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The average (arithmetic mean) of 17 numbers is j. If two of the numb
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14 May 2021, 09:53
14 May 2021, 09:53
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The average (arithmetic mean) of 17 numbers is ๐. If two of the numbers are ๐ and ๐, what is the average of the remaining 15 numbers in terms of ๐, ๐ and ๐?
A. \(\frac{k+m}{17}\)
B. \(17j+k+m\)
C. \(\frac{16j-k-m}{17}\)
D. \(\frac{17j-k-m}{15}\)
E. \(\frac{17(k-m)-j}{15}\)
A. \(\frac{k+m}{17}\)
B. \(17j+k+m\)
C. \(\frac{16j-k-m}{17}\)
D. \(\frac{17j-k-m}{15}\)
E. \(\frac{17(k-m)-j}{15}\)
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Re: The average (arithmetic mean) of 17 numbers is j. If two of the numb
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12 Nov 2021, 11:02
12 Nov 2021, 11:02
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Carcass wrote:
The average (arithmetic mean) of 17 numbers is ๐. If two of the numbers are ๐ and ๐, what is the average of the remaining 15 numbers in terms of ๐, ๐ and ๐?
A. \(\frac{k+m}{17}\)
B. \(17j+k+m\)
C. \(\frac{16j-k-m}{17}\)
D. \(\frac{17j-k-m}{15}\)
E. \(\frac{17(k-m)-j}{15}\)
A. \(\frac{k+m}{17}\)
B. \(17j+k+m\)
C. \(\frac{16j-k-m}{17}\)
D. \(\frac{17j-k-m}{15}\)
E. \(\frac{17(k-m)-j}{15}\)
Let F = the sum of the 15 numbers that don't include k and m
So, the average of the 15 numbers \(= \frac{F}{15}\)
The average (arithmetic mean) of 17 numbers is j.
We can write: \(\frac{F + k + m}{17} = j\)
What is the average of the remaining 15 numbers in terms of j, k and m?
Take: \(\frac{F + k + m}{17} = j\)
Multiply both sides of the equation by 17 to get: \(F + k + m= 17j\)
Subtract k and m from both sides: \(F= 17j - k - m\)
Divide both sides by 15: \(\frac{F}{15}= \frac{17j - k - m}{15}\)
Answer: D
General Discussion
Re: The average (arithmetic mean) of 17 numbers is j. If two of the numb
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14 May 2021, 19:31
14 May 2021, 19:31
1
The average (arithmetic mean) of 17 numbers is ๐.
Total = 17๐
If two of the numbers are ๐ and ๐.
Hence, total of 15 nos = \(17๐ - ๐ - ๐\)
Average = \(\frac{17๐ - ๐ - ๐}{15}\)
Answer D
Total = 17๐
If two of the numbers are ๐ and ๐.
Hence, total of 15 nos = \(17๐ - ๐ - ๐\)
Average = \(\frac{17๐ - ๐ - ๐}{15}\)
Answer D
