C++ CodeConsider the LinkedList class and the Node class that we s.pdf

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C++ Code Consider the LinkedList class and the Node class that we saw in lecture. The LinkedList class basically represents a singly linked list. Now, for this lab, consider a doubly linked list, i.e. each node has a next and previous pointers to the next and previous node, respectively. Modify the Node class to reflect this change and implement the following methods for DoublyLinkedList class: addFirst addLast addAtIndex removeFirst removeLast removeAtIndex Note: We saw how to implement these methods for a singly linked list and the code is added as an attachment. C++ Code ***************************************************************************** ****************** ***************************************************************************** ****************** C++ Code Solution #include #include using namespace std; template class Node { public: T element; // Element contained in the node Node *next; // Pointer to the next node Node *previous; Node(T element) // Constructor { this->element = element; next = NULL; } }; template class LinkedList { public: LinkedList(); void addFirst(T element); void addLast(T element); T removeFirst() ; T removeLast(); void addAtIndex(int index, T element); T removeAtIndex(int index); void traverseList(); private: Node *head; Node *tail; int size; }; template LinkedList::LinkedList() { head = NULL; tail = NULL; size = 0; } template void LinkedList::addFirst(T element) { Node *newNode = new Node(element); newNode->next = head; if(head != NULL) { head->previous = newNode; } newNode->previous = NULL; head = newNode; size++; if (tail == NULL) tail = head; } template void LinkedList::addLast(T element) { if (tail == NULL) { head = tail = new Node(element); } else { tail->next = new Node(element); tail = tail->next; } size++; } template void LinkedList::addAtIndex(int index, T element) { if (index == 0) addFirst(element); else if (index >= size) addLast(element); else { Node *current = head; for (int i = 1; i < index; i++) current = current->next; Node *temp = current->next; current->next = new Node(element); (current->next)->next = temp; size++; } } template T LinkedList::removeFirst() { if (size == 0) exit(1); else { Node *temp = head; head = head->next; size--; T element = temp->element; delete temp; return element; } } template T LinkedList::removeLast() { if (size == 0) exit(1); else if (size == 1) { Node *temp = head; head = tail = NULL; size = 0; T element = temp->element; delete temp; return element; } else { Node *current = head; for (int i = 0; i < size - 2; i++) current = current->next; Node *temp = tail; tail = current; tail->next = NULL; size--; T element = temp->element; delete temp; return element; } } template T LinkedList::removeAtIndex(int index) { if (index < 0 || index >= size) exit(1); else if (index == 0) return removeFirst(); else if (index == size - 1) return removeLast(); else { Node *previous = head; for (int i = 1; i < index; i++) { previous = previous->next; } Node *current = previous-.

Education

Node(T element) // Constructor
{
this->element = element;
next = NULL;
}
};
template
class LinkedList
{
public:
LinkedList();
void addFirst(T element);
void addLast(T element);
T removeFirst() ;
T removeLast();
void addAtIndex(int index, T element);
T removeAtIndex(int index);
void traverseList();
private:
Node *head;
Node *tail;
int size;
};
template
LinkedList::LinkedList()
{
head = NULL;
tail = NULL;
size = 0;
}
template
void LinkedList::addFirst(T element)
{
Node *newNode = new Node(element);
newNode->next = head;

if(head != NULL) {
head->previous = newNode;
}
newNode->previous = NULL;
head = newNode;
size++;
if (tail == NULL)
tail = head;
}
template
void LinkedList::addLast(T element)
{
if (tail == NULL)
{
head = tail = new Node(element);
}
else {
tail->next = new Node(element);
tail = tail->next;
}
size++;
}
template
void LinkedList::addAtIndex(int index, T element)
{
if (index == 0)
addFirst(element);
else if (index >= size)
addLast(element);
else
{
Node *current = head;
for (int i = 1; i < index; i++)
current = current->next;
Node *temp = current->next;
current->next = new Node(element);

