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Elina [12.6K]
3 years ago
14

What’s your fave tv show?

Computers and Technology
2 answers:
kompoz [17]3 years ago
7 0

Answer:

the vampire diareas

Explanation:

uranmaximum [27]3 years ago
7 0

Answer: friends

Explanation:✌

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Does anybody know how to use these showers?
GrogVix [38]
Twist the knob I'm guessing , maybe ask someone who lives there .. lol
6 0
3 years ago
Read 2 more answers
Assume that you have created a class named DemoCar. Within the Main() method of this class, you instantiate a Car object named m
marysya [2.9K]

Answer:

It can be static, but it shouldn't, given the way it is invoked.

It can also be protected but <u>not</u> private.

Explanation:

The question is confusing, since the ComputeMpg() should be a public non-static member. Also, is this Java or C#?

5 0
3 years ago
The hostel in which you plan to spend the night tonight offers very interesting rates, as long as you do not arrive too late. Ho
faust18 [17]

Answer:

Answered below

Explanation:

// Python implementation

incrementAmount = 5

basePrice = 10

price = 0

arrivalHour = int(input("Enter arrival hour 0-12: ")

if arrivalHour < 0 or arrivalHour > 12:

print ("Invalid hour")

if arrivalHour == 0:

price = basePrice

else:

price = arrivalHour * incrementAmount

totalPrice = price + basePrice

if totalPrice > 53:

totalPrice = 53

print ("Your bill is $totalPrice")

4 0
3 years ago
Write a method named removeDuplicates that accepts a string parameter and returns a new string with all consecutive occurrences
Nitella [24]

Answer:

//Method definition

//Method receives a String argument and returns a String value

public static String removeDuplicates(String str){

       //Create a new string to hold the unique characters

       String newString = "";

       

       //Create a loop to cycle through each of the characters in the

       //original string.

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

           // For each of the cycles, using the indexOf() method,

           // check if the character at that position

           // already exists in the new string.

           if(newString.indexOf(str.charAt(i)) == -1){

               //if it does not exist, add it to the new string

               newString += str.charAt(i);

           }  //End of if statement

       }   //End of for statement

       

       return newString;   // return the new string

   }  //End of method definition

Sample Output:

removeDuplicates("bookkeeeeeper") => "bokeper"

Explanation:

The above code has been written in Java. It contains comments explaining every line of the code. Please go through the comments.

The actual lines of codes are written in bold-face to distinguish them from comments. The program has been re-written without comments as follows:

public static String removeDuplicates(String str){

       String newString = "";

       

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

           if(newString.indexOf(str.charAt(i)) == -1){

               newString += str.charAt(i);

           }

       }

       

       return newString;

   }

From the sample output, when tested in a main application, a call to removeDuplicates("bookkeeeeeper") would return "bokeper"

7 0
3 years ago
Write code which takes a sentence as an input from the user and then prints the length of the first word in that sentence.
Afina-wow [57]

import java.util.Scanner;

public class U2_L3_Activity_Four {

   public static void main(String[] args) {

       Scanner scan = new Scanner(System.in);

       System.out.println("Enter a sentence.");

       String sent = scan.nextLine();

       int count = 0;

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

           char c = sent.charAt(i);

           if (c != ' '){

               count++;

       }

           else{

               break;

           }

       

   }

       System.out.println("The first word is " + count +" letters long");

   

   }

}

We check to see when the first space occurs in our string and we add one to our count variable for every letter before that. I hope this helps!

7 0
3 years ago
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