D doesn't make much sense, it's valid, but not what we're looking for. B also has a correlation with D, same reasoning applies. As for A, it seems pretty legit, but I don't believe that's what we're looking for.
Choice C is the most obvious one though. We're talking about a network and as may or may not know it's a wireless one in a manner of speaking. A <span>couple of computers in the network that have trouble maintaining a signal will indefinitely lead to failure of a network since both the links and nodes of certain computer systems are incapable of maintaining a signal. </span>
Answer:
<u>720</u> possible PIN can be generated.
Explanation:
To calculate different number of orders of digits to create password and PIN, we calculate permutation.
Permutation is a term that means the number of methods or ways in which different numbers, alphabets, characters and objects can arranged or organized. To calculate the permutation following formula will be used:
nPr = n!/(n-r)!
there P is permutation, n is number of digits that need to be organize, r is the size of subset (Number of digits a password contains)
So in question we need to calculate
P=?
where
n= 10 (0-9 means total 10 digits)
r= 3 (PIN Consist of three digits)
So by using formula
10P3 = 10!/(10-3)!
=10!/7!
= 10x9x8x7!/7!
= 10x9x8
= 720
Answer:
B. The process will be terminated by the operating system.
Explanation:
When Round Robin is used for CPU scheduling, a time scheduler which is a component of the operating system is used in regulating the operation. A time limit is set for each of the processes to be run.
So, when a process fails to complete running before its time elapses, the time scheduler would log it off and return it to the queue. This queue is in a circular form and gives each of the processes a chance to run its course.
Stack is LIFO data structure (Last In First Out) where the last element entered in stack will be the last one to be out of stack. It has three operations: push() : used to insert an element in stack, pop() : used to delete an element from the stack, top() : used to return the top of the stack i.e. the newest member of the stack. All these operations will take place at the top.
<u>Explanation:</u>
Now, looking at the program, x and y are initialized the values of 2 and 3 respectively. The stack pushes 8 onto the stack making it the first member of the stack. Then the value of x which is 2 is pushed onto the stack. Next, (x+5) = (2+5) = 7 is pushed onto the stack.
Pop() is used to delete hence 7 is popped out from the stack. top() is assigned to y which is 2 in this case and again 2 is popped out from the stack. Now, (x+y) = (2+2) = 4 is pushed onto the stack. And the top() is assigned to x which is 4. 4 is again popped out from the stack. Hence the value of x is 4.