(current->next)->next = temp;
size++;
}
}
template
T LinkedList::removeFirst()
{
if (size == 0)
exit(1);
else
{
Node *temp = head;
head = head->next;
size--;
T element = temp->element;
delete temp;
return element;
}
}
template
T LinkedList::removeLast()
{
if (size == 0)
exit(1);
else if (size == 1)
{
Node *temp = head;
head = tail = NULL;
size = 0;
T element = temp->element;
delete temp;
return element;
}
else
{
Node *current = head;

for (int i = 0; i < size - 2; i++)
current = current->next;
Node *temp = tail;
tail = current;
tail->next = NULL;
size--;
T element = temp->element;
delete temp;
return element;
}
}
template
T LinkedList::removeAtIndex(int index)
{
if (index < 0 || index >= size)
exit(1);
else if (index == 0)
return removeFirst();
else if (index == size - 1)
return removeLast();
else {
Node *previous = head;
for (int i = 1; i < index; i++)
{
previous = previous->next;
}
Node *current = previous->next;
previous->next = current->next;
size--;
T element = current->element;
delete current;
return element;
}
}
template
void LinkedList::traverseList()

{
Node *temp = head;
while(temp !=NULL) {
cout<element<<" ";
temp = temp->next;
}
}
int main() {
LinkedList *myList = new LinkedList();
myList->addFirst(10);
myList->addFirst(20);
myList->addLast(30);
myList->removeAtIndex(1);
myList->traverseList();
delete myList;
}

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Foofy obtained a significant Fobt from an experiment with five level.pdf

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Explain how water interacts with the following types of compounds to influence their water solubility. 1.Compounds that can hydrogen bond 2. Compounds that are polar but cannot hydrogen bond 3. Compounds that are non-polar 4. Compounds that dissociate into cations and anions. Thank you so much. Solution Compounds that can hydrogen bond Water tends to disassociate into H+ and OH- ions. In this disassociation, the oxygen retains the electrons and only one of the hydrogens, becoming a negatively charged ion known as hydroxide. Pure water has the same number (or concentration) of H+ as OH- ions. Acidic solutions have more H+ ions than OH- ions. Basic solutions have the opposite. An acid causes an increase in the numbers of H+ ions and a base causes an increase in the numbers of OH- ions. Water, which not only dissolves many compounds but also dissolves more substances than any other liquid, is considered the universal solvent. A polar molecule with partially-positive and negative charges, it readily dissolves ions and polar molecules. Water is therefore referred to as a solvent: a substance capable of dissolving other polar molecules and ionic compounds. The charges associated with these molecules form hydrogen bonds with water, surrounding the particle with water molecules. This is referred to as a sphere of hydration, or a hydration shell, and serves to keep the particles separated or dispersed in the water. Compounds that dissociate into cations and anions.-When ionic compounds are added to water, individual ions interact with the polar regions of the water molecules during the dissociation process, disrupting their ionic bonds. Dissociation occurs when atoms or groups of atoms break off from molecules and form ions. Consider table salt (NaCl, or sodium chloride): when NaCl crystals are added to water, the molecules of NaCl dissociate into Na+ and Cl– ions, and spheres of hydration form around the ions. The positively-charged sodium ion is surrounded by the partially-negative charge of the water molecule\'s oxygen; the negatively-charged chloride ion is surrounded by the partially-positive charge of the hydrogen in the water molecule. Compounds that are polar and non-polar- -Since many biomolecules are either polar or charged, water readily dissolves these hydrophilic compounds. Water is a poor solvent, however, for hydrophobic molecules such as lipids. Nonpolar molecules experience hydrophobic interactions in water: the water changes its hydrogen bonding patterns around the hydrophobic molecules to produce a cage-like structure called a clathrate. This change in the hydrogen-bonding pattern of the water solvent causes the system\'s overall entropy to greatly decrease, as the molecules become more ordered than in liquid water. Thermodynamically, such a large decrease in entropy is not spontaneous, and the hydrophobic molecule will not dissolve. -Polar substances tend to dissolve well in other polar substances, but not nonpolar substances, whil.

Explain how water interacts with the following types of compounds to.pdf

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Explain how the financial reporting of gains and losses on assets differs from the financial reporting of unrealized gains and losses on martketable securities. Explain how the financial reporting of gains and losses on assets differs from the financial reporting of unrealized gains and losses on martketable securities. Solution Particulars Gains and Losses on Assets unrealized gains and losses on martketable securities Actual or Not Actual It is actual or real in nature. It is differrence between the values of securities Income Tax It is taxable / deductible under income tax It is not considered under income tax Cash flow There is cash inflow There is no any cash flow Realisation It is realised gain or loss It may be realised or not. It is unrealised.

