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qaws [65]
2 years ago
15

G(x) 2x/3+3 what is the value of g(-30)

SAT
2 answers:
evablogger [386]2 years ago
6 0
G(-30)=2(-30)/3+3
g(-30)=(-60/3)+3
g(-30)=-20+3
g(-30)=-17
9966 [12]2 years ago
3 0
G(-30)=2(-30)/3+3
-60/3+3
-20+3
-17
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A single-platter disk has rotation speed 7200 rpm, number of tracks on one side of platter is 30,000, number of sectors per trac
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Answer:

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 O. 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

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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

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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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2 years ago
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Answer:

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