Write code that prints: Ready! numVal ... 2 1 Start! Your code should contain a for loop. Print a newline after each number and after each line of text Ex: numVal

Answer:

lookup value

Explanation:

Answer:

4096 VLANs

Explanation:

A VLAN (virtual LAN) is a group of devices on one or more LAN connected to each other without physical connections. VLANs help reduce collisions.

An 802.1Q Ethernet frame header has VLAN ID of 12 bit VLAN field. Hence the maximum number of possible VLAN ID is 4096 (2¹²).  This means that a switch supporting the 802.1Q protocol can have a maximum of 4096 VLANs

A lot of VLANs ID are supported by a switch. The maximum number of VLANs that can be configured on a switch supporting the 802.1Q protocol is 4,094 VLANS.

But that of 0x000 and 0xFFF are set apart. This therefore makes or leaves it as 4,094 possible VLANS limits. Under IEEE 802.1Q, the maximum number of VLANs that is found on an Ethernet network is 4,094.

Learn more about VLANs from

https://brainly.com/question/25867685

Answer:

See solution below

See comments for explanations

Explanation:

import java.util.*;

class Main {

 public static void main(String[] args) {

   //PrompT the User to enter a String

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

   //Receiving the string entered with the Scanner Object

   Scanner input = new Scanner (System.in);

   String string_input = input.nextLine();

   //Print out string entered by user

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

   //Call the first method (GetNumOfCharacters)

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

   //Call the second method (OutputWithoutWhitespace)

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

   }

 //Create the method GetNumOfCharacters

   public static int GetNumOfCharacters (String word) {

   //Variable to hold number of characters

   int noOfCharactersCount = 0;

   //Use a for loop to iterate the entire string

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

     //Increase th number of characters each time

     noOfCharactersCount++;

   }

   return noOfCharactersCount;

 }

 //Creating the OutputWithoutWhitespace() method

 //This method will remove all tabs and spaces from the original string

 public static String OutputWithoutWhitespace(String word){

   //Use the replaceAll all method of strings to replace all whitespaces

   String stringWithoutWhiteSpace = word.replaceAll(" ","");

   return stringWithoutWhiteSpace;

 }

}

Answer:

the clock rate of the first machine need to be 1.7 GHz

Explanation:

Given:

CPI of A = 1.9

CPI of B = 1.1

machine, B executes the same program with 20% less instructions and with a CPI of 1.1 at 800MHz

To find:

In order for the two machines to have the same performance, what does the clock rate of the first machine need to be

Solution:

CPU execution time = Instruction Count * Cycles per Instruction/ clock rate

CPU execution time = (IC * CPI) / clock rate

(IC * CPI) (A) / clock rate(A) =  (IC * CPI)B / clock rate(B)

(IC * 1.9) (A) / clock rate(A) = (IC * (1.1 * (1.0 - 0.20)))(B) / 800 * 10⁶ (B)

Notice that 0.20 is basically from 20% less instructions

(IC * 1.9)  / clock rate = (IC * (1.1 * (1.0 - 0.20))) / 800 * 10⁶

(IC * 1.9)  / clock rate =  (IC*(1.1 * ( 0.8))/800 * 10⁶

(IC * 1.9)  / clock rate =  (IC * 0.88) / 800 * 10⁶

clock rate (A) =  (IC * 1.9) / (IC * 0.88) / 800 * 10⁶

clock rate (A) =  (IC * 1.9) (800 * 10⁶) /  (IC * 0.88)

clock rate (A) = 1.9(800)(1000000)  / 0.88

clock rate (A) =  (1.9)(800000000)  / 0.88

clock rate (A) = 1520000000  / 0.88

clock rate (A) = 1727272727.272727

clock rate (A) = 1.7 GHz

Answer:

a task and a calendar entry

Explanation:

ITS RIGHT

Answer:

its A) a task and a calendar entry

Explanation:

correct on e2020

Answer:

The solution is provided in the explanation section.

Detailed explanation is provided using comments within the code

Explanation:

import java.util.*;

public class Main {

//The Bubble sort method

public static void bb_Sort(int[] arr) {  

   int n = 10; //Length of array  

   int temp = 0; // create a temporal variable  

   for(int i=0; i < n; i++){  

         for(int j=1; j < (n-i); j++){  

           if(arr[j-1] > arr[j]){  

               // The bubble sort algorithm swaps elements  

               temp = arr[j-1];  

               arr[j-1] = arr[j];  

               arr[j] = temp;  

             }  

         }            

         }  

        }

 public static void main(String[] args) {

   //declaring the array of integers

   int [] array = new int[10];

   //Prompt user to add elements into the array

   Scanner in = new Scanner(System.in);

   //Use for loop to receive all 10 elements

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

     System.out.println("Enter the next array Element");

     array[i] = in.nextInt();

   }

   //Print the array elements before bubble sort

   System.out.println("The Array before bubble sort");

   System.out.println(Arrays.toString(array));

   //Call bubble sort method

   bb_Sort(array);  

   System.out.println("Array After Bubble Sort");  

   System.out.println(Arrays.toString(array));

 }

}

Answer:

DNS

Explanation:

Hope this helps

Answer:

ok so u have take the 5 and put 6

Explanation:

5-5. Ethernet network design: UTP connections from client PC and server to workgroup switches, 900m fiber optic link between switches, 800 Mbps communication.

5-6. Additional client connects to the same switch, UTP connection, maintains existing fiber optic link, 800 Mbps communication with the server.

5-5. For connecting a single client PC to a single server, both located 900 meters apart and requiring communication at 800 Mbps, the following Ethernet network design can be implemented:

- Client PC and server connect to their respective workgroup switches via UTP.

- Use fiber optic cables for the 900-meter transmission link between the switches.

- Install switches at the client PC and server locations.

- Ensure that the switches support at least 1 Gbps Ethernet speeds to accommodate the required transmission speed.

5-6. In addition to the previous design, for adding another client next to the first client:

- Connect both clients to the same switch.

- Use UTP cables to connect the second client to the switch.

- Ensure the switch supports 1 Gbps Ethernet speeds.

- Maintain the existing fiber optic transmission link between the switches.

- The second client can also communicate with the server at the required 800 Mbps transmission speed.

Learn more about Network Design at:

https://brainly.com/question/7181203

#SPJ2

since this has already been answered, who is your favorite SMP character?

mines Wilbur/ Ghostbur

some people will disagree with me but jshlatt is one of my favorite characters on the dream smp . But my all time favorite characters is ALL of Wilbur's characters

public class Drive {

   int miles;

   int gas;

   String carType;

   public String getGas(){

       return Integer.toBinaryString(gas);

   }

   public Drive(String driveCarType){

       carType = driveCarType;

   }

   public static void main(String [] args){

       System.out.println("Hello World!");

   }

}

I'm pretty new to Java myself, but I think this is what you wanted. I hope this helps!