, is added? Why C NaCl is added? Why d.
Heat is added? Why? Consider the folewing equilibrium foe nitous acid, HNOs,a weak acid In
which direction will the equilibrium shift if a NaOH is added? b. NaNO, is added HCl is added?
d. The acid solution is made more dilute? Complete and balance the following equations and
then wrae balanced net ionic equations. 4· 3. On the busis of LeChitelier\'s Principle, explain
why Ag-C0 disos when HNO, is added Previous Next Dashboard Calendar To DoNotifications
Inbox
Solution
Answer:
1. Le Chtelier\'s principle states that if a dynamic equilibrium is disturbed by changing the
conditions, the position of equilibrium shifts to counteract the change to re-establish a new
equilibrium
2.a) H2SO4 is added:
the equilibrium shifts to the left because as there are more sulfate ions are added.
b)BaCl2 is added
The equilibrium shifts to the left because as there are more Ba^+2 are added.
c) NaCl is added
There is no shift in the equilibrium. As there are no Na^+1 and Cl^-1 are present in the
equilibrium mixture
d) Heat is added
The equilibrium will shift to the right. As the forward reaction is endothermic, when the heat is
added, the reaction will favours the product
3. a) NaOH is added
The equilibrium will shift to the left because as there are more OH^-1 ions added, it will react
with H+ present in the right side and forms more water.
b)NaNO2 is added
The equilibrium will shift to the left because the more NO2^-1 ions added, it forms more HNO2
by reacts with H3O^+1.
c) HCl is added
The equilibrium will shift to the right because more H+ is added in the equilibrium mixture
d) The acid solution is made more dilute
The equilibrium will shift to the right as there are more water is added.
4 a) AgNO3 (aq) + HCl (aq) <<>> AgCl (s) + HNO3 (aq)
Net ionic equation: Ag^+1 (aq) + Cl^-1 (aq) >> AgCl (s)
b) NH3(aq) + HCl (aq) <<<>> NH4Cl (s)
Net equation: NH3 (aq) + H^+1 (aq) + Cl^-1 (aq) > NH4Cl(s)
c) Na2CO3 (aq) + 2HNO3 (aq) <<>> H2O(l) + 2NaNO3 (aq) + CO2 (g)
Net ionic equation: CO3 ^-2 (aq) + 2H+ (aq) >> H2O (l) + CO2 (g)
5. Ag2CO3 (s) + 2HNO3 (aq) <<<>>> 2AgNO3 (aq) + H2O (l) + CO2(g)
Ag2CO3 is a solid . When adding nitric acid to silver carbonate, it forms soluble silver nitrate..
Prove that the T_i-property is a topological property for i = 0So.pdf
Prove that the T_i-property is a topological property for i = 0
Solution
Hausdorff spaces do not have in general the homotopy type of any finite space. Recall that a
topological space
Xsatisfies the T1- separation axiom if for any two distinct points x, y X there exist open setsu
and V such that
x U,y V , y / U, x / V . This is equivalent to saying that the points are closed in X. AllHausdorff
spaces are
T1,but the converse is false..
4. Refer to the table of Gini coefficients in the Added Dimension box.pdf
4. Refer to the table of Gini coefficients in the Added Dimension box entitled \"The Gini
Coefficient\" in your textbook to answer the following questions: Is income generally more or
less equally distributed in the United States compared to income distribution in other countries?
a. b. Look also at Figure 18-4 in your text. In terms of per capita income, how does the United
States compare to other countries? c. What is more important, the distribution of income or the
level of income?
Solution
a) Income is less equally distributed in US as compared to other countries, which has increased
significantly since the 1970s after several decades of stability. In other words, there has been an
increase in the share of nation\'s income received by higher income households. The U.S. ranks
behind than 70% of countries who have a more equal income distribution than the US.
b) Compared to the other nations per capita income of US is much better than majority of the
other countries (more than its peers or more than other advanced countries), it is more than
Germany, France, United Kingdom, and Italy. US (per capita PPP) is ranked at 9th position
according to world bank and US (per capita nominal) ranked at 7th position.
c) The distribution of income is of enormous importance than the income level. As it influences
the structure of the society while determining the poverty level of it for any given level of GDP.
