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For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >
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30 Aug 2017, 23:21
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For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a > z. Which of the following is the median of the six integers? A. z−a B. (x+y)/2 C. (a+z)/2 D. z E. (c−y)/2
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Re: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >
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30 Aug 2017, 23:31
Bunuel wrote: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a > z. Which of the following is the median of the six integers?
A. z−a B. (x+y)/2 C. (a+z)/2 D. z E. (c−y)/2 y>x>a>z>b>c (a+z)/2 C
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Re: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >
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31 Aug 2017, 00:09
x>a and a>z Thus, x>a>z x>a>z and y>x Thus, y>x>a>z y>x>a>z and z>b Thus, y>x>a>z>b y>x>a>z>b and b>c Thus, y>x>a>z>b>c Since there are 6 items in the series Median= \(\frac{(3rd item+4th item)}{2}\)= \(\frac{(a+z)}{2}\) Answer: C. Kudos please if you like my explanation!
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Re: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >
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31 Aug 2017, 03:58
Given that :
x>a b>c z>b y>x a>z
If we combine all 5 given equations we can write : y>x>a>z>b>c
Now we have 6 terms.. so median will be average of center 2 terms, ie average of 3rd and 4th term ( if no. of terms odd, median is center term, if no. of terms even median is average of center 2 terms)
our series is : y>x>a>z>b>c So center 2 terms are : a and z
Median = (a+z)/2
Answer : C



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For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >
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31 Aug 2017, 16:34
Bunuel wrote: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a > z. Which of the following is the median of the six integers?
A. z−a B. (x+y)/2 C. (a+z)/2 D. z E. (c−y)/2 x > a b > c z > b y > x a > z To find the greatest and least variables in a sequence, look for the variables that show up only one time on list above, e.g.. Here, those variables are y (the greatest) and c (the smallest). Start on one end, high or low: y > x. Now find x on left side. x > a. Find "a" on left side, etc. Reverse the process on the low end. c is the smallest number. c < b. Look on right side for b. b < z, etc. c < b < z < a < x < y For an even number of terms, the median is the average of the middle two: \(\frac{a + z}{2}\) Answer C
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Re: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >
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02 Sep 2017, 07:20
Bunuel wrote: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a > z. Which of the following is the median of the six integers?
A. z−a B. (x+y)/2 C. (a+z)/2 D. z E. (c−y)/2 Using the given information, we can determine the following: c < b < z < a < x < y Thus, the median is (z + a)/2. Answer: C
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Re: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a > &nbs
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