Answer:
Swap daemon
Explanation:
Swap daemon manages the physical memory by moving process from physical memory to swap space when more physical memory is needed. The main function of the swap daemon is to monitor processes running on a computer to determine whether or not it requires to be swapped.
The physical memory of a computer system is known as random access memory (RAM).
A random access memory (RAM) can be defined as the internal hardware memory which allows data to be read and written (changed) in a computer.Basically, a random access memory (RAM) is used for temporarily storing data such as software programs, operating system (OS),machine code and working data (data in current use) so that they are easily and rapidly accessible to the central processing unit (CPU).
Additionally, RAM is a volatile memory because any data stored in it would be lost or erased once the computer is turned off. Thus, it can only retain data while the computer is turned on and as such is considered to be a short-term memory.
There are two (2) main types of random access memory (RAM) and these are;
1. Static Random Access Memory (SRAM).
2. Dynamic Random Access Memory (DRAM).
Out-of-band management is required to establish a new network switch and configure its IP address for the first time.
<h3>What is Out-of-Band management?</h3>
Out-of-Band management is known to be a form of services that makes sure that a tech do not require to be sent onsite and it is one where remediation can be done in a remote place.
Hence, Out-of-band management is required to establish a new network switch and configure its IP address for the first time.
See full question below
Which of the following is required to establish a new network switch and configure its IP address for the first time?
A. Client-to-site VPN
B. Site-to-site VPN
C. Out-of-band management
D. In-band management
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The time that will be used to perform a multiplication using the approach is 28 time units.
<h3>How to calculate time taken?</h3>
From the information given, the following can be noted:
A = 8 (bits wide)
B = 4 time units.
The multiplication will be performed based on the adder stack. Since A = 8, then, A - 1 = 8 - 1 = 7.
Now, the time taken will be:
= 7 × B
= 7 × 4tu
= 28 tu
In conclusion, the correct option is 28 time units.
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