A) \[(m-n)\]
B) \[mn\]
C) \[(m+n)\]
D) \[m/n\]
Correct Answer: B
Solution :
[b] Total of \[(m+n)\] numbers \[=(m{{n}^{2}}+n{{m}^{2}})=mn\,(m+n)\] Average of \[(m+n)\] numbers \[=\frac{mn\,(m+n)}{(m+n)}=mn\] |
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