the everage of 6 number is 30 if one number is excluded ,the everage of remaning number is 29 ,the excluded number is
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Answered by
1
Let, The sum of 6 number's be 'x'.
ie (x/6)=30.
→x=30×6=180 →(1)
Given, The average of 5 numbers is 29
Let that excluded no: be 'n'.
ie (x-n)/5=29
→x-n=145
→x-145=n
From (1). we have x=180
→n=180-145=35.
Hence,the excluded number is 35.
Hope it helps...
Regards,
Leukonov.
ie (x/6)=30.
→x=30×6=180 →(1)
Given, The average of 5 numbers is 29
Let that excluded no: be 'n'.
ie (x-n)/5=29
→x-n=145
→x-145=n
From (1). we have x=180
→n=180-145=35.
Hence,the excluded number is 35.
Hope it helps...
Regards,
Leukonov.
Answered by
0
Hi there!
Here's the answer:
•°•°•°•°•°<><><<><>><><>°•°•°•°•°•
![Average =\frac{Sum\: of\: observations}{No.\:of\: Observations} Average =\frac{Sum\: of\: observations}{No.\:of\: Observations}](https://tex.z-dn.net/?f=Average+%3D%5Cfrac%7BSum%5C%3A+of%5C%3A+observations%7D%7BNo.%5C%3Aof%5C%3A+Observations%7D)
Given,
No. of observations = 6
Average = 30
Let sum be S
=>![30 = \frac{S}{6} 30 = \frac{S}{6}](https://tex.z-dn.net/?f=30+%3D+%5Cfrac%7BS%7D%7B6%7D)
=> S = 60×3
=> S = 180 -----------(1)
When No. of observations = 5[One observation (or No.) is excluded]
Let excluded No. = k
Average = 29
New Sum = (S-k)
=>![29 = \frac{S-k}{5} 29 = \frac{S-k}{5}](https://tex.z-dn.net/?f=29+%3D+%5Cfrac%7BS-k%7D%7B5%7D)
=> S-k = 29×5 = 145
From(1), S = 180
=> k = 180 - 145 = 35
•°• Excluded No. = 35
•°•°•°•°•°<><><<><>><><>°•°•°•°•°•
©#£€®$
:)
Hope it helps
Here's the answer:
•°•°•°•°•°<><><<><>><><>°•°•°•°•°•
Given,
No. of observations = 6
Average = 30
Let sum be S
=>
=> S = 60×3
=> S = 180 -----------(1)
When No. of observations = 5[One observation (or No.) is excluded]
Let excluded No. = k
Average = 29
New Sum = (S-k)
=>
=> S-k = 29×5 = 145
From(1), S = 180
=> k = 180 - 145 = 35
•°• Excluded No. = 35
•°•°•°•°•°<><><<><>><><>°•°•°•°•°•
©#£€®$
:)
Hope it helps
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