Answered: Analyze the code below: dogs ["KK",…

Analyze the code below: dogs ["KK", "Cyber", "Blake", "Pinky", "Taki", "Chubby"] a = len (dogs) b= dogs index("Pinky") C = dogs.index("Blake") What will be the value of a? a. 6. b. 2 C. 3 d. 5

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter12: Points, Classes, Virtual Functions And Abstract Classes

Section: Chapter Questions

Problem 17SA

See similar textbooks

Similar questions

Dont post from internet. Which type of polymorphism is depicted in the following code? void PlayerName(string Coach) { cout << "In the team since " << Coach << endl; } void PlayerName(Coach Coach) { cout << "In the team since " << Coach.GetDetails() << endl; } void PlayerName(Players* playerPtr) { cout << "In the team since " << playerPtr->GetDetails() << endl; } int main() { vector<Players*> memberList; Players* playerPtr; Coach* coachPtr; memberList.push_back(playerPtr); memberList.push_back(coachPtr); for (int i = 0; i < memberList.size(); ++i) { PlayerName(memberList.at(i)); } } 1)Static 2)Runtime 3)No polymorphism 4)Compile-time.
Using Dr-Racket, what is the purpose of the number function?(define (number n)(local [(define (abstract int)(* int 11))] (build-list n abstract))) A) Create a list of the first n multiples of 11 starting at 11. B) Create a list of the first n multiples of 11 starting at 0. C) Create a list of the first n multiples of 11 starting at 1. D) Create a list of the first 11 multiples of n starting at 1. E) Create a list of the first 11 multiples of n starting at 0.F) Create a list of the first 11 multiples of n starting at n.
Please, I want to modify the code so that the user can add the employee's name, number and specialization, and use for loob He must add more than one employee. When he finishes adding the employee’s data, he asks the user: Do you want 1-Print the data 2-Add a new employee 3-Exit class Doctor:"""Represents a Doctor""" #initializer with specialization default to "general"def __init__(self, Id, name, specialization="general"):#attributesself.Id = Idself.name = nameself.specialization = specialization.lower() #initializing salary to 25000, basic salaryself.salary = 25000 #incrementing salary based on specialization#if specialization is pediatric, increasing salary by 10%if specialization == "pediatric":self.salary += self.salary*10/100 #if specialization is dental, increasing salary by 15%elif specialization == "dental":self.salary += self.salary*15/100 #str() functiondef __str__(self):return (f"Id: {self.Id} \n"f"Name: {self.name} \n"f"Specialization: {self.specialization} \n"f"Salary:…
Question: Change this code in java just with changing abstraction and exception handling #include <stdio.h>#include <string.h>#include <stdlib.h> typedef struct Guests{char name[50];int status;int nights;float room_cost;float srv;float cost;} G; void options(G guest[]){int option;printf("\n\nMenu\n-----------\n");printf("1. Check in\n");printf("2. Services\n");printf("3. Invoice\n");;printf("4. Check out\n");printf("5. Exit\n");printf("\nPlease Choose a Number According to the Option: ");scanf("%d", &option);switch(option){case 1:printf("\nCheck in\n-----------\n");check_in(guest);break; case 3:printf("\nInvoice\n-----------\n");invoice(guest);break; case 2:printf("\nRequest\n-----------\n");request(guest);break; case 4:printf("\nCheck out\n-----------\n");check_out(guest);break; case 5:exit(0);break;}} void check_in(G guest[]){int i, choice, nights;printf("Rooms:\n\n");for(i=0;i<4;i++){printf("Room No.%d\n", i+1);printf("Room Cost: %.2f/night\n",…
INT_MIN = -32767 def cut_rod(price): """ Returns the best obtainable price for a rod of length n and price[] as prices of different pieces """ n = len(price) val = [0]*(n+1) # Build the table val[] in bottom up manner and return # the last entry from the table for i in range(1, n+1): max_val = INT_MIN for j in range(i): max_val = max(max_val, price[j] + val[i-j-1]) val[i] = max_val return val[n] # Driver program to test above functionsarr = [1, 5, 8, 9, 10, 17, 17, 20].
