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If the average (arithmetic mean) of k and k + 5 is b and if th
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14 May 2021, 09:55
14 May 2021, 09:55
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If the average (arithmetic mean) of ๐ and ๐ + 5 is ๐ and if the average of ๐ and ๐ โ 9 is ๐, what is the average of ๐ and ๐?
A. ๐ โ 2
B. ๐ โ 1
C. ๐
D. ๐ + 1 /2
E. 2๐
A. ๐ โ 2
B. ๐ โ 1
C. ๐
D. ๐ + 1 /2
E. 2๐
If the average (arithmetic mean) of k and k + 5 is b and if th
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14 May 2021, 19:36
14 May 2021, 19:36
1
If the average of ๐ and ๐ + 5 is ๐
\(๐ = \frac{๐ + ๐ + 5}{2}\)
If the average of ๐ and ๐ โ 9 is ๐
\(๐ = \frac{๐ + ๐ - 9}{2}\)
Average of ๐ & ๐ = ( \(\frac{2๐ + 5}{2} + \frac{2๐ - 9}{2}\) ) * \(\frac{1}{2}\)
= \(๐ - 1\)
Answer B
\(๐ = \frac{๐ + ๐ + 5}{2}\)
If the average of ๐ and ๐ โ 9 is ๐
\(๐ = \frac{๐ + ๐ - 9}{2}\)
Average of ๐ & ๐ = ( \(\frac{2๐ + 5}{2} + \frac{2๐ - 9}{2}\) ) * \(\frac{1}{2}\)
= \(๐ - 1\)
Answer B
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If the average (arithmetic mean) of k and k + 5 is b and if th
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09 Jul 2021, 06:49
09 Jul 2021, 06:49
2
Theory
If the average (arithmetic mean) of ๐ and ๐ + 5 is ๐
=> k + (k+5) = b * 2 [As there are 2 values]
=> 2k + 5 = 2b
=> b = \(\frac{2k+5}{2}\) = k + \(\frac{5}{2}\)
If the average of ๐ and ๐ โ 9 is ๐
=> k + (k-9) = c * 2 [As there are 2 values]
=> 2k - 9 = 2c
=> c = \(\frac{2k-9}{2}\) = k - \(\frac{9}{2}\)
What is the average of ๐ and ๐
Average of b and c = \(\frac{b+c}{2}\) = (\(\frac{1}{2})\)*(k + \(\frac{5}{2}\) + k - \(\frac{9}{2}\))
= (\(\frac{1}{2})\)*(2k - \(\frac{4}{2}\))
= (\(\frac{1}{2})\)*(2k - 2) = k - 1
So, Answer will be B
Hope it helps!
To learn more about Statistics watch the following video
- โก Average = Sum of all the Values / Total Number of Values
โก Sum of All the values = Average * Total Number of Values
If the average (arithmetic mean) of ๐ and ๐ + 5 is ๐
=> k + (k+5) = b * 2 [As there are 2 values]
=> 2k + 5 = 2b
=> b = \(\frac{2k+5}{2}\) = k + \(\frac{5}{2}\)
If the average of ๐ and ๐ โ 9 is ๐
=> k + (k-9) = c * 2 [As there are 2 values]
=> 2k - 9 = 2c
=> c = \(\frac{2k-9}{2}\) = k - \(\frac{9}{2}\)
What is the average of ๐ and ๐
Average of b and c = \(\frac{b+c}{2}\) = (\(\frac{1}{2})\)*(k + \(\frac{5}{2}\) + k - \(\frac{9}{2}\))
= (\(\frac{1}{2})\)*(2k - \(\frac{4}{2}\))
= (\(\frac{1}{2})\)*(2k - 2) = k - 1
So, Answer will be B
Hope it helps!
To learn more about Statistics watch the following video
