Answer:
Yes
Explanation:
Based on this scenario it can be said that Yes, it is ok for you to view this information. The main reason for this being that the information is already in the public domain. Even though you are cleared you still do not have the authority to view private information from the companies projects because you do no longer work for the company but since it has been leaked and already viewed by the entire world then it is ok for you to view as well. Anything made public on the internet can be viewed by anyone around the world.
Answer: False
Explanation:
Yes, the given statement is false as, a software process basically describe the phases by express their frequency and order. But it is also define the products that are deliverable in the projects.
The phases in the software process basically describe the various steps and task which is necessary to make plans and schedules. Project deliverable also include project scheduling for determining the each phases.
Answer:
A snapshot is most similar to incremental backup scheme
<u>Definition</u>
A backup process in which successive copies of data contain the portion that has changed since last copy was made.
<u>Explanation:</u>
Snapshot are similar to incremental backup because only those blocks that are changed after your most recent snapshot has been saved on device.
This is the reason, snapshots are similar to incremental backup schemes.
Answer:
open source software
Explanation:
An open source software is any software that is released under an open source license, meaning that anybody can access its source code, see how it was made, and as the question says, "sutdy, change and improve it".
This is opposite to proprietary or closed source software where you can have a license to use the software but you can't see or change it's "building blocks" (i.e. code)
Answer:
Follows are the explanation of the choices:
Explanation:
Following are the Pseudocode for selection sort:
for j = 0 to k-1 do:
SS = i
For l = i + 1 to k-1 do:
If X(l) < X(SS)
SS= l
End-If
End-For
T = X(j)
X(j) = X(SS)
X(SS) = T
End-For
Following are the description of Loop invariants:
The subarray A[1..j−1] includes the lowest of the j−1 components, ordered into a non-decreasing order, only at beginning of the iteration of its outer for loop.
A[min] is the least amount in subarray A[j.. l−1] only at beginning of the each loop-inner iterations.
Following are the explanation for third question:
Throughout the final step, two elements were left to evaluate their algorithm. Its smaller in A[k-1] would be placed as well as the larger in A[k]. One last is the large and medium component of its sequence because most and the last two components an outer loop invariant has been filtered by the previous version. When we do this n times, its end is a repetitive, one element-sorting phase.
Following is the description of choosing best-case and worst-case in run- time:
The body the if has never been activated whenever the best case time is the list is resolved. This number of transactions are especially in comparison also as a procedure, that will be (n-1)(((n+2)/2)+4).
A structure iterator at every point in the worst case that array is reversed, that doubles its sequence of iterations in the inner loop, that is:(n−1)(n+6) Since both of them take timeΘ(n2).
