Explanation & answer:
Four pair UTP (unshielded twisted pair) cables are common cables with RJ45 connectors. Most RJ45 connectors have a snap to lock the male connector in place.
ST and SC connectors are for fibre-optics.
The appropriate answer is d. mail merge. Mail merge uses a database of addresses that are used to create pre-addressed mailing labels that are generally used when sending letters to a very large group. This type of application is used by utility companies or any other organizations that requires mass mailings. Mail merge is found in the Microsoft Word application. Excell spreadsheets can also be used to complete tasks similar to that of mail merge.
Answer:
Explanation:
Following are the Semaphores:
Customers: Counts waiting customers;
Barbers: Number of idle barbers (0 or 1)
mutex: Used for mutual exclusion.
Cutting: Ensures that the barber won’t cut another customer’s hair before the previous customer leaves
Shared data variable:
count_cust: Counts waiting customers. ------------copy of customers. As value of semaphores can’t access directly.
// shared data
semaphore customers = 0; semaphore barbers = 0; semaphore cutting = 0; semaphore mutex = 1;
int count_cust= 0;
void barber() {
while(true) { //shop is always open
wait(customers); //sleep when there are no waiting customers
wait(mutex); //mutex for accessing customers1
count_cust= count_cust-1; //customer left
signal(barbers);
signal(mutex);
cut_hair();
}
}
void customer() {
wait(mutex); //mutex for accessing count_cust
if (count_cust< n) {
count_cust= count_cust+1; //new customer
signal(customers); signal(mutex);
wait(barbers); //wait for available barbers get_haircut();
}
else { //do nothing (leave) when all chairs are used. signal(mutex);
}
}
cut_hair(){ waiting(cutting);
}
get_haircut(){
get hair cut for some time; signal(cutting);
}
B. Metropolitan area network
Answer:
0+1=1
1+1=2
1+2=3
2+3=5
3+5=8
5+8=13
Explanation:
// C++ program to print
// first n Fibonacci numbers
#include <bits/stdc++.h>
using namespace std;
// Function to print
// first n Fibonacci Numbers
void printFibonacciNumbers(int n)
{
int f1 = 0, f2 = 1, i;
if (n < 1)
return;
cout << f1 << " ";
for (i = 1; i < n; i++) {
cout << f2 << " ";
int next = f1 + f2;
f1 = f2;
f2 = next;
}
}
// Driver Code
int main()
{
printFibonacciNumbers(7);
return 0;
}
