Btw..which dtadium do u want cricket or football!?
Answer:
The NEA’s focus on building new performing arts centers led to an increased production of arts.
#1 is D #2 is B #3 is A #4 is D #5 is B
Answer:
b) Bounded Waiting
Explanation:
int currentThread = 1;
bool thread1Access = true;
bool thread2Access = true;
thread1 { thread2 {
While (true) {
While (true)
{
while(thread2Access == true)
{
while(thread1Access == true)
{
If (currentThread == 2) {
If (currentThread == 1)
{
thread1Access = false; thread2Access = false;
While (currentThread == 2);
While (currentThread == 1);
thread1Access = true; thread2Access = true;
} }
/* start of critical section */ /* start of critical section */
currentThread = 2 currentThread = 1
… ...
/* end of critical section */ /* end of critical section */
thread1Access = false; thread2Access = false;
… ...
} }
} }
} }
It can be seen that in all the instances, both threads are programmed to share same resource at the same time, and hence this is the bounded waiting. For Mutual exclusion, two threads cannot share one resource at one time. They must share simultaneously. Also there should be no deadlock. For Progress each thread should have exclusive access to all the resources. Thus its definitely the not the Progress. And hence its Bounded waiting.
Answer:
import java.util.Arrays;
import java.util.Scanner;
public class num4 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("How many numbers? ");
int n = in.nextInt();
int []intArray = new int[n];
//Entering the values
for(int i=0; i<intArray.length;i++){
System.out.println("Enter the numbers");
intArray[i]=in.nextInt();
}
System.out.println(Arrays.toString(intArray));
int min =intArray[0];
for(int i =0; i<intArray.length; i++){
if(min>intArray[i]){
min = intArray[i];
}
}
System.out.println("The Minimum of the numbers is "+min);
}
}
Explanation:
- Using Java programming language
- Prompt the user for the number of values
- Using Scanner class receive and store in a variable
- Create an array of size n
- Using an for loop continuously ask the user to enter the integers
- Print the array of integers
- Using another for loop with an if statement, find the smallest element in the array of numbers
- Output the the smallest number
