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zimovet [89]
3 years ago
6

Which two forms of financial aid require the student to bear the cost of college education?

SAT
2 answers:
Arada [10]3 years ago
7 0

direct loans and work study programs. is the answer for PLATO

Aliun [14]3 years ago
5 0

Answer:

A. Direct loans and  C. Work-study programs is the correct answer.

Explanation:

Direct loans and Work-study programs are the two forms of financial aid require the student to bear the cost of college education.

direct loans help the student to provide cash that they could confer and be cleared off based on the time.Direct Loan Program provide funding to the students to help them to pay their course cost.

The work-study program assists students where the student gets money and in return, they do work for them with no cash to be given for them.

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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
nexus9112 [7]

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

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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