I think B and the second one C
Answer:
The order of the efficiencies is as following:-
10,000 < 2n < nlog(n) < n5 < n!.
Explanation:
10,000 is constant time whatever will be the size of the problem the efficiency will remain the same.
2n this efficiency is linear it will grow proportionally as the size of the problem increases.
nlog(n) this efficiency is is a bit greater than 2n though it will grow faster than 2n but slower than n2 as the size of the problem increases.
n5 this efficiency is very poor.It is growing very rapidly as the size of the problem increases.
n! is the worst efficiency of them all.
n!=n*(n-1)*(n-2)*(n-3)*(n-4)*.......2*1.
It will grow beanstalk in jack and the beanstalk.
Answer:
Explanation:
The following Java code creates both the Advance class and Student Advance class. The StudentAdvance class extends the Advance class and uses its constructor to always be up to date even when there are changes in the Advance class. Due to technical difficulties I have attached the code as a txt file below.
Answer:
Dell is a globally integrated company
Explanation:
A company that designs its strategy, management, and operations in pursuit of a new goal; which is the integration of production and value delivery worldwide is known to be a globally integrated company.
From the question, Dell was described as an American corporation that deals with computer technology with its worldwide sourcing and fully merged production and marketing system. Thus, Dell is a globally integrated company.
Answer:
It can be repaired with disk utility / recovery algorithms
In UNIX it can be done by scanning
In FAT file scan the FAT looking for free entries
Explanation:
When the crash happens, it is not a problem for UNIX and FAT file system
It can be repaired with disk utility/recovery algorithms.
The recovery algorithm is a list of all blocks in all files and takes a compliment as new free file.
In UNIX scanning can be done at all I-nodes.
In FAT file problem cannot occur because there is no free list .If there was any problem than it would have to be done to recover it to scan the FAT looking for free entries.
