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In the list above, if n is an integer between 1 and 10, inclusive, the
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16 Sep 2018, 22:28
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2 4 6 8 n 3 5 7 9 In the list above, if n is an integer between 1 and 10, inclusive, then the median must be A. 4 or 5 B. 5 or 6 C. 6 or 7 E. n E. 5.5
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In the list above, if n is an integer between 1 and 10, inclusive, the
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16 Sep 2018, 23:02
Afc0892 wrote: 2+4+6+8+3+5+7+9+n = 44+n
If n is 1 then median is 45/9 = 5
If n is 10 then median is 54/9 = 6
B is the answer. Afc0892Your method is perfect if n must equal 1 OR 10, because then the integers are consecutive and median = mean. But median \(\neq\) mean if n = 2,3,4,5,6,7,8, or 9 n is an integer between 1 and 10, inclusive. If n = 2, mean = \(\frac{46}{9}=5.1111\) If n = 7, mean = \(\frac{51}{9}= 5.67\) Just an issue with the method (absent a mention of extremes and why they work)
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Re: In the list above, if n is an integer between 1 and 10, inclusive, the
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16 Sep 2018, 22:36
2+4+6+8+3+5+7+9+n = 44+n
If n is 1 then median is 45/9 = 5
If n is 10 then median is 54/9 = 6
B is the answer.



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Re: In the list above, if n is an integer between 1 and 10, inclusive, the
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16 Sep 2018, 23:05
generis wrote: Afc0892 wrote: 2+4+6+8+3+5+7+9+n = 44+n
If n is 1 then median is 45/9 = 5
If n is 10 then median is 54/9 = 6
B is the answer. Afc0892Your method is perfect if n must equal 1 OR 10, because then the integers are consecutive and median = mean. But median \(\neq\) mean if n = 2,3,4,5,6,7,8, or 9 n is an integer between 1 and 10, inclusive. If n = 2, mean = \(\frac{46}{9}=5.1111\) If n = 7, mean = \(\frac{51}{9}= 5.67\) Just an issue with the method (absent a mention of extremes and why they work) Got it. Thanks



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Re: 2,4,6,8,n,3,5,7,9 In the list above, if n is an integer
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26 Nov 2019, 13:03
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Re: 2,4,6,8,n,3,5,7,9 In the list above, if n is an integer
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