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If the average (arithmetic mean) of 31, 41, and p is between
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09 Dec 2017, 18:21
09 Dec 2017, 18:21
Expert Reply
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Question Stats:
88% (01:30) correct
11% (01:28) wrong based on 18 sessions
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If the average (arithmetic mean) of 31, 41, and p is between 29 and 47, inclusive, what is the least possible value of \((p - 7)^2 =\)
Drill 2
Question: 6
Page: 525
Question: 6
Page: 525
Show: :: OA
64
Re: If the average (arithmetic mean) of 31, 41, and p is between
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15 Dec 2017, 07:24
15 Dec 2017, 07:24
1
I think there's a typo in the question. It should be \((p - 7)^2\).
Solution:
To get the least possible value of p, we need p to be as close to 7 as possible.
Of course, because p is contingent on the average, we should form an equation for that:
\(\frac{31 + 41 + p}{3} = X\)
\(p = 3(X) - 72\)
X can be anywhere from 29 to 47 inclusive, but 29 gives the closest possible value to 7, which is 15. The answer is then (15-7)^2 = 64.
Solution:
To get the least possible value of p, we need p to be as close to 7 as possible.
Of course, because p is contingent on the average, we should form an equation for that:
\(\frac{31 + 41 + p}{3} = X\)
\(p = 3(X) - 72\)
X can be anywhere from 29 to 47 inclusive, but 29 gives the closest possible value to 7, which is 15. The answer is then (15-7)^2 = 64.
Re: If the average (arithmetic mean) of 31, 41, and p is between
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15 Dec 2017, 09:54
15 Dec 2017, 09:54
Expert Reply
You are right!
Fixed it! Thanks
Fixed it! Thanks
