Calculate eigenvector centrality and degree of vertices for .pdf

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Calculate eigenvector centrality and degree of vertices for students data. This is the code I have for loading the students graph. M thood the grophel fille [ G text { = nx.read_graphml("C2/Users/atifa/Onebrive/Desktop/students.graphml") } ] moras the graph nx.draw (G) M aprint the number of nodes and edges print("Wunber of nodes is:", nx. nunber_of_nodes (6) ) print("Number of edges 1s:", nx, number_of_edges(G)) Number of nodes ist 291 Number of edpes is: 1396.

$Calculate eigenvector centrality and degree of vertices for students data. This is the code I have for loading the students graph. M thood the grophel fille [ G text { = nx.read_graphml("C2/Users/atifa/Onebrive/Desktop/students.graphml") } ] moras the graph nx.draw (G) M aprint the number of nodes and edges print("Wunber of nodes is:", nx. nunber_of_nodes (6) ) print("Number of edges 1s:", nx, number_of_edges(G)) Number of nodes ist 291 Number of edpes is: 1396$

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Can someone please help with this :) Now that you have completed the Punnete squaies you can ule this information to pecdict the Probability of specific outcomes in the F2, 2. Each of the four boxes represents a one in four (or 25% ) chance or probabilicy of a genorype. What is the genotypic ratio for this cross? 3. For each of the above genotypes, give the phenorype that would be expressod in the F2 generation. 4. What is the phenotypic ratio for this cross? SECTION 5: DIHYBRID CROSSES UP to this point, you have considered genetic crosses that involved a single trait controlled by 2 single gene (monohybrid cross). Mendel also made crosses that involved two genes, each located on a different pair of homologous chromosomes. This type of cross is called a dihybrid cross and illustrates Mendel's Law of Independent Assortment. Mendel demonstrated that when two traits are determined by genes located on two different chromosomes, each pair of alleles segregates independently of one another during gamete formation. The different possible allele combinations that result from meiosis are illustrated in Figure 8-3. Two equallv probable arranaements of chromosomes at metaphase l:Step 9: Phenotypic ratio of F2 Tongue roller, Taster Tongue roller, Non-taster Non-roller, Taster Non-roller, Non-taster.

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Can someone solve the TODO parts of the following problem #include <iostream> #include <fstream> #include <sstream> #include<string> #include "Queuecpp.h" #include "Songcpp.h" #include "RequestCpp.h" //GIVEN int main() { LinkedList<Song> list; Queue<Request> q; populateSongDataBase(list); // fill the Songs Data Base into Linked List std::cout<<"*************************************n"; list.printList(); // std::cout<<"*************************************n"; populateRequests(q); // fill the request q std::cout<<"n"; std::cout<<"---------------PRINTING QUEUE-------------------n"; q.printList(); std::cout<<"------------------------------------n"; processQueue(list,q); // process the requests std::cout<<"*************************************n"; list.printList(); std::cout<<"*************************************n"; } //GIVEN void populateRequests(Queue<Request> &q){ } //GIVEN std::string playSong(LinkedList<Song>&list, Song s){ } //GIVEN void printTopTenSongs(LinkedList<Song>& list){ } //TODO void addThisSong(LinkedList<Song> &list, Song s){ //TODO /* Use this code if (found>=0) {std::cout<<"ADDED THIS SONG "<<found+1<<" " ;list.printSong(found+1);} else std::cout<<"SORRY CANNOT FIND SONG n"; */ } //TODO void removeThisSong(LinkedList<Song> &list, Song s){ /* Use this code if (found>=0) {std::cout<<"SORRY CANNOT REMOVE THIS SONG - STILL FOUND AT "<<found;} else{ std::cout<<"REMOVED THIS SONG n"; std::cout<<"THE SONG AT CHARTPOSITION "<<chartPos<<" IS NOW "; list.printSong(chartPos); std::cout<<"n"; */ } } //TODO void moveThisSong(LinkedList<Song> &list, Song s, int pos){ int chartPos=list.get(s); std::cout<<"MOVING SONG "<<s.toString()<<"n"; list.move(s,pos); std::cout<<"ITS NEW CHART POSITION n"; list.printSong(pos); std::cout<<"THE SONG AT PREVIOUS POSITION "<<chartPos+1<< " IS NOW ";list.printSong(chartPos+1); } //GIVEN void processRequest(LinkedList<Song> &list,std::string action, std::string title,std::string singer,int chartPos){ Song s(title,singer,chartPos); char ch =action[0]; switch(ch){ case 'P' : std::cout<<playSong(list,s)<<"n";break; case 'A' : addThisSong(list,s);break;; case 'R' : removeThisSong(list,s);break; case 'M' : moveThisSong(list,s,chartPos);break; case 'T' : printTopTenSongs(list);break; } } //TODO void processQueue(LinkedList<Song> &list ,Queue<Request>& q){ int requestNumber=1; //Create an iterator for the Queue to iteratate through requests std::cout<<"-------------------------------------------------------------------------n"; // call processRequest std::cout<<"-------------------------------------------------------------------------n" std::cout<<"-------------------------------------------------------------------------n"; } //GIVEN void populateSongDataBase(LinkedList<Song> &list){ std::string line=""; std::string token=""; std::string title=""; std::string singer=""; int chartPos=-1; try { std::ifstream input("SongsData.txt"); if (input.fail()) throw new std::string("FILE OPEN ERROR EXCEPTION n"); else{ while (getli.