do some changes in code and make it unique #include <stdio.h>#include <stdlib.h>#include<string.h>//declaring functionsvoid firstFit(int [], int , int [],int );void bestFit(int [], int , int [],int );void worstFit(int [], int , int [],int );//starting programint main() {//declare partitions and processint partitions [] = {110, 450, 100, 250, 500};int processes [] = {212, 417, 112, 426};//getting their sizesint size_partitions = sizeof(partitions )/sizeof(partitions [0]);int size_processes = sizeof(processes )/sizeof(processes [0]);printf("Partitions size: ") ;for (int i=0; i<size_partitions; i++){printf("%d\t" ,partitions[i] );}//index partprintf("\nPartitions index: " );for (int i=0; i<size_partitions; i++){printf("%d\t" ,(i+1)) ;}printf( "\n" );// calling functionsfirstFit(partitions , size_partitions, processes , size_processes);bestFit(partitions , size_partitions, processes , size_processes);worstFit(partitions , size_partitions, processes ,…
write a python code named get_total_cases() takes the a 2D-list (similar to database) and an integer x from this set {0, 1, 2} as input parameters. Here, 0 represents Case_Reported_Date, 1 represents Age_Group and 2 represents Client_Gender (these are the fields on the header row, the integer value represents the index of each of these fields on that row). This function computes the total number of reported cases for each instance of x in the text file, and it stores this information in a dictionary in this form {an_instance_of_x : total_case}. Finally, it returns the dictionary and the total number of all reported cases saved in this dictionary. (Suppose we want to know the total number of cases reported on each date, so use x = 0.) >>> result, total_cases = get_total_cases(database, 0) >>> display_dict(result) 2021-05-19: 8 2021-05-20: 2 2021-05-21: 1 2021-05-22: 1 >>> print(total_cases)
Q1 #include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…
Need help making a java file that combines both linearSearch and binarySearch •Both search methods must use the Comparable<T> interface and the compareTo() method.•Your program must be able to handle different data types, i.e., use generics.•For binarySearch, if you decide to use a midpoint computation formula that is different fromthe textbook, explain that formula briefly as a comment within your code. //code from textbook //linearSearch public static <T> boolean linearSearch(T[] data, int min, int max, T target) { int index = min; boolean found = false; while (!found && index <= max) { found = data [index].equals(target); index++; } return found; } //binarySearch public static <T extends Comparable<T>> boolean binarySearch(T[] data, int min, int max, T target) { boolean found = false; int midpoint = (min + max)/2; if (data[midpoint].compareTo(target)==0)…
TODO 1 Obtain all the indexes with labels equal 2 with np.where and the label array y. Keep the results in two_class_idx also index np.where() at 0 # TODO 1.1 two_class_idx = print(f"two_class_idx output: \n {two_class_idx}") try: print(f"two_class_idx shape: {two_class_idx.shape}") except Exception: pass todo_check([ (isinstance(two_class_idx, np.ndarray),f'two_class_idx is not an NumPy array! two_class_idx is currently a {type(two_class_idx)}'), (np.all(two_class_idx == np.array([3,9,11,12,16])),'two_class_idx does not contain the correct location values') ])
Program this in java the state and variable display must be:Unsorted Partition Reference Index: 0, Traversing Index: 7, Current Traversing Index Value: mango, Next Index Value: plum, Swapping Condition: FalseCurrent Array: ['apple', 'avocado', 'orange', 'banana', 'strawberry', 'pineapple', 'plum', 'mango'] The input:["apple", "avocado", "orange", "banana", "strawberry", "pineapple", "plum", "mango"]Note:The sorted partition is on the left side. Then, the order is increasing based on the number of vowels of a string.the output must be like this:['plum', 'apple', 'strawberry', 'mango', 'orange', 'banana', 'avocado', 'pineapple']
In this final submission, you will build on checkpoint B to load the database and DNA sequence from files. There will be two databases and several sequences that will be available for download below. Example of the database is the file small.txt: name,AGATC,AATG,TATC Alice,2,8,3 Bob,4,1,5 Charlie,3,2,5 Example of the sequence is the file 1.txt: AAGGTAAGTTTAGAATATAAAAGGTGAGTTAAATAGAATAGGTTAAAATTAAAGGAGATCAGATCAGATCAGATCTATCTATCTATCTATCTATCAGAAAAGAGTAAATAGTTAAAGAGTAAGATATTGAATTAATGGAAAATATTGTTGGGGAAAGGAGGGATAGAAGG Implement/modify the following functions: Modify the function readData, which will now take an additional parameter: a string representing the filename containing the database of individuals and their STR counts. It will also return a bool indicating if opening the file was successful or not: bool readData(string filename, vector<string>& nameSTRs, vector<string>& nameIndividuals, vector<vector<int>>& STRcounts) Update the function…