5. In determining the centroid, the distribution is depended on ___.pdf

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5. In determining the centroid, the distribution is depended on: ______ A. Geometry only B. Geometry and density of material C. Geometry, density of material and presence of gravity D. Density of material and presence of gravity Solution As centroid is a geometrical property defined as average position of all the points in a shape provided it has a unifom density and a unifom gravitational field. So the answer is (C) Geometry, density of material and presence of gravity.

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What is the role of intuition in decision-making? Solution Intuition is ability to acquire the knowledge, information and skills without any forceful activity; intuition is something which is come from inner sense that provoked itself. Decision- making process always influence by the intuition because decision making is a kind of a practices which is not founded in every one but mostly founded in leaders ,managers and businessmen. Intuition always linked with emotions, intuition linked to knowledge, intuition always fast and processing of information. Role of intuition in decision-making as follows-.

What is the role of intuition in decision-makingSolutionIntui.pdf

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What is the difference between a HTML element’s id attribute and name attribute? Solution The form input element id has nothing to do with the data contained within the element. Input element ids are for hooking the element with JavaScript and CSS. The name attribute is used in the HTTP request sent by your browser to the server as a variable name associated with the data contained in the value attribute. In other words, By using id attribute we can validate and manipulate the value of an element at client side. By using name attribut, we can validate and manipulate the value of an element at server side. In java script, by using id attribute we can get the value of input element. document.getElementById(\"id\").value; In server side, by using name attribute we can get the value of input element. request.getParameter(\"name\");.

What is the difference between a HTML element’s id attribute and nam.pdf

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What happens when you\'re Running Scripts without a console in Linux/UNIX. Solution The script will run as a background process that is daemon process. These daemon processes doesnot interact with the user. Console is used to send text output and receive user input and is under the direct control of the user. However daemon process is run in the background by forking a child process and the parent is exited immediately..

What happens when youre Running Scripts without a console in Linux.pdf

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True/False A virtual private network is a way to use the Internet to create a connection between a remote user and a central location. IPsec encrypts the packet data but not the header information. L2TP uses IPsec for its encryption. Kerberos is an authentication protocol used for logging into Windows and uses a ticket granting system that sends an encrypted ticket to the user\'s machine. A digital signature is used to guarantee who sent a message. This is referred to as authenticity. Public key encryption is fast becoming the most widely used type of encryption because there are no issues to deal with concerning distribution of keys. PGP involves private key encryption. The most widely used symmetric key algorithm is Advanced Encryption Standard. Cipher text is encrypted text. Binary numbers are made up of 0s and 1s. Solution Answer: 1) True explanation: => A VPN, or Virtual Private Network, allows you to create a secure connection to another network over the Internet. 2)true explanation: =>IPsec includes Encapsulating Security Payload (ESP) to encrypt the data and provide confidentiality 3) TRUE explanation: => L2TP does not encrypt the tunnel itself but often uses IPsec for encryption. 5) True explanation: => Digital signatures are the public-key primitives of message authentication. => A valid digital signature gives a recipient reason to believe that the message was created by a known sender. 6) True explanation: => Public key encryption in turn provides a solution to key distribution and data transmission issues. Used together, performance and key distribution are improved without any sacrifice in security. 7) False. 8) True explanation: => Encryption algorithms that use the same key for encrypting and for decrypting information are called symmetric- key algorithms. =>The symmetric key is also called a secret key because it is kept as a shared secret between the sender and receiver of information. 9) true explanation: =>An encryption function takes cleartext and a key as input and returns ciphertext. 10) TRUE explanation: =>A binary code represents text or computer processor instructions using the binary number system\'s two binary digits, 0 and 1..

TrueFalse A virtual private network is a way to use the Internet to.pdf

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The \"total physical units accounted for\" is the sum of the units completed and transferred out during the period plus the True ? False Solution Total physical units accounted for = Units of completed and transferred out+Ending work in process inventory Units So above statement that the total physical units accounted for is the sum of the units completed and transferred out during the period plus the units in beginning wip inventory is False So above statement is False.

