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kumpel [21]
2 years ago
8

A local supermarket organised two weeks of promotions to boost sales. The average daily sales before the campaign is $10 000, wh

ich becomes $11 000 during the promotion (average over 2 weeks). Which analysis below is not useful for evaluating whether the increase in sales is statistically significant?.
SAT
1 answer:
mamaluj [8]2 years ago
8 0

The analysis that is not useful for evaluating whether the increases in sales are statistically significant is to monitor sales for a longer period of time.

Statistical analysis helps to use different methods to properly collect data, and therefore future predictions and solutions. This is important to achieve organizational goals.

From the complete information, analysis that is not useful for evaluating whether the increases in sales are statistically significant is to monitor sales for a longer period of time.

Learn more about sales on:

brainly.com/question/14727688

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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

tracks traversed. So the seek time for 29,999 tracks traversed is

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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

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

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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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