Explain how the financial reporting of gains and losses on assets.pdf

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Explain two ways in which microbiologists have overcome the difficulty of viewing microbes in wet preparations Solution First we have to find two ways in which microbiologists have overcome the difficulty of viewing microbes in wet preparations are a bacteria which has the intrinsic ability of movement in the surrounding medium in which it remains suspended is a motile bacteria. A bacteria which does not have the intrinsic ability of movement in the surrounding medium in which it remains suspended is a non-motile bacteria. In wet mount though the shape and size of bacteria can be observed motility may be hampered as the suspension is pressed between the slide and the cover slip. Motility test is performed for clear observation of the motility of bacteria, besides their shape and size. The hanging drop is observed under a microscope using oil immersion objective its cells can be seen to have erratic movement in the surrounding medium. If it is non motile its cells remain static in the medium without any movement or may show brownian movement resulting from the bombardment by the water molecules in the medium..

Explain two ways in which microbiologists have overcome the difficul.pdf

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During the preparation of the bank reconciliation for Building Concepts Co., Joel Kimmel, the assistant controller, discovered that Lone Peak National Bank incorrectly recorded a $3,290 check written by Building Concepts Co. as $329. Joel has decided not to notify the bank but wait for the bank to detect the error. Joel plans to record the $2,961 error as Other Income if the bank fails to detect the error within the next three months. Discuss whether Joel is behaving in a professional manner. Solution Joel is not behaving in a professional manner, Joel record differece amount ($2961) as other income, this treatment is wrong because that amount can not be include in income of Building concepts Co..

During the preparation of the bank reconciliation for Building Conce.pdf

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DNA Paternity Testing Simulation Post-lab Questions A)Why do different individuals such as siblings have different restriction enzyme recognition sites? B) What is the function of probes in DNA praternity analysis? C) Why is there more than one single locus probe used in an actual paternity DNA test? D)Why do we not use probes in this DNA paternity simulation and still obtain results? Solution A)Why do different individuals such as siblings have different restriction enzyme recognition sites? The different individuals such as siblings have different restriction enzyme recognition because the size of the gene fragments are mainly relies upon the “distance between the recognition sites”. This is due to polymorphism i.e. single nucleotide polymorphism (SNPs) with intervening repetitive base sequences in siblings. The paternal parents are goiung to inherit genetic variability into the sibling due to gametes reduction division & crossing over B) What is the function of probes in DNA praternity analysis? The functions of DNA probes are meticulously to identify “length differences between these intervening repetitive gene sequences via isotopic & non-isotopic labeling by the use of “single stranded DNA finally these DNA probes are going to complement and hybridize (attach) to single stranded DNA due to SNP during paternity analysis. C) Why is there more than one single locus probe used in an actual paternity DNA test? More than single locus DNA probe or Multiple- Locus probes are used to identify “multiple intervening repetitive DNA sequences” located on many chromosomes result in generation of 20 to 30 bands so that is possible to avoid chances of same pattern due to the multi-band patterns,.

DNA Paternity Testing Simulation Post-lab QuestionsA)Why do differ.pdf

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Describe how the lung cancer travels to the first site of metastasis. •Given your primary tumor site, how did lung cancer cells of the primary tumor spread to the initial secondary site? How does lung cancer cell get access to the body during metastasis? What pathway did your cancer cells’ take? Solution answer: metastasis: delopment of secondry malignant growth at a distance from primary site of cancer lung cancer tumor travel process: primary tumor willl be prolification of the clone and invesion of vessel through the metastasis clone evolves and its transport by circulation then its leads to the embolisam finally new tumor formation at the site of metestasis.

Describe how the lung cancer travels to the first site of metastasis.pdf

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Define the following engineering terms: 1. engineering ethics 2. Whistleblowing 3. Responsibility of engineers 4. recent developments 5.general principles 6 obligation of society 7. conduct 8. case studies and key individuals Solution 1. Engineering Ethics - Engineering Ethics is the study of moral issues and decisions confronting individuals and organizations engaged in engineering.The field examines and sets the obligations by engineers to society, to their clients, and to the profession. 2. Whistleblowing - The duty of an engineer to report to the concerned authority about a possible risk to others when a client or employer fails to follow the engineer\'s directions. 3. Responsibility of engineers - The responsibility of an engineer to commit to serving society, attending to the welfare and progress of the majority. The engineer should reject any project that is intended to harm the general interest, thus avoiding a situation that might be hazardous or threatening to the environment, life, health, or other rights of human beings. 4. recent developments - Recent efforts to uphold engineering ethics include addressing of bribery and political corruption by several professional societies and business groups around the world while getting ready to face challenges like offshoring, sustainable development, and environmental protection. 5. general principles - General guidelines set by various engineering professional societies which tend to include 6. obligation of society - The paramount value recognized by engineers is the safety and welfare of the public. All the major engineering societies around the world. These societies include Institute of Electrical and Electronics Engineers, Institution of Civil Engineers, American Society of Mechanical Engineers, Institute of Industrial Engineers etc. 7. conduct - Several ethical issues faced by the engineers include considerations of business conduct. The issues faced are relationships with clients, ensuring legal compliance by clients, treatment of confidential or proprietary information, conflict of interest etc. 8. case studies and key individuals - The study of engineering failures in the last century indicates that not all engineering failures involve ethical issues. Many failures involve the failure of the design process or management culture. The following mentioned failures include ethical as well as technical issues..