The total physical units accounted for is the sum of the units co.pdf

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The problem can be found below. I would like someone to run this. I don’t get it because there’s an error. Please, fix this because I don’t know what is happening. What version of python should I use? Please, if you don’t know what is happening let someone else to do it. Bp Project untitled PycharmProjects/untitled 3 import sys ServerPort 80 new server Socket socket (AF INET, S0CK STREAM) Illi External Libraries 6 #Prepare a sever socket 7 #Fill in start server Socket.bind server Port)) server Socket.listen (1) print the web server is up on port server Port 10 #Fill in end 11 while True: Establish the connection 13 print \"Ready to serve... 14 connectionSocket add J server Socket. accept() 15 16 try: message connectionSocket.recv(1024) 17 print message message split 01, message. split 1] 18 filename 3 message. split [1] print filename \'II filename [1:] 20 open (filename (1:1) 21 output data f.read 22 print outputdata 23 Send one HTTP header line into socket Fill in start 25 connection Socket.send(\' InHTTP/1.1 200 0KNnNn 26 connectionSocket.send(outputdata) 27 Fill in end 28 Send the content of the requested file to the client 29 for i in range (0, len(outputdata)): 30 connection Socket.send(outputdata [i]) 31 connection Socket.close 32 33 except IOError w Send response message for file not found 34 Fill in start 35 connectionSocket.send NnHTTP/1.1 404 Not Found nAn 36 connectionSocket.send \'InHTTP/1.1 404 Not Found n\ \') 37 38 39 /Library/Frameworks/Python, framework/Versions/2.7/bin/python2.7 /Users. File User S untitled new py\", line 33 except IOError: SyntaxError: invalid syntax Process finished with exit code 1 Solution Update your code as below : #!/bin/python import sys import socket serverPort=80 serverSocket = socket.socket(socket.AF_INET,socket.SOCK_STREAM) serverSocket.bind((\'\',serverPort)) serverSocket.listen(1) while True : connectionSocket, addr = serverSocket.accept() try : message = connectionSocket.recv[1024] filename = message.split()[1] f = open(filename[1:]) outputdata = f.read() connectionSocket.send(\'\ HTTP/1.1 200 OK\ \ \') connectionSocket.send(outputdata) for i in range(0,len(outputdata)): connectionSocket.send(outputdata[i]) connectionSocket.close() except IOError : #indentation is not correct here connectionSocket.send(\'\ HTTP/1.1 404 Not Found\ \ \') serverSocket.close();.

The problem can be found below. I would like someone to run this. I .pdf

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The probability that a teacher will give an unannounced test during any class is 1/5. If a student is absent twice ,then probability that misses atleast one test is ? Solution Given the probability of teacher will give an unannounced test during any class is 1/5. The probability of not missing test 1-1/5=4/5 probability of missing test in only one day 2(4/25)=8/25 probability of missing test on both the day (1/5)*(1/5)=1/25 probability that misses atleast one test is (8/25)+(1/25)=9/25 probability that misses atleast one test is 9/25.

The probability that a teacher will give an unannounced test duri.pdf

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The Orlando MedicalOrlando Medical Corporation financial statements? follow: LOADING... ?(Click the icon to view the consolidated balance? sheets.) LOADING... ?(Click the icon to view the consolidated income? statements.) Orlando MedicalOrlando Medical?\'s inventory turnover during fiscal year 20142014 was A. 8484 times. B. ?$35 comma 16435,164. C. very slow. D. 138.0138.0 times. Orlando Medical Corporation Consolidated Statements of Financial Position December 31, (In Millions) 2014 2013 Assets: Current assets Cash and cash equivalents $4,344 $4,237 Accounts and notes receivable 3,402 2,409 Short-term investments 855 504 Inventories, at cost 435 404 Prepaid expense and other current assets 1,638 1,220 Total current assets 10,674 8,774 Property and equipment, net 1,560 933 Investments 6,804 5,300 Other non-current assets 301 160 Total assets $19,339 $15,167 Liabilities and stockholder\'s equity: Current liabilities Accounts payable $7,703 $6,002 Accrued and other liabilities 3,787 3,011 Total current liabilities 11,490 9,013 Long-term debt 305 306 Other non-current liabilities 1,687 1,177 Total liabilities 13,482 10,496 Stockholders’ equity Preferred stock and capital in excess of $0.02 par value; shares issued and outstanding: none - - Common stock and capital in excess of $0.05 par value; shares authorized: 6,000; shares issued: 2,112 and 1,852, respectively 7,804 7,001 Treasury stock, at cost; 184 and 124 shares, respectively (6,333) (4,401) Retained earnings 4,516 2,162 Other comprehensive loss (98) (49) Other (32) (42) Total stockholders’ equity 5,857 4,671 Total liabilities and stockholders’ equity $19,339 $15,167 PrintDone Orlando Medical Corporation Consolidated Statements of Income Year ended December 31, (In Millions, Except per Share Amounts) 2014 2013 2012 Net revenue $42,667 $35,613 $31,124 Cost of goods sold 35,164 29,270 25,478 Gross profit 7,503 6,343 5,646 Operating expenses: Selling, general, and administrative 3,747 3,275 2,600 Research, development, and engineering 584 560 534 Special charges - - 503 Total operating expenses 4,331 3,835 3,637 Operating income 3,172 2,508 2,009 Investment and other income (loss), net 195 194 (78) Income before income taxes 3,367 2,702 1,931 Income tax expense 1,100 912 475 Net income $2,267 $1,790 $1,456 Earnings per common share: Basic $1.24 $0.94 $0.41 Solution Answer : A. 84 times => Inventory t/o = COGS/average inventoryYear20142014Cost of Goods Sold3516435164Average Inventory(435+404)/2419.5Inventory Turnover83.82.

