Question

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11
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The mean values of x and y for the regression equations 5x+2y-104=0
and 3xty-60 = 0.
A) 16 and 12
B) 10 and 12
C) 12 and 16
D 11 and 15​

Answer 1

Answer:

$\boxed{\bf \: A) \: \: 16 \: and \: 12 \: }$

Step-by-step explanation:

Given regression lines are

$\sf \: 5x + 2y - 104 = 0 - - - (1)$

and

$\sf \: 3x + y - 60 = 0 - - - (2)$

We know, the point of intersection provides the mean values of x and y.

So, to evaluate the value of x and y, we use method of elimination.

So, on multiply equation (2) by 2, we get

$\sf \: 6x + 2y - 120 = 0 - - - (3)$

On Subtracting equation (1) from (3), we get

$\sf \: x - 16 = 0$

$\implies\sf \: x = 16$

On substituting the value of x in equation (2), we get

$\sf \: 3 \times 16 + y - 60 = 0$

$\sf \: 48 + y - 60 = 0$

$\sf \: y - 12 = 0$

$\implies\sf \: y = 12$

As the point of intersection of two regression lines 5x + 2y - 104 = 0 and 3x + y - 60 = 0 is (16, 12)

$\implies\bf \: \overline{x} \: = \: 16, \: \overline{y} \: = \: 12 \:$

Hence,

$\implies\boxed{\sf \: \bf \: \overline{x} \: = \: 16, \: \overline{y} \: = \: 12 \: }$