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frutty [35]
3 years ago
7

It’s been six months since the disk crash at CSM Tech Publishing, and the owner is breathing a little easier because you install

ed a fault-tolerant solution to prevent loss of time and data if a disk crashes in the future. Business is good, and the current solution is starting to run low on disk space. In addition, the owner has some other needs that might require more disk space, and he wants to keep the data on separate volumes (what he calls drives). He wants a flexible solution in which drives and volumes aren’t restricted in their configuration. He also wants to be able to add storage space to existing volumes easily without having to reconfigure existing drives. He has the budget to add a disk storage system that can contain up to 10 HDDs. a. Which Windows feature can accommodate these needs, and how does it work?
Computers and Technology
1 answer:
Semmy [17]3 years ago
5 0

Answer:

Storage Spaces

Explanation:

The feature that would best accommodate his needs would be the Storage Spaces feature that has been implemented in the Windows and Windows Server systems. This feature basically takes various hard drives and combines them together, this ultimately acts as copies of each other which protect the data of each one in the unfortunate case that one of the drives tends to fail. This feature also allows you to add more drives to the existing ones with very minimal effort without restricting configuration. Which would solve the needs that are needed in this scenario.

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2 years ago
Public places where you can wirelessly connect to the internet are known as wifi ________.
AleksAgata [21]

<span>Answer - Hotspots</span>

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2 years ago
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You also learn in class that one kilobyte of computer memory will hold about 2/3 of a page of typical text (without formatting).
Travka [436]

Answer:

The flashdrive can hold 35389 400-pages-books

Explanation:

If \frac{2}{3} of a page occupies 1 kB of memory, we can calculate how much memory a book will take

2/3kB=1 page\\ x kB=400 pages\\266.67 kB=400 pages

Now that we know that a book average file size is about 266,67 kB, we calculate how many of them can a 9 GB flash drive hold.

To do the calculation, we have to know how many kilobytes are in 9 gigabytes.

There is 1024 kilobytes in a megabyte, and 1024 megabytes in a gigabyte, so:

Memory_{in _kilobytes}^{} =1024\frac{kilobytes}{megabytes} *1024\frac{megabytes}{gigabytes}*9=9437184 kilobytes

Finally, knowing the average file size of a book and how much memory in kilobytes the 9 GB flash drive holds, we calculate how many books can it hold.

Books=\frac{Flash drive memory}{Filesize of a book} =\frac{9437184kilobytes}{266.67\frac{kilobytes}{book} } =35389,4 books

The flashdrive can hold 35389 400-pages-books, or 14155776 pages of typical text.

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3 years ago
What would happen if computers only had input peripherals and a CPU?
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2 years ago
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