Answer:
The answer is "Option b".
Explanation:
The active directory can manage the data with active directory customers but a snap-on machine. It is used in scrolling the list, as the earlier replies demonstrate. It will appear on the screen if there is no other startup software installed on a computer.
This snap-on desktop is only a component, that allows its simultaneous use of different systems or devices on the very same system. It also turns the objects more or less into the pieces of the whole.
Answer:
1-rollover, 2-straight-through, 3-crossover are the correct answer of this question .
Explanation:
Rollover , Straight-through and Crossover are the types of UTP cables that can be used in connected to the device.
- Rollover refers to a slider on a Website page which allowing the user to engage with the Website page.
- In Straight-through a network patch panels which connects a machine to a node on the web.
- A Crossover Ethernet used during direct communication of portable devices. It combines two similar-type computers.
Answer: Variety
Explanation:
According to the given question, Haley is dealing with the variety of the organization products posts on the social media networking as Haley's employer wants to comping all the data or information related to the products into the database system.
- The database is one of the type of management system that manages all the database that are shared by the customers or users.
- The main important function of the database is that it organize the data more accurately and properly in the system.
- The database management system (DBMS) handle all the data in the system more effectively from the variety of the users.
Therefore, Variety is the correct answer.
Answer:
Check the explanation
Explanation:
1) f(n) = O( 1 ), since the loops runs a constant number of times independent of any input size
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
2) f(n) = O( log n! ), the outer loop runs for n times, and the inner loop runs log k times when i = k,ie the total number of print will be – log 1 + log2 +log3 +log4+…...+ log n = log (1 . 2 . 3 . 4 . ……. . n ) =log n!
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
Note : Log (m *n) = Log m + Log n : this is property of logarithm
3) f(n) = 
, since both outer and inner loop runs n times hence , the total iterations of print statement will be : n +n+n+…+n
for n times, this makes the complexity – n * n = n2
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
