The ampacity of a No. 12 aluminum wire with RHW insulation installed in a raceway that has 19 other wires is

Write the proghrams for the following:

belka [17]

Answer:

a)

#include <iostream>

using namespace std;

int main() {

bool a,b,c;

cin>>a>>b;

if(a^b)//X-OR operator in C++.

c=true;

else

c=false;

cout<<c;

return 0;

}

b)

#include <iostream>

using namespace std;

int main() {

bool a,b,c,d;

cin>>a>>b>>c;

if((a^b)^c)//X-OR operator in C++.

d=true;

else

d=false;

cout<<d;

return 0;

}

Explanation:

The above written programs are in C++.There is an operator (^) called X-OR operator in C++.It returns true if the number of 1's are odd and returns false if the number of 1's are even.

In the if statement I have user X-OR operator(^) to find the result and storing the result in another boolean variable in both the questions.

5 0

3 years ago

Complete the getNumOfCharacters() method, which returns the number of characters in the user's string. We encourage you to use a

photoshop1234 [79]

Answer:

The Java code is given below with appropriate comments

Explanation:

//Program

//Import the necessary libraries

import java.util.Scanner;

//Define the class

public class CountStringCharacters

{

//Start the main method

public static void main(String[] args)

{

//Prompt the user to enter the input line

Scanner scan = new Scanner(System.in);

System.out.println("Enter a sentence or phrase: ");

String s = scan.nextLine();

//Take the number of characters and keep the count

int numOfCharacters = getNumOfCharacters(s);

System.out.println("You entered: "+s);

System.out.println("Number of characters: "+numOfCharacters);

//Call the method outputWithoutWhitespace

outputWithoutWhitespace(s);

}

//The method getNumOfCharacters

public static int getNumOfCharacters(String s)

{

int numOfCharCount = 0;

for(int i=0; i<s.length();i++)

{

numOfCharCount++;

}

return numOfCharCount;

}

//Definition of the method outputWithoutspace

public static void outputWithoutWhitespace(String s)

{

String str = "";

for(int i=0; i<s.length();i++)

{

char ch = s.charAt(i);

if(ch != ' ' && ch != '\t')

str = str + ch;

}

System.out.println("String with no whitespace: "+str);

}

5 0

3 years ago

Program 3.Study the code carefully and write out the line where there is error, debug it and write out the correct code

stich3 [128]

Answer:

Several errors:

0 where ( ) was intended (probably OCR problem)
space between As and Double
Single quote in stead of double quote in text string
$ in stead of & near Math.sqrt
missing spaces around "is" for proper formatting

5 0

3 years ago

To define constructors and member functions outside of a class's original scope, the operator can be used.

Artyom0805 [142]

Answer: Scope resolution operator(::)

Explanation: A member function and the constructor can be called within the function easily but for the execution of the these components outside the class , a special operator is required to call the functions. The scope resolution operator(::) preceding with the name of class is thus used for defining of the function outside class.This operator maintains the cope of the function and constructor outside the class.

8 0

2 years ago

Write a function analyze_text that receives a string as input. Your function should count the number of alphabetic characters (a

laiz [17]

Answer:

def analyze_text(sentence):

count = 0

e_count = 0

for s in sentence:

s = s.lower()

if s.isalpha():

count += 1

if s == "e":

e_count += 1

return "The text contains " + str(count) + " alphabetic characters, of which " + str(e_count) + " (" + str(e_count*100/count) + "%) are ‘e’."

Explanation:

Create a function called analyze_text takes a string, sentence

Initialize the count and e_count variables as 0

Create a for loop that iterates through the sentence

Inside the loop, convert all letters to lowercase using lower() function. Check each character. If a character is a letter, increment the count by 1. If a character is "e", increment the e_count by 1.

Return the count and e_count in required format

5 0

2 years ago