A program that is required to three (3) numbers. calculate and print their total is given below:
<h3>The Program</h3>
import java.util.Scanner;
public class SumAndAverage {
public static void main(String[ ] args) {
System.out.println("Enter three numbers - ");
// Accepting and finding sum of numbers.
int sum = 0;
Scanner sc = new Scanner(System.in);
for (int i = 0; i < 3; i++)
sum += sc.nextInt( );
// Printing sum and average.
System.out.println("Sum - " + sum);
System.out.println("Average - " + (sum / 3f));
}
}
The output would request three different numbers, then add them up, and display the output of the sum and then display or print their total
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Answer:
See the attached pictures for detailed answer.
Explanation:
The cache contain 8K number of blocks.each block has 16 Byte worth of data. Each of these bytes are addressable by block offset. Now, each of these blocks are addressable and their address is provided by line offset or set offset. If block number is given in Decimal format then the block to which that block will be mapped is given by expression
mappedToBlock = (bNumber)mod213
TAG and valid but consititues something called cache controller. TAG holds the physical address information because the TAG in physical address get divided into TAG and line offset in cache address. Valid bit tells status of blocks. If it is 1 then data blocks are referred by CPI else not.
Answer: Input is the answer.
Answer:
406.
Explanation:
continuously dividing the base 10 or decimal number by 8 till the decimal number does not becomes zero and the result is the remainders in reverse.
Remainder on 262 / 8 = 6.
Remainder on 32 / 8 = 0.
Remainder on 4 / 8 = 4.
Octal number=406.
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