Answer:
import java.util.Scanner;
public class BarChart {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
//take input from user
System.out.println("Enter Score");
int score=sc.nextInt();
int count=score/10;
int i=1;
//print horizontal bar
//if you want to print vertical bar then simply change into print which is replace by println
while(i<=count)
{
System.out.print("*");
i++;
}
}
}
Explanation:
They code and learn how to go in the inner and deeper parts of the computer
IFLY is probably what it was.
Answer:
1. 2588672 bits
2. 4308992 bits
3. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
Explanation:
1. Number of bits in the first cache
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^15 (1+14+(32*2^1)) = 2588672 bits
2. Number of bits in the Cache with 16 word blocks
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^13(1 +13+(32*2^4)) = 4308992 bits
3. Caches are used to help achieve good performance with slow main memories. However, due to architectural limitations of cache, larger data size of cache are not as effective than the smaller data size. A larger cache will have a lower miss rate and a higher delay. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
Answer:
otp
Explanation:
don't know bro ask to your teacher
