Let, coordinate of point A' is (x,y).
Since, A' is the symmetric point A(3, 2) with respect to the line 2x + y - 12 = 0.
So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.
Now, their center is given by :
Also, product of slope will be -1 .( Since, they are parallel )
x = 2y - 1
So,
Also, C satisfy given line :
Also,
Therefore, the symmetric points is .
Answer:
The third quartile is 56.45
Step-by-step explanation:
The given parameters are;
The first quartile, Q₁ = 30.8
The median or second quartile, Q₂ = 48.5
The mean, = 42.0
Coefficient of skewness = -0.38
The Bowley's coefficient of skewness (SK) is given as follows;
Plugging in the values, we have;
Which gives;
-0.38×(Q₃ - 30.8) = Q₃ + 30.8 - 2 × 48.5
11.704 - 0.38·Q₃ = Q₃ - 66.2
1.38·Q₃ = 11.704 + 66.2 = 77.904
Q₃ = 56.45
The third quartile = 56.45.