And also because of the fact that income distribution among the population provides a method of
measurement employed in the determination of the dispersion of incomes..
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Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
His expertise extends across a diverse spectrum of reporting, database, and web development applications, underpinned by an exceptional grasp of data storage and virtualization technologies. His proficiency in application testing, database administration, and data cleansing ensures seamless execution of complex projects.
What sets Denis apart is his comprehensive understanding of Business and Systems Analysis technologies, honed through involvement in all phases of the Software Development Lifecycle (SDLC). From meticulous requirements gathering to precise analysis, innovative design, rigorous development, thorough testing, and successful implementation, he has consistently delivered exceptional results.
Throughout his career, he has taken on multifaceted roles, from leading technical project management teams to owning solutions that drive operational excellence. His conscientious and proactive approach is unwavering, whether he is working independently or collaboratively within a team. His ability to connect with colleagues on a personal level underscores his commitment to fostering a harmonious and productive workplace environment.
Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
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Any substance (other than food) that is used to prevent, diagnose, treat, or relieve symptoms of a
disease or abnormal condition
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Implement the Queue ADT using array – based approach. Using C++ prog.pdf
- 1. Implement the Queue ADT using array – based approach. Using C++ programming Language #include "QueueArray.h" template QueueArray::QueueArray(int maxNumber) { } template QueueArray::QueueArray(const QueueArray& other) { } template QueueArray& QueueArray::operator=(const QueueArray& other) { } template QueueArray::~QueueArray() { } template void QueueArray::enqueue(const DataType& newDataItem) throw (logic_error) { } template DataType QueueArray::dequeue() throw (logic_error) { DataType temp; return temp; } template void QueueArray::clear() { } template bool QueueArray::isEmpty() const
- 2. { return false; } template bool QueueArray::isFull() const { return false; } template void QueueArray::putFront(const DataType& newDataItem) throw (logic_error) { } template DataType QueueArray::getRear() throw (logic_error) { DataType temp; return temp; } template int QueueArray::getLength() const { return -1; } //-------------------------------------------------------------------- template void QueueArray::showStructure() const // Array implementation. Outputs the data items in a queue. If the // queue is empty, outputs "Empty queue". This operation is intended // for testing and debugging purposes only. { int j; // Loop counter if ( front == -1 ) cout << "Empty queue" << endl; else { cout << "Front = " << front << " Back = " << back << endl;
- 3. for ( j = 0 ; j < maxSize ; j++ ) cout << j << "t"; cout << endl; if ( back >= front ) for ( j = 0 ; j < maxSize ; j++ ) if ( ( j >= front ) && ( j <= back ) ) cout << dataItems[j] << "t"; else cout << " t"; else for ( j = 0 ; j < maxSize ; j++ ) if ( ( j >= front ) || ( j <= back ) ) cout << dataItems[j] << "t"; else cout << " t"; cout << endl; } } QueueArray.h ___----------------------------------------------------------------------------- #ifndef QUEUEARRAY_H #define QUEUEARRAY_H #include #include using namespace std; #include "Queue.h" template class QueueArray : public Queue { public: QueueArray(int maxNumber = Queue::MAX_QUEUE_SIZE); QueueArray(const QueueArray& other); QueueArray& operator=(const QueueArray& other); ~QueueArray(); void enqueue(const DataType& newDataItem) throw (logic_error); DataType dequeue() throw (logic_error); void clear();
- 4. bool isEmpty() const; bool isFull() const; void putFront(const DataType& newDataItem) throw (logic_error); DataType getRear() throw (logic_error); int getLength() const; void showStructure() const; private: int maxSize; int front; int back; DataType* dataItems; }; #endif Solution QueueArray.cpp #include "QueueArray.h" template QueueArray::QueueArray(int maxNumber) { maxSize = maxNumber; this->dataItems = new DataType[maxSize]; front = -1; back = -1; } template QueueArray::QueueArray(const QueueArray& other) { dataItems = new DataType[maxSize]; for (int i = 0; i <= top; i++) { dataItems[i] = other.dataItems[i]; } }
