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olganol [36]
2 years ago
11

Under which menu would you be able to see the number of continuous track minutes available on each mounted hard drive at the cur

rent session's sample rate and bit depth?
Computers and Technology
1 answer:
Crazy boy [7]2 years ago
3 0

Answer:

looooollll

Explanation:

yooooooooo u good

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Which one of these is the most commonly used hardware interface for attaching peripherals to a microcomputer?
Tanzania [10]
Hey there!

The most commonly used hardware interface for attaching peripherals to a microcomputer is the universal series bus, or USB, port. Many computers have at least 2 to 4 USB ports that allow for multiple devices to be plugged in at once. In terms of wired devices as opposed to wireless, this is the main way that a mouse, a keyboard, a smartphone (for charging and syncing information), and many other devices are connected to computers. 

Hope this helped you out! :-)
4 0
3 years ago
Which audio format is used to create chiptunes?
Scilla [17]
Chiptune, also known as chip music or 8-bit music, is synthesized electronic music which is made for programmable sound generator suns chops used in vintage computers,consoles , and arcade machines.
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3 years ago
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Talk to you about... :)
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3 years ago
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

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A program in which students learn a specific skill for a particular job is called a
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