Question

If x and y are related as 3x+4y=20 and the quartile deviation of x is 12,then the quartile deviation of y is

Answer 1

3x+4y= 20

3(12)+4y=20

36+4y=20

4y=20-36=16

4y=16

y=16/4

y= 4

$\frac{16}{4}$

Answer 2

Answer:

Quartile division of $y$ is $9$.

Step-by-step explanation:

Quartile division is defined as half the difference between upper and lower quartile.

It is given that

$3x+4y=20$

$4y=20-3y$

$y=\frac{20}{4}-\frac{3y}{4}$

Now, if $y=a+bx$

Quartile division of $y=$mode$b*$QD of $x$

Hence, $b=\frac{-3}{4}$

mode $b=\frac{3}{4}$

If QD of $x=12$

QD of $y=\frac{3}{4}*12=9$

#SPJ2