- 5. template QueueArray& QueueArray::operator=(const QueueArray& other) { if (maxSize < other.maxSize) { delete[] dataItems; dataItems = new dataType[other.maxSize]; } maxSize = other.maxSize; top = other.top; for (int i = 0; i <= top; i++) { dataItems[i] = other.dataItems[i]; } return *this; } template QueueArray::~QueueArray() { this->clear(); } template void QueueArray::enqueue(const DataType& newDataItem) throw (logic_error) { if (this->isFull()) throw logic_error("Queue is Full."); if (this->isEmpty()) { dataItems[back + 1] = newDataItem; back = (++back) % maxSize; front++; } else if (back == 7 && !isFull()) {
- 6. back = 0; dataItems[back] = newDataItem; } else { back = (++back) % maxSize; dataItems[back] = newDataItem; } } template DataType QueueArray::dequeue() throw (logic_error) { if (this->isEmpty()) throw logic_error("Queue is empty. "); if (getLength() == 1) { DataType temp = dataItems[this->front]; clear(); return temp; } else { DataType temp; temp = dataItems[this->front]; front++; return temp; } } template void QueueArray::clear() { front = -1; back = -1; } template bool QueueArray::isEmpty() const { return (front == -1);
- 7. } template bool QueueArray::isFull() const { return (getLength() == this->maxSize); } template void QueueArray::putFront(const DataType& newDataItem) throw (logic_error) { if (isFull()) throw logic_error("Queue is Full."); else { if (isEmpty()) { front = 0; back = 0; } else { front--; if (front < 0) front = (maxSize - 1); } dataItems[front] = newDataItem; } } template DataType QueueArray::getRear() throw (logic_error) { int cursor = -1; if (isEmpty()) { throw logic_error("Queue is Empty."); } else { cursor = back; if (front == back) {
- 8. front = -1; back = -1; } else { --back; if (back < 0) back = (maxSize - 1); } return dataItems[cursor]; } } template int QueueArray::getLength() const { if (front > back) return (back - front + maxSize + 1); else return back - front + 1; } //-------------------------------------------------------------------- template void QueueArray::showStructure() const // Array implementation. Outputs the data items in a queue. If the // queue is empty, outputs "Empty queue". This operation is intended // for testing and debugging purposes only. { int j; // Loop counter if (front == -1) cout << "Empty queue" << endl; else { cout << "Front = " << front << " Back = " << back << endl; for (j = 0; j < maxSize; j++) cout << j << "t";
- 9. cout << endl; if (back >= front) for (j = 0; j < maxSize; j++) if ((j >= front) && (j <= back)) cout << dataItems[j] << "t"; else cout << " t"; else for (j = 0; j < maxSize; j++) if ((j >= front) || (j <= back)) cout << dataItems[j] << "t"; else cout << " t"; cout << endl; } } QueueArray.h // QueueArray.h #ifndef QUEUEARRAY_H #define QUEUEARRAY_H #include #include using namespace std; #include "Queue.h" template class QueueArray : public Queue { public: QueueArray(int maxNumber = Queue::MAX_QUEUE_SIZE); QueueArray(const QueueArray& other); QueueArray& operator=(const QueueArray& other); ~QueueArray(); void enqueue(const DataType& newDataItem) throw (logic_error); DataType dequeue() throw (logic_error); void clear(); bool isEmpty() const; bool isFull() const;
- 10. void putFront(const DataType& newDataItem) throw (logic_error); DataType getRear() throw (logic_error); int getLength() const; void showStructure() const; private: int maxSize; int front; int back; DataType* dataItems; }; #endif Queue.h #ifndef QUEUE_H #define QUEUE_H #include #include using namespace std; #pragma warning( disable : 4290 ) //-------------------------------------------------------------------- template class Queue { public: static const int MAX_QUEUE_SIZE = 8; virtual ~Queue(); virtual void enqueue(const DataType& newDataItem) throw (logic_error) = 0; virtual DataType dequeue() throw (logic_error) = 0; virtual void clear() = 0; virtual bool isEmpty() const = 0; virtual bool isFull() const = 0; #if LAB7_TEST2 virtual void putFront(const DataType& newDataItem) throw (logic_error) = 0; virtual DataType getRear() throw (logic_error) = 0; #endif #if LAB7_TEST3 virtual int getLength() const = 0;
- 11. #endif virtual void showStructure() const = 0; }; template Queue::~Queue() // Not worth having a separate class implementation file for the destuctor {} #endif // #ifndef QUEUE_H