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Can someone please help me implement the "addSong" function in the code below? Thank you! // You are given a partially completed program that creates a list of songs, like a music repository. // Each song has this information: song's id, song's name, singer's name, genre of the song, and its published year. // The struct 'musicRepository' holds information of one song. Genre is enum type. // An array of structs called 'list' is made to hold the list of songs. // You should not modify any of the given code, the return types, or the parameters, you will risk of getting compilation error. // You are not allowed to modify main (). // You can use string library functions. #include <stdio.h> #include <stdlib.h> #include <string.h> #pragma warning(disable : 4996) // for Visual Studio Only #define MAX_SONGS 20 #define MAX_SONG_NAME_LENGTH 40 #define MAX_SINGER_NAME_LENGTH 40 typedef enum { unclassified = 0, Pop, Rock, Reggae, Country, Blues, Balad, } genreType; // enum type struct musicRepository { // struct for song details unsigned int ID; char songName[MAX_SONG_NAME_LENGTH]; char singerName[MAX_SINGER_NAME_LENGTH]; genreType genre; unsigned int year; }; struct musicRepository list[MAX_SONGS]; // declare the list of songs int numSongs = 0; // the number of songs currently stored in the list (initialized to 0) // Given functions void initializeRepository(struct musicRepository repo[], int size); void printRepository(struct musicRepository repo[], int numSongs); void flushSngIn(); // functions that need to be implemented by you int addSong(struct musicRepository repo[], char* songName, char* singerName, char* genre, unsigned int year, int numSongs); struct musicRepository* searchSong(struct musicRepository repo[], char* songName, int numSongs); void printSong(struct musicRepository repo[], int ID); int editSong(struct musicRepository* repo, int numSongs, char* songName); int deleteSongByName(struct musicRepository* repo, char* songName, int numSongs); // This function takes in an array of musicRepository structures and the size of the array as parameters. It then loops through each index of the array and sets the values of id, year, genre, name, and singer to their respective initial values. // NOTE that strcpy is used to copy an empty string to name and singer, rather than setting them equal to "". This is because arrays in C are not assignable, so the strcpy function must be used to copy the empty string into the array. void initializeRepository(struct musicRepository repo[], int size) { for (int i = 0; i < size; i++) { repo[i].ID = 0; repo[i].year = 0; repo[i].genre = 0; strcpy(repo[i].songName, ""); strcpy(repo[i].singerName, ""); } } // This function takes in the musicRepository array and iterates through each element. If the ID of the song is not 0, it prints the details of the song. void printRepository(struct musicRepository repo[], int numSongs) { printf("n--- Song Repository ---n"); for (int i = 0; i < numSongs; i++) { if (repo[i].ID != 0) .