The Orlando MedicalOrlando Medical Corporation financial statements.pdf

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The IP network is a virtual network and must rely on a link layer network to transport a frame. What elements support getting an IP frame over an Ethernet LAN? Solution The data link layer (L2) is the second layer of the seven-layer OSI model of computer networking. This layer is the protocol layer that transfers data between adjacent network nodes in a wide area network (WAN) or between nodes on the same local area network (LAN) segment. The data link layer, also sometimes termed as the link layer, is where many wired and wireless local area networking (LAN) technologies primarily function. Ethernet is the most widely used local area network (LAN) in metropolitan area networks (MAN). Ethernet is a link layer protocol in the TCP/IP stack. It describes how networked devices can format data for transmission to other network devices on the same network segment, and how to put that data out on the network connection. It was commercially introduced in 1980 and first standardized in 1983 as IEEE 802.3, and has since been refined to support higher bit rates and longer link distances. The data link layer is divided into two sublayers: logical link control (LLC) and media access control (MAC).By separating LLC and MAC functions, interoperability of different network technologies is made easier. Data Link Layer Functions: Given below are the key tasks performed at the data link layer: Logical Link Control (LLC): Logical link control refers to the functions required for the establishment and control of logical links between local devices on a network. It provides services to the network layer above it and hides the rest of the details of the data link layer to allow different technologies to work seamlessly with the higher layers. Most local area networking technologies use the IEEE 802.2 LLC protocol. Media Access Control (MAC): This refers to the procedures used by devices to control access to the network medium. Since many networks use a shared medium, it is necessary to have rules for managing the medium to avoid conflicts. Ethernet uses the CSMA/CD method of media access control, while Token Ring uses token passing. Data Framing: The data link layer is responsible for the final encapsulation of higher-level messages into frames that are sent over the network at the physical layer. Addressing: The data link layer is the lowest layer in the OSI model that is concerned with addressing: labeling information with a particular destination location. Each device on a network has a unique hardware address or MAC address imbined with the hardware i.e present in network interface card (NIC) to uniquely identify the device. Error Detection and Handling: The data link layer handles errors that occur at the lower levels of the network stack. For example, a cyclic redundancy check (CRC) field is often employed to allow the station receiving data to detect if it was received correctly. The hardware address is used for the MAC sublayer addressing. Ethernet uses the.

The IP network is a virtual network and must rely on a link layer ne.pdf

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Table 1 Endowment of Labor and Capital US 100 20 Canada 10 Workers Machines 1. Based on Table 1, according to the Heckscher-Ohlin Theorem, Canada exports should be goods that A) intensively use labor input. B) intensively use capital input. C) use capital and labor in about equal proportions D) use either labor or capital input, depending on the good. 2. Suppose that Brazil is capital abundant and Chile is natural resource abundant. If timber is natural resource intensive and computers are capital intensive, then according to the Stolper Samuelson Theorem, the incomes of the owners of trade with Brazil begins A) capital B) labor C) natural resources D) It is impossible to determine which will be favored. are likely to rise in Chile after Solution 1) According to Heckscher Ohlin Theorem, A country will export the commodity in which it uses its intensively endowed factor. Since Canada has more machine per worker compared to US. Thus Canada is more capital intensive and will export the good that intensively uses capital input. ans is B. 2) Since Brazil is capital abundant, and computer are capital intensive then Brazil will export computer and Chile will export Timber. Thus income of natural resources are likely to rise in Chile Thus ans is C.