Define the following engineering terms1. engineering ethics2. W.pdf

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Data structuresUsing java language and develop a prot.pdf

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You have been cut! While you wage war in the streets of Gotham for your life, your cells wage war against incoming invaders. Place the following statement concerning the fates of unwanted invaders/immigrants in your body in order: 1.The invader is cut apart while alive and digested in the harsh environment of your new super- max detention/recycling center. Die invader, Die! 2.Contact with one of your cells triggers phagocytosis and a signal trough transmembrane proteins triggers the cytoskeleton. 3.The vesicle containing the bacterium is pinched off. 4.A bacterium that has stormed the outter walls of your body (the skin through the cut) is recognized and tagged as foreign. 5.Other vesicles containing hydrolases, enzmes that depolymerize proteins, carbohydrates, lipids, and nucleic acids, as well as other proteins that act to increase the pH of the larger vesicle that contains the invader arrive on the cell\'s microtubule transportation system and fuse. 6.Kinesin walks its vesicle down the microtubule highways into the cytoplasm. 7.Actin polymerizes to form rods that push out the membrane to form fingers or pseudopods that begin to envelop the bacterium. 8.The vesicle containing the bacterium is moved by kinesin. Solution 4A bacterium that has stormed the outter walls of your body (the skin through the cut) is recognized and tagged as foreign. 2Contact with one of your cells triggers phagocytosis and a signal trough transmembrane proteins triggers the cytoskeleton. 7Actin polymerizes to form rods that push out the membrane to form fingers or pseudopods that begin to envelop the bacterium. 3 The vesicle containing the bacterium is pinched off. 8 The vesicle containing the bacterium is moved by kinesin 6 Kinesin walks its vesicle down the microtubule highways into the cytoplasm 5 .Other vesicles containing hydrolases, enzmes that depolymerize proteins, carbohydrates, lipids, and nucleic acids, as well as other proteins that act to increase the pH of the larger vesicle that contains the invader arrive on the cell\'s microtubule transportation system and fuse 1 The invader is cut apart while alive and digested in the harsh environment of your new super- max detention/recycling center..

You have been cut! While you wage war in the streets of Gotham for y.pdf

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Why has Mexico encountered such a significant collapse from an economic perspective? Solution From an economic perspective Mexico had a significant collapse due to two major factors: 1. The first one was a collapse in export demand 2. The second one being the susequent constraints that it had in accessing the International Financial markets. Due to the above factors Mexico started feeling a significant negative effect. As the negative effects increased, the country\'s fiscal position weakened. This was mainly due to the fall in oil revenues. Additionally, a drop in international energy prices brought about by the global recession along with a sharp decline in domestic oil production attributed to the crisis. A downward rigidity in public expenditure was observed which along with the above mentioned factors raised concerns about the sustainability of fiscal policy.All these factors played a major spoilsport and generated doubts about the country’s capacity to finance the growth in the current account deficit for 2009. The Mexican economy was forced to adjust to a new external environment that was characterised by lower foreign currency revenues with a limited access to external borrowing..

Why has Mexico encountered such a significant collapse from an econom.pdf

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Which of the following is/are true of BOTH natural selection and genetic drift? I. Promotes traits that increase survival or reproduction II. Depends on genetic mutations as the source of variation III. Can lead to evolutionary change Question 5 options: I and III I only I, II, and III II and III I and II I and III I only I, II, and III II and III I and II Solution answer will be I,II and III, because in a population genetic drift and natural selection always play a role together..

