I think it can't.
The one that get high needs met rating are the pages that satisfy the user's search. The search engine result page that shown at the top of the result doesn't necessarily fulfil this criteria
hope this helps
I usually include my teachers name in the upper right hand corner along with the date my name and the subject.
Answer:
Option b: Instruct the user on how to prevent the problem.
Explanation:
As a system verifier, the main job is not only to spot the error but also to ensure the errors are flagged and some remedies can be recommended. The most direct way to avoid the computer user having the same problem is to give instruction the user on how to prevent the problem. The instruction can be written in a document with a proper step by step guideline that is easy to follow. This can save computer user a lot of times to tackle the same problem that has been addressed before.
Answer:
The correct answer to the following question will be "Run-length encoding (RLE)".
Explanation:
RLE seems to be useful for replicated information, swapping it with a qualify as well as a copy of something like a repeat element.
- Optimized dictionary strategies construct a table of sequences, then substitute appearances of chords with simpler codes.
- This is a straightforward type of data compression, where data runs become stored as an individual data count as well as a value rather than as the initial run.
Answer:
a. 100.
b. 31500.
Explanation:
So, we are given the following data which is going to help in solving this particular question.
The time required to execute (on the average) = 1/2 microsecond , a page fault takes of processor time to handle = 250 microseconds and the disk time to read in the page = 10 milliseconds.
Thus, the time taken by the processor to handle the page fault = 250 microseconds / 1000 = 0.25 milliseconds.
The execution time = [ 1/2 microseconds ]/ 1000 = 0.0005 milliseconds.
The number of Pages sent in a second by the disc = 1000/10 milliseconds = 100.
Assuming U = 1.
Hence, the disc transfer time = [2/3 × 1 } + [ 1/3 × 0.25 milliseconds + 15 ] × 2.
=0.667 + 15.083.
= 15.75 millisecond.
Average number of instruction = 15.75/0.0005 = 31500.