Can someone please help me implement the addSong function .pdf

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Can someone please help me complete the "add_song" function below. Thank you! // Each song has this information: song's id, song's name, singer's name, genre of the song, and its published year. // The struct 'musicRepository' holds information of one song. Genre is enum type. // An array of structs called 'list' is made to hold the list of songs. // You should not modify any of the given code, the return types, or the parameters, you will risk of getting compilation error. // You are not allowed to modify main (). // You can use string library functions. #include <stdio.h> #include <stdlib.h> #include <string.h> #pragma warning(disable : 4996) // for Visual Studio Only #define MAX_SONGS 20 #define MAX_SONG_NAME_LENGTH 40 #define MAX_SINGER_NAME_LENGTH 40 typedef enum { unclassified = 0, Pop, Rock, Reggae, Country, Blues, Balad, } genreType; // enum type struct musicRepository { // struct for song details unsigned int ID; char songName[MAX_SONG_NAME_LENGTH]; char singerName[MAX_SINGER_NAME_LENGTH]; genreType genre; unsigned int year; }; struct musicRepository list[MAX_SONGS]; // declare the list of songs int numSongs = 0; // the number of songs currently stored in the list (initialized to 0) // This function takes in an array of musicRepository structures and the size of the array as parameters. It then loops through each index of the array and sets the values of id, year, genre, name, and singer to their respective initial values. // NOTE that strcpy is used to copy an empty string to name and singer, rather than setting them equal to "". This is because arrays in C are not assignable, so the strcpy function must be used to copy the empty string into the array. void initializeRepository(struct musicRepository repo[], int size) { for (int i = 0; i < size; i++) { repo[i].ID = 0; repo[i].year = 0; repo[i].genre = 0; strcpy(repo[i].songName, ""); strcpy(repo[i].singerName, ""); } } // This function takes in the musicRepository array and iterates through each element. If the ID of the song is not 0, it prints the details of the song. void printRepository(struct musicRepository repo[], int numSongs) { printf("n--- Song Repository ---n"); for (int i = 0; i < numSongs; i++) { if (repo[i].ID != 0) { printf("ID: %dn", repo[i].ID); printf("Name: %sn", repo[i].songName); printf("Singer: %sn", repo[i].singerName); if (repo[i].genre == 0) printf("Genre: unclassifiedn"); else if (repo[i].genre == 1) printf("Genre: Popn"); else if (repo[i].genre == 2) printf("Genre: Rockn"); else if (repo[i].genre == 3) printf("Genre: Reggaen"); else if (repo[i].genre == 4) printf("Genre: Countryn"); else if (repo[i].genre == 5) printf("Genre: Bluesn"); else if (repo[i].genre == 6) printf("Genre: Baladn"); printf("Year: %dn", repo[i].year); printf("n"); } } } // Problem 2 // struct to hold information about a song struct Song { int id; char name[MAX_SONG_NAME_LENGTH]; char singer[MAX_SINGER_NAME_LENGTH]; genreType genre; int year; struct Song* next; }; // Function declarati.

Can someone please help me complete the add_song function .pdf

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can someone please assist me with analysis of this case atudy VAKRANGEE: ANATOMY OF STOCK VALUATION 1 Poga Gupti and Zahid Jamar wrote this case solby to provide matorial for chas discussion. The authors do not intond to iflustrate either oflective or innflloctive handing of a managonial sihution. The authors may have disguised certain names and other idantifying information to protect confidentilith. This publication may not be transmitted, pholocopied, digitinod, or othorwise reproduced in any form or by any moans without the pormissian of the copynight holder. Reproduction of this material is not covered under autherization by any reproduction rights organization. To arder copies or nequest pemission to reproduce materials, contact Ney Pubbishing. Avey Business School, Westem Cogyright 92020, Ney Business School Foundation Vorsion: 20200703 On April 20, 2018, Priyankar Khandelwal was furious as he watched the stock value from his fund seemingly evaporate right in front of his eyes. Khandelwal, the fund manager at Principal First, had been managing the mid-sized information technology (IT) fund for the previous five years. Khandelwal asked his assistant to put in a sell order to cover his long position 2 in Vakrangee Limited (Vakrangee) at a substantial loss. He had bought the stock in the beginning of October 2017. The stock hit a high in Jamuary 2018 but had subsequently been sliding (see Exhibit 1). In his fund, Khandelwal had allocated a substantial amount to Vakrangee in the initial investment. Based on the current market price of the stock, there was a net loss of 169 million 3 and a loss of 592 million unrealized from the top for the fund. 4 This loss caused the net asset value of the fund to fall by 35 per cent year to date, because Khandelwal had not hedged his position. Khandelwal's clients were irate, and more than 25 per cent of his firm's investors had withdrawn their money from the fund. Principal First had also reprimanded him and put him on notice. Khandelwal wondered where he went wrong in his analysis of Vakrangee. He had done the valuation exercise himself, checked and cross-checked his assumptions, looked at industry and credit reports, and had found it to be a good stock with good future prospects. What had he missed? What could he have done better? PRINCIPAL FIRST Principal First was a Mumbai-based boutique investment management and investment advisory firm. The investment objective of the firm was to deliver notable risk-adjusted retums consistently while ensuring capital preservation. The firm was founded in 2005 and had been an carly investor in the Indian alternative 'This case was awarded an honourable mention in the the ISB-lvey Global Case Compettion 2018. The prize was sponsored by ISB. 2 A long position was the buying of a stock, commodity, or currency with the expectation that it (would] rise in value; "Long Position (Long)," Investopedia, March 20, 2020, accessed April 6, 2010, www. investopedia.com/terms/Vi.

