find the average of 32/5+53/10+21/5 and 77/10
Answers
Answered by
11
32/5+21/5+53/10+77/10
53/5+130/10
26.5+13
39.5
53/5+130/10
26.5+13
39.5
Answered by
24
Hi,
Sum of the terms
= 32/5 + 53/10 + 21/5 + 77/10
= ( 64 + 53 + 42 + 77 ) / 10
= 236/10 ---( 1 )
Number of terms = 4
Mean = ( 1 ) / ( 2 )
= ( 236 / 10 ) / 4
= ( 236 ) / ( 10 × 4 )
= 59/10
= 5.9
Therefore ,
Required mean = 5.9
I hope this helps you.
:)
Sum of the terms
= 32/5 + 53/10 + 21/5 + 77/10
= ( 64 + 53 + 42 + 77 ) / 10
= 236/10 ---( 1 )
Number of terms = 4
Mean = ( 1 ) / ( 2 )
= ( 236 / 10 ) / 4
= ( 236 ) / ( 10 × 4 )
= 59/10
= 5.9
Therefore ,
Required mean = 5.9
I hope this helps you.
:)
ABHAYSTAR:
Awesome answer sir !
Similar questions