Which of the following isare true of BOTH natural selection and gen.pdf

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Which of the following information-management systems uses artificial intelligence? A. Polyinstantiation B. Known signature scanning C. Application programming interface D. Knowledge discovery in databases Solution D. Knowledge Discovery in Databases (KDD) is an artificial intelligence method used to identify useful patterns in data; as such, it provides a type of automatic analysis. Answer A is incorrect because polyinstantiation is a technique used to prevent inference violations. Answer B is incorrect because known signature scanning is a method used to detect computer viruses. Answer C is incorrect because the application programming interface (API) is in no way associated with artificial intelligence..

Which of the following information-management systems uses artificia.pdf

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What steps and tools are used for Pen Testing? Solution Steps for Pen Testing: 1:Attack Surface analysis 2:Threat Modelling 3:Test Configuration 4:automatic and manual scans 5:validation of vulnerabilities 6:false positive elimination 7:proof of concept 8:Check for the service/version running on the particular port. 9:Search the vulnerability in the service/version. 10:Exploit the target using tools. Tools used for Pen Testing 1:Metasploit 2:Wireshark 3:W3af 4:core Impact 5:Back track.

What steps and tools are used for Pen TestingSolutionSteps fo.pdf

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which event in this life cycle is comparable to fertilizarion? 9. Distinguish among the various groups of fungi as outlined in the table below: Asexual Spores Fruiting Body Structure (microscopic) Giving sexual stage) Rise to Sexually Produced Spore (large; asexually produced spores from sporangium. Zygomycota Ascomycota Basidiomycota Solution Structure Fruiting body Asexual spores Zygomycetes Chitin,chitosan,polyglucornic acid No fleshy fruiting body reproduce by conjugation, sporangiospores, conidia Ascomycetes Chitin,both alpha and beta Glucans,glactomannoproteins ascus asexually produced spores from sporangium, ascospores, conidia Basisiomycetes Chitin,beta Glucans basidia contain basidium, the spore forming bodies, basidiospores,conidia.

which event in this life cycle is comparable to fertilizarion .pdf

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what is the value of nursing education Solution The main value of nursing education is to deliver efficient ethical based health care to the patients suffering from various devastating disease finally to implement clinical protocol & clinical trials on the patients’ with moral & ethical aspects to collect epidemiological clinical data. As a clinical nurse leader is to obtain a better clinical outcome is main value of \"nursing education\" by respecting patent’s autonomy. The main characteristics are to develop a strategic plan to implement a “do” and “step “policy to active organizational goals, allocation of resources as a clinical nurse leader. The other aspects of nursing education is to implement the treatment protocols, epidemiological studies of disease, and disorders within time- frame using interdisciplinary teams and according ethical based informatics. The responsibilities are allocated to each RN, LPN to obtain an efficient health out come from patient and within the community. Nursing education is crucial to inform the fellow nurses about the current organizational philosophy of nursing in clinical services. Finally, collecting the patient data and monitoring, analysis & evaluation is essential using nursing informatics to develop future prospects to prevent diseases. Nursing education has specific impact on the role of health care workers and clinicians to maintain efficient patient’s electronic health care records using information technology tools. In order to automate the aspects of acquisition of patient health information and management.

what is the value of nursing educationSolutionThe main value o.pdf

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What is the exact output of the following code#include stdio.h.pdf

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Using a Fabry-Perot interferometer, we can Observe small objects. Detect fine optical spectrum lines. Examine optical quality of lens. Measure light speed in different media. Solution ANSWER is : B) Detect fine optical spectrum lines It consists of two highly reflective and strictly parallel plates called an etalon. Because of the high reflectivity of the plates of the etalon, the successive multiple reflections of light waves diminish very slowly in intensity and form very narrow, sharp fringes. These may be used to reveal hyperfine structures in line spectra..

Using a Fabry-Perot interferometer, we can Observe small objects. D.pdf

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TRUE of falseA researcher found 95 confident interval for the aver.pdf

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The major type of interactive forces between molecules of HCl are: ionic covalent hydrogen bonding dipole-dipole attractions dispersion forces Solution Dipole - Dipole attraction is the major type of interactive forces between HCl molecules. these interaction are basically the electrostatic interaction between the permanent dipoles in the mlecules. this interaction occurs when the partially negatively charged part of one molecule (in case of HCl it is Cl- ) interacts with partially positive charge of the other molecule (H+ in case of HCl)..

The major type of interactive forces between molecules of HCl are i.pdf

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