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T (triple) = Three Users provided this score. D (double) = T

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T (triple) = Three Users provided this score. D (double) = T [#permalink] New post 07 Jun 2018, 09:24
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Attachment:
stores.jpg
stores.jpg [ 22.44 KiB | Viewed 280 times ]


T (triple) = Three Users provided this score. D (double) = Two Users provided this score.

What was the average (arithmetic mean) rating for Readily available in stores among the items that were rated 8 for Durability?

A. 3

B. 4

C. 5

D. 6

E. 7
[Reveal] Spoiler: OA

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Re: T (triple) = Three Users provided this score. D (double) = T [#permalink] New post 07 Jun 2018, 23:17
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Carcass wrote:
Attachment:
stores.jpg


T (triple) = Three Users provided this score. D (double) = Two Users provided this score.

What was the average (arithmetic mean) rating for Readily available in stores among the items that were rated 8 for Durability?

A. 3

B. 4

C. 5

D. 6

E. 7


Just want to see whether my solution is correct.

Here for 8 durability , we have 8, 4 and 3 as the values.

Then to know AM = 8+4+3 / 3 = 15/3 = 5.

Answer C. I hope T and D not have any influence for this question, if yes then please do let me know.
Re: T (triple) = Three Users provided this score. D (double) = T   [#permalink] 07 Jun 2018, 23:17
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T (triple) = Three Users provided this score. D (double) = T

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