Hello , If x is the average (arithmetic mean) of m and 9, y is the average of 2m and 15, and z is the average of 3m and 18, what is the average of x, y, and z in terms of m?
thanks
Answers
Answered by
4
x= ( m+9)/2
y= ( 2m +15)/2
z= (3m +18)/2
Average of x,y,z
x+y+z)/3
= m+9 + 2m +15 + 3m +18/6
= 6 m + 42)/6
= m +7
✌✌✌Dr.Dhruv✌✌✌✌
y= ( 2m +15)/2
z= (3m +18)/2
Average of x,y,z
x+y+z)/3
= m+9 + 2m +15 + 3m +18/6
= 6 m + 42)/6
= m +7
✌✌✌Dr.Dhruv✌✌✌✌
Answered by
5
x=m+9/2
y=2m+15/2
z=3m+18/2
===
===
Average of x,y and z
x+y+z/3
(m+9/2)+(2m+15/2)+(3m+18/2)/3
L.C.M is 2
===
===
→(m+9+2m+15+3m+18/2)/3
→(6m+42/2)/3
→{6(m+7)/2}/3
→3(m+7)/3
→(m+7)
Hence m+7 is answer
y=2m+15/2
z=3m+18/2
===
===
Average of x,y and z
x+y+z/3
(m+9/2)+(2m+15/2)+(3m+18/2)/3
L.C.M is 2
===
===
→(m+9+2m+15+3m+18/2)/3
→(6m+42/2)/3
→{6(m+7)/2}/3
→3(m+7)/3
→(m+7)
Hence m+7 is answer
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