The resiliency technique which would provide the aforementioned capabilities is: D. Full backups.
An operating system (OS) can be defined as a system software which is pre-installed on a computing device, so as to manage or control software application, computer hardware and user processes.
In this scenario, a manufacturing company cannot migrate its several one-off legacy information systems (IS) to a newer operating system (OS), due to software compatibility issues.
Resiliency can be defined as a measure of the ability of a network, server, storage system, computing system or data center, to recover quickly and continue operating when it experience adverse conditions such as:
In Computers and Technology, there are four (4) main resiliency technique and these include:
I. Redundancy.
II. RAID 1+5.
III. Virtual machines.
IV. Full backups.
Full backup is a resiliency technique which create backups of the systems for recovery and it allows operating system (OS) patches to be installed on computer systems.
Read more: brainly.com/question/17586013
<span>The RAID level 2 could give high
information exchange rates and would be straightforward contrasted with other
Raid levels. Be that as it may, it has a high cost and would need a high rate move
required with a specific end goal to legitimize this cost.</span>
Can u plz give me a picture
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec
Answer:
- <u>Partial Copy Sandbox,</u>
- <u>Full Sandbox,</u>
- <u>Developer k Pro Sandbox</u>
Explanation:
Note that, cloud services are made possible by computer programming, and <em>soundboxes are used by Software developers to test new programming code before implementation</em>.
Therefore, for UC to successfully test its Service Cloud implementation with real Sales Cloud data this three soundboxes are best suitable.
