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marusya05 [52]
2 years ago
5

Write an if-else statement that displays 'Speed is normal' if the speed variable is within the range of 24 to 56. If the speed v

ariable’s value is outside this range, display 'Speed is abnormal'.
Computers and Technology
1 answer:
kari74 [83]2 years ago
4 0

Answer:

import java.util.Scanner;

public class Speed{

int speed;

public Speed(int speed){

this.speed = speed;

}

public void checkSpeed(){

if(speed >= 24 || speed <= 56){

System.out.println("Speed is normal");

}

else

System.out.println("Speed is abnormal");

}

public static void main(String...args){

Scanner input = new Scanner(System.in);

int userSpeed = 0;

System.out.println("Enter a speed: ");

userSpeed = input.nextInt();

Speed obj1 = new Speed(userSpeed)

obj1.checkSpeed();

}

Explanation:

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(NumberFormatException)Write the bin2Dec(String binaryString) method to convert a binary string into a decimal number. Implement
tresset_1 [31]

The following program will be used that prompts the user to enter a binary number as a string and displays decimal equivalent of the string.

<u>Explanation:</u>

NumberFormat.java

import java.util.Scanner;

public class NumberFormat{

public static void main(String[] args) {

      Scanner scan = new Scanner(System.in);

      try{

      System.out.println("Enter a binary string:");

      String s = scan.next();

      int integerValue = bin2Dec(Integer.parseInt(s));

      System.out.println("The decimal value of "+s+" is "+integerValue);

      }

      catch(NumberFormatException e){

          System.out.println(e);

      }

  }  

  public static int bin2Dec(int binaryNumber) throws NumberFormatException{

  int decimal = 0;

  int p = 0;

  try{

  while(true){

  if(binaryNumber == 0){

  break;

  } else {

  int temp = binaryNumber%10;

  decimal += temp*Math.pow(2, p);

  binaryNumber = binaryNumber/10;

  p++;

  }

  }

  }

  catch(NumberFormatException e){

      throw new NumberFormatException("Invalid binary number");

  }

  return decimal;

  }

}

Output:

Enter a binary string:

1011000101

The decimal value of 1011000101 is 709

5 0
2 years ago
I am doing keyboarding keyboarding is very boring and yeah
Bingel [31]

Answer:

yes I agree with you

Explanation:

I have been keyboarding for the past 7 hours.

7 0
2 years ago
Read 2 more answers
The __________ contains a list of all the resources owned by the library.
Ira Lisetskai [31]

The catalog contains a list of all the resources owned by the library.

- Mabel <3

5 0
2 years ago
Open "Wireshark", then use the "File" menu and the "Open" command to open the file "Exercise One.pcap". You should see 26 packet
murzikaleks [220]
Idk idk idk idk idk idk
8 0
2 years ago
Write a program that reads an integer and displays, using asterisks, a filled diamond of the given side length. For example, if
zaharov [31]

Answer:

  1. import java.util.Scanner;
  2. public class Main {
  3.    public static void main(String[] args) {
  4.        Scanner input = new Scanner(System.in);
  5.        int n = input.nextInt();
  6.        int l = 1;
  7.        for(int i=1; i <= n; i++){
  8.            for(int j=1; j<= l; j++){
  9.                System.out.print("*");
  10.            }
  11.            System.out.println();
  12.            l = l + 2;
  13.        }
  14.        l = l - 4;
  15.        for(int i=1; i < n; i++){
  16.            for(int j=1; j <= l; j++){
  17.                System.out.print("*");
  18.            }
  19.            System.out.println();
  20.            l = l-2;
  21.        }
  22.    }
  23. }

Explanation:

The solution code is written in Java.

Firstly use a Scanner object to accept an input integer (Line 5-6).

Create the first set of inner loops to print the upper part of the diamond shape (Line 8-14). The variable l is to set the limit number of asterisk to be printed in each row.

Next, create another inner loops to print the lower part of the diamond shape (Line 17-23). The same variable l is used to set the limit number of asterisk to be printed in each row.

3 0
2 years ago
Read 2 more answers
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