Table 1 Endowment of Labor and Capital US 100 20 Canada 10 Workers Ma.pdf

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State if you agree or disagree with the statement made and why? Lesson #1 - Discussion Forum – Linkage between Public Administration and Information Technology The article “Computing and Public Organizations” by Kraemer and King is an interesting look back at the prevailing questions and concerns thirty years ago about the potential impact of the use of computers in public organizations. The main point of the authors\' discussion is whether a wide-spread organizational use of computers would affect the centralized or decentralized nature of an organization. While the authors conclude that any move, either way, would involve more factors than just the use of computers, they do seem to come down on the side of computer use increasing the centralization of organizational decision making. When considering the impact of computer use on the centralization of decision making in public organizations as it stands today, thirty years after the article was written, it is still difficult to make a definitive statement on whether organizations have become more or less centralized due to the use of information technology. In some cases, data gathering both with and without IT is so decentralized throughout an agency that a desired level of centralization is never achieved. A case in point is the Veterans Administration, where data from field offices were either inefficiently or never sent to regional offices that could make service-related decisions, an outcome of decentralization that had very negative consequences for the veterans that are supposed to be the focus of the agency. Other government agencies also suffer from having so many points of entry to a centralized system that the public is uncertain how to access the information they need. An example would be the military’s health insurance program, which has three or four portals that each accomplish a different task within the overall program. A strong and well-organized linkage between public agencies and information technology is a deciding factor in how beneficial the use of technology is to public organizations. Solution I disagree with the fact the IT helps in the centralization of decision making as well as with the point that it is difficult to determine whether IT helps in decentralization or centralization. I believe IT has helped in the decentralization of decision making especially in public administration. The banking sector is a very good example. Centralization is where the distance- between where the problems arise and where the solution is decided- is greater. In decentralization, this distance is smaller and decisions are solved closer to the place where the problems arise. Decision making depends upon many factors- like access to information, knowledge, experience, skills, abilities, talent etc. With the inflow of technology into the hands of everyone, data collection, gathering, assortment and analysis has become easier. Data is converted into meaningful information and then into.

State if you agree or disagree with the statement made and whyLes.pdf

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subject.....project management. Tools and Processes Based on the progressive case study (PCS) project, you will need to develop a list of tools and processes that you would use as the contractor of record for opening the new division office. •Will you use subcontractors for different parts of the project? •How will you maintain contact communication? •How will you track progress, milestones, and setbacks? •How will you monitor resources and budget? •What reporting systems will you use? You are not expected to address all of these areas, though this is not an exhaustive list. Choose one or two components and describe the tools and processes you will use, then explain the rationale for using these tools. You may choose to use a certain tool as that is where your experience is, or it is more cost effective for example. not less than 150 words A project team requires individuals from the cross-functional area who can come and work together on different aspects of the project. In this regard, a project based organizational structure will be developed that will be flat in nature. The first parameter of the selection of the team member is the ability to work in the respective areas of the project and or the past experience of working on similar projects. The second parameter is the ability to manage a dual line of command (line and project) while working on the project. The third parameter is the propensity to work in a team and complement each other so that project work runs smoothly. The fourth parameter is the ability to take decisions with the given authority so that project work is not delayed due to the decision-making process. Besides, there should be guidelines related to the scope, objectives, time-based milestones as well as closure of the project. It should be based on the ethical principles. Regarding the challenges, the first challenge is the procurement of resources to build capacity as no any department or function wants to release the best employees. The second challenge is the building of the clarity in command so that any ambiguity is eliminated. The third challenge is the timely availability of resources and funds other than HR to start the project. The next challenge is to build the cooperation based team culture in the project team Solution TRACKING PROGRESS, MILESTONES, AND SETBACKS: A contractor needs to keep a close check on the progress of the project on which he is working. Project management is a difficult and very important task and requires skills and expertise of the contractor so as to make sure that the project is working as planned for. Keeping a track on the progress of the project is must from time to time and making sure that the milestones are achieved properly and if in case of any setbacks, recording why it happened. Following are the list of the tools that can be used to track the progress, milestones and setback of the project: MONITORING RESOURCES AND BUDGET: For successful completion of every project it is .

