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Suppose that the average (arithmetic mean)
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21 May 2021, 03:50
21 May 2021, 03:50
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Suppose that the average (arithmetic mean) of ๐, ๐, and ๐ is โ, the average of ๐, ๐, and ๐ is ๐, and the average of ๐ and ๐ is ๐. What is the average of ๐ and ๐?
A. \(h-j+k\)
B. \(\frac{3h+3j-2k}{2}\)
C. \(\frac{3h-3j+2k}{2}\)
D. \(\frac{3h-3j+2k}{5}\)
E. \(\frac{3h-3j+2k}{8}\)
A. \(h-j+k\)
B. \(\frac{3h+3j-2k}{2}\)
C. \(\frac{3h-3j+2k}{2}\)
D. \(\frac{3h-3j+2k}{5}\)
E. \(\frac{3h-3j+2k}{8}\)
Re: Suppose that the average (arithmetic mean)
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21 May 2021, 05:30
21 May 2021, 05:30
1
a+b+c= 3h -----(1)
b+c+d= 3j -----(2)
d+e= 2k -----(3)
(1) - (2) a-d= 3h-3j ----(4)
(4) + (3) a+e= 3h-3j+2k
average of ๐ and ๐ is = (a+e)/2 = (3h-3j+2k)/2
option c is correct
b+c+d= 3j -----(2)
d+e= 2k -----(3)
(1) - (2) a-d= 3h-3j ----(4)
(4) + (3) a+e= 3h-3j+2k
average of ๐ and ๐ is = (a+e)/2 = (3h-3j+2k)/2
option c is correct
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Re: Suppose that the average (arithmetic mean)
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09 Jul 2021, 06:43
09 Jul 2021, 06:43
1
Theory: Average = \(\frac{Sum Of All The Values}{Total Number Of Values}\)
=> Sum = Average* Total Number of Values
the average (arithmetic mean) of ๐, ๐, and ๐ is โ
=> a + b + c = h * 3 [as there are three values]
=> a + b + c = 3h ...(1)
the average of ๐, ๐, and ๐ is ๐
=> b + c + d = j * 3 [as there are three values]
=> b + c + d = 3j ...(2)
the average of ๐ and ๐ is ๐
=> d + e = k * 2 [as there are two values]
=> d + e = 2k ...(3)
What is the average of ๐ and ๐?
(1) - (2) + (3) will give us
a + b + c - (b + c + d) + d + e = 3h - 3j + 2k
=> a + b - b + c - c - d + d + e = 3h - 3j + 2k
=> a + e = 3h - 3j + 2k
Average of a + e = \(\frac{a + e}{2}\) = \(\frac{3h - 3j + 2k}{2}\)
So, answer will be C
Hope it helps!
To learn more about Statistics watch the following video
=> Sum = Average* Total Number of Values
the average (arithmetic mean) of ๐, ๐, and ๐ is โ
=> a + b + c = h * 3 [as there are three values]
=> a + b + c = 3h ...(1)
the average of ๐, ๐, and ๐ is ๐
=> b + c + d = j * 3 [as there are three values]
=> b + c + d = 3j ...(2)
the average of ๐ and ๐ is ๐
=> d + e = k * 2 [as there are two values]
=> d + e = 2k ...(3)
What is the average of ๐ and ๐?
(1) - (2) + (3) will give us
a + b + c - (b + c + d) + d + e = 3h - 3j + 2k
=> a + b - b + c - c - d + d + e = 3h - 3j + 2k
=> a + e = 3h - 3j + 2k
Average of a + e = \(\frac{a + e}{2}\) = \(\frac{3h - 3j + 2k}{2}\)
So, answer will be C
Hope it helps!
To learn more about Statistics watch the following video
