Question

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?​

Answer 1

Answer:

$\huge{\boxed{\rm{\red{answer=m+7}}}}$

Step-by-step explanation:

Given that:-

The average of m and 9=x

=>(m+9)/2=x

x=(m+9)/2-----(1)

The average of 2 m and 15=y

=>(2m+15)/2=y

y=(2m+15)/2-----(2)

The average of 3m and 18=z

=>(3m+18)/2=z

z=(3m+18)/2-----(3)

The average of x,y,z=

=>Sum of observations/No.of observations

=>(x+y+z)/3

from(1),(2)&(3)

=>[{(m+9)/2}+{(2m+15)/2}+{(3m+18)/2}]/3

=>[(m+9+2m+15+3m+18)/2]/3

=>[(6m+42)/2]/3

=>(6m+42)/(2×3)

=>(6m+42)/6

=>6(m+7)/6

=>m+7

The average of x,y,z= m+7

Answer 2

Answer:

$\huge\fbox\pink{AnsWer=m+7}$

Step-by-step explanation:

Given that:-

The average of m and 9=x

=>(m+9)/2=x

x=(m+9)/2-----(1)

The average of 2 m and 15=y

=>(2m+15)/2=y

y=(2m+15)/2-----(2)

The average of 3m and 18=z

=>(3m+18)/2=z

z=(3m+18)/2-----(3)

The average of x,y,z=

=>Sum of observations/No.of observations

=>(x+y+z)/3

from(1),(2)&(3)

=>[{(m+9)/2}+{(2m+15)/2}+{(3m+18)/2}]/3

=>[(m+9+2m+15+3m+18)/2]/3

=>[(6m+42)/2]/3

=>(6m+42)/(2×3)

=>(6m+42)/6

=>6(m+7)/6

=>m+7

The average of x,y,z= m+7

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