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can do this in java please thanks in advance The code that y.pdf

can do this in java please thanks in advance The code that y.pdf

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can assume that any sufficient statistic found is also a com.pdf

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Can anyone help me with 2.1 , 2.3 ,2.5, 2.7 In C programming 2.1 Elephant *new_elephant(const char name[], enum type which_type, unsigned int id, unsigned short weight, float trunk_length) Project #5 discussed some approaches for writing structure functions in C. Another option not mentioned then is for a function to return a pointer to a structure that the function has itself dynamically allocated. Using this approach a function can return a pointer to an initialized structure without requiring the caller to declare a structure or to allocate memory for : structure. It is common for functions to dynamically allocate a structure and return a pointer to it after suitably initializing its fields. This function will use this approach. The function should just return NULL without doing anything else if its array parameter is NULL. Otherwise it should return a pointer to a dynamically-allocated Elephant structure that has the parameters' values stored in its fields. The name field of the newly created Elephant structure has no fixed or maximum length, so this function now must allocatea string of exactly the right size to store the name parameter (however long it may be), copy the value of the name parameter to the allocated string, and make the name field of the allocated structure point to it. To emphasize: the new elephant's name field must point to new memory that this function allocates, not just point to the parameter name that was passed in. In other words, this function should not just be doing pointer aliasing. Use string library functions for all manipulations of the elephant's name, not loops (or recursion), or you will lose considerable credit. 2.2 unsigned short get_type(const Elephant * const el_ptr, enum type *const type_ptr) This function should "return" to the caller the type of its parameter elephant, but it should also return 0 if either parameter is NULL. However, since the first constant in an enum has the numeric value 0 , if the function only returned a value, the caller would not be able to distinguish between it returning 0 to indicate that either parameter was NULL, or it returning the enum value AFRICAN. What the function should do instead is to return 0 or 1 depending upon its parameters' validity, and, if the parameters are valid, to store the elephant's type into the variable that type ptr points to. So if either parameter is NULL the function should return 0 without changing anything else, while otherwise it should return 1 after storing its parameter elephant's type into the variable that its second parameter points to. 2.3 const char *get_name(const Elephant *const el_ptr) This function should return NULL if it parameter is NULL, or if its parameter is not NULL but its parameter's name field is NULL. Otherwise it should return its parameter elephant's name, but it should not just return a pointer to the name field of the Elephant structure itself. Instead, it should allocate enough memory to store the name, copy th.

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Campus Solution is an education management consulting company located in Austin Texas. Dr. Judy Marconi (retired President of a local college) and her friend Mr. Thomas Irwin formed the company in 1990. The company specializes in providing housing campus management solutions for small, medium, and large colleges and universities. The system can be added to any existing student management system and configured according to the processes and policies of the college and/or university. Vision of the company: Provide a customizable intuitive campus housing management software solution for delightful customer experiences. Mission of the company: To build a robust platform-independent secure end-to-end software product for managers and students to save them time and provide the best living experiences while in college. Most of the colleges and universities provides safe living spaces for their students in dormitories/dorms in an around campus. Managing students and providing a variety of services require proper processes and record keeping of all the students. There are large number of students (freshman to senior) living in, leaving, or changing dormitory every year which requires proper management. It creates a lot of confusion and inconvenience for the dormitory management personnel to deal with. The workload is large, and the processing time is long and requires a lot of human and material resources. There should be a proper process of reviewing applications, processing them and approving student requests. There is a process of proper matching of students, sometimes according to the student choices, but it is not always possible. Students pay their dorm fees in various ways. Some students pay them monthly, but some pay them annually. In addition, there are request for dormitory equipment repairs and other student accommodation requests. Your team is tasked with creating a proper dormitory management system to handle all the tasks. Due to the limiting time and resources availability, your job would be identifying the possible components, and built some part of the software platform. Follow the SDLC life cycle process and complete the process. Your submission would be in two different phases. I have chose Move in/ Move out for my component and on a web-based platform. Deliverables phase 1: Group members Scope (identify the problem and focus on one chosen component) Project Schedule Workflow diagrams /Activity diagrams Event table Couple of fully developed description of events from event table.

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