subject.....project management.Tools and Processes Based on the pr.pdf

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Read and discuss the following three articles: 1. ACAs Performance Based Healthcare Standards ACAsPerformanceBasedHealthCareStandards.pdf 2. Road to Accreditation RoadToAccreditation.pdf 3. JCAHO Accreditation and Quality of Care for Acute Myocardial Infarction JCAHO accreditation and quality of care for acute myocardial infarction.pdf Have an open discussion about these articles. Share your thoughts For example, here are some questions to answer and discuss: Does accreditation impact quality? Are there less errors in hospitals that are accredited? What is the value of accreditation? Do quality concerns initiate changes in staff behavior? Should accreditation be based on results? Read and discuss the following three articles: 1. ACAs Performance Based Healthcare Standards ACAsPerformanceBasedHealthCareStandards.pdf 2. Road to Accreditation RoadToAccreditation.pdf 3. JCAHO Accreditation and Quality of Care for Acute Myocardial Infarction JCAHO accreditation and quality of care for acute myocardial infarction.pdf Have an open discussion about these articles. Share your thoughts For example, here are some questions to answer and discuss: Does accreditation impact quality? Are there less errors in hospitals that are accredited? What is the value of accreditation? Do quality concerns initiate changes in staff behavior? Should accreditation be based on results? Solution ACAs Performance Based Healthcare Standards- This booklet is intended to assist anyone dealing with or affected by Health and Employment issues. It is one of a series of booklets and handbooks designed to give impartial advice on employment matters to employers, employees and their representatives. Legal information is provided for guidance only and should not be regarded as an authoritative statement of the law, which can only be made by reference to the particular circumstances which apply. It may, therefore, be wise to seek legal advice. Acas is committed to building better relationships in the workplace and offers training to suit you. From a two-hour session on the key points of new legislation or employing people to courses specially designed for people in your organisation. Click here to find out about training sessions in your area. We also offer hands-on practical help and support to tackleissues in your business with you. This might be through one of our wellknown problem-solving services or a programme we have worked out together to put your business firmly on track for effective employment relations. road to accrediation - Accreditation is usually a voluntary program, sponsored by a non-governmental organization (NGO), in which trained external peer reviewers evaluate a healthcare organization\'s compliance and compare it with pre-established performance standard Accreditation is a quality assurance process designed to ensure an educational program meets a national standard. It serves to support and encourage program responsiveness to the ra pidly changing environmental .

Read and discuss the following three articles 1. ACAs Perform.pdf

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Question 1. Retrieve the first and last names and department numbe.pdf

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Polysaccharides, lipids, and proteins are similar in that they A) are synthesized from monomers by dehydration reactions. B) are synthesized as a result of peptide bond formation between monomers. C) are decomposed into their subunits by dehydration reactions. D) are synthesized from monomers by the process of hydrolysis. E) all contain nitrogen in their monomer building blocks. Solution The answer is A. Are synthesized from monomers by dehydration reactions. Peptide bond is present in polypeptides only hence the option b is wrong. They are formed by dehydration decomposed by hrdration hence options c and d are wrong. N molecule is not present in all polysac hence option E is wrong..

Polysaccharides, lipids, and proteins are similar in that they A) ar.pdf

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Please make the complete program, Distinguish between header files a.pdf

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Part C You are a graduate student in the lab of a famous fly geneticist. You need to analyze a batch of mutant flies that were recently created in her laboratory to identify the single gene that is most likely mutated in each of the flies. Drag the mutations on the left to the appropriate blanks on the right to complete the sentences. Help Reset knirps (a gap gene) hunchback (a gap gene) hedgehog (a segment-polarity gene) wingless (a segment-polarity gene) Kruppel (a gap gene) even-skipped (a pair-rule gene) 1. Mutant fly F has predominantly lost abdominal structures, which is likely the result of a mutation in . 2. The wings are missing in mutant fly L, which is likely the result of a mutation in . 3. Thoracic and abdominal structures are missing in mutant fly Q. The gene most likely responsible for this mutation is . 4. Segment-sized sections of every other segment are missing in mutant fly Z. One likely candidate gene is . 5. Mutant fly X has lost the head and thorax. The gene most likely responsible for this mutation is . 6. The mutation observed in mutant fly O resulted in defects within the anterior or posterior regions of each segment. The gene that most likely caused this mutation is . Help Reset knirps (a gap gene) hunchback (a gap gene) hedgehog (a segment-polarity gene) wingless (a segment-polarity gene) Kruppel (a gap gene) even-skipped (a pair-rule gene) 1. Mutant fly F has predominantly lost abdominal structures, which is likely the result of a mutation in . 2. The wings are missing in mutant fly L, which is likely the result of a mutation in . 3. Thoracic and abdominal structures are missing in mutant fly Q. The gene most likely responsible for this mutation is . 4. Segment-sized sections of every other segment are missing in mutant fly Z. One likely candidate gene is . 5. Mutant fly X has lost the head and thorax. The gene most likely responsible for this mutation is . 6. The mutation observed in mutant fly O resulted in defects within the anterior or posterior regions of each segment. The gene that most likely caused this mutation is . Solution Answer: 1. Mutant fly F has predominantly lost abdominal structures, which is likely the result of a mutation in knirps (a gap gene). Explaination: The knirps gene is a type of gap gene (which is responsible for the regulation of transcription for the product resulting to the development of embryo). The mutation in the gap gene results to the gaps in the embryo, resulting to the missing of many adjacent segments in phenotype. And the mutation in knirps gene resulting to the lack of posterior segments in the fly phenotype. 2. The wings are missing in mutant fly L, which is likely the result of a mutation in wingless (a segment-polarity gene). Explaination: The wingless gene is type of Segment polarity gene, which is responsible for the regulation of transcription for the product resulting to the development of anterior or posterior regions of each segment. And the mutation in wingless g.

Part CYou are a graduate student in the lab of a famous fly geneti.pdf

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Part A Calculations for Solution Preparation d. Calculate the mass o.pdf

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In java. Write a class called GenDoubleLinkedList which is a generic double linked list. This link list is similar to the single linked list that was shown in class except that each node in addition to having data and nextLink (which was originally called link) now has prevLink. The class GenDoubleLinkedList needs to have the following: Internal class ListNode which hasInstance Variables data of type T nextLink of type ListNode prevLink of type ListNode Constructors Default Parameterized Instance Variables head of type ListNode which always points to the beginning of the linked list current of type ListNode which moves around pointing to one of the nodes Constructor A default constructor that initializes head to an empty ListNode and sets current to point at the head. Methods goToNext – This moves the current node forward in the list by one node. It doesn’t move forward if that node is null goToPrev – This moves the current node backwards in the list by one node. It doesn’t move backwards if that node is null. getDataAtCurrent – returns the data at the current node as long as the current isn’t null setDataAtCurrent – takes in a parameter and sets the data at the current node to that value as long as current is not null insertNodeAfterCurrent – creates a new node based on data that is passed in by a parameter and puts that node after the current position deleteCurrentNode – removes the current node from the list by resetting the links showList – prints out the contents of the list line-by-line inList – returns a true or false value based on whether or not the data passed in via a parameter is in the list Solution package com.quad.generic; public class GenDoubleLinkedList { private ListNode head; private ListNode tail; private int size; public int getSize() { return size; } /** * */ public GenDoubleLinkedList() { this.head = null; this.size = 0; } @SuppressWarnings(\"hiding\") class ListNode{ private T data; private ListNode nextLink; private ListNode prevLink; /** * */ public ListNode() { nextLink=null; prevLink=null; data=null; } /** * @param data * @param nextLink * @param prevLink */ public ListNode(T data, ListNode nextLink, ListNode prevLink) { super(); this.data = data; this.nextLink = nextLink; this.prevLink = prevLink; } public T getData() { return data; } public void setData(T data) { this.data = data; } public ListNode getNextLink() { return nextLink; } public void setNextLink(ListNode nextLink) { this.nextLink = nextLink; } public ListNode getPrevLink() { return prevLink; } public void setPrevLink(ListNode prevLink) { this.prevLink = prevLink; } } public void goToNext(){ ListNode temp = head; while(temp != null){ temp = temp.getNextLink(); } } public void goToPrev() { ListNode temporary = tail; while(temporary != null){ temporary = temporary.getPrevLink(); } } public T getDataAtCurrent (int index) { if (index < 0 || index >= size) { throw new IndexOutOfBoundsException(); } else { ListNode current = head; for (int i = 0; i < index; i.

In java.Write a class called GenDoubleLinkedList which is a generi.pdf

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Let G and H be disjoint events in some sample space . The probability that G occurs is P(G) and H occurs is P(H). a. Describe the event G U H. b. What is P(G U H) in terms of P(G) and P(H)? c. Describe the event G intersection H. d. What is the probability of the event G intersection H? Solution G U H can be denoted by G +H as G and H are disjoint events P(G U H) = P(G) + P(H) G H ={ € S : € G and € H} P(G H) = (null set).

Let G and H be disjoint events in some sample space . The probabilit.pdf

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