Given the plane 3x + 5y - z = 9, give an equation of a plane parallel to the given plane, and a plane orthogonal to the given plane.

parallel: 3x +5y -z = 0perpendicular: 5x -3y = 0

Step-by-step explanation:

A parallel plane will have the same normal vector (the coefficients of x, y, z), so will differ only in the constant. Changing the constant from 9 to anything else gives the equation of a parallel plane.

A perpendicular plane will have a normal vector that is perpendicular to the normal vector of the given plane. That is, the dot-product of the normal vectors will be zero. There are an infinite number of possible solutions. One of them is ...

  5x -3y = 0

Its normal vector is <5, -3, 0> and the dot-product of that with the normal vector of the given plane is ...

  <5, -3, 0> · <3, 5, -1> = (5)(3) +(-3)(5) +(0)(-1) = 15 -15 +0 = 0

Any plane whose coefficients a, b, c satisfy 3a+5b-c = 0 will be a normal plane.

Step-by-step explanation:

a)

Given that ; X ~ N ( µ = 65 , σ = 4 )P ( 64 < X < 66 )

From application of normal distribution ;

Z = ( X - µ ) / σ, Z = ( 64 - 65 ) / 4,  Z = -0.25Z = ( 66 - 65 ) / 4,  Z = 0.25

Hence, P ( -0.25 < Z < 0.25 ) = P ( 64 < X < 66 ) = P ( Z < 0.25 ) - P ( Z < -0.25 ) P ( 64 < X < 66 ) = 0.5987 - 0.4013

P ( 64 < X < 66 ) = 0.1974

b) X ~ N ( µ = 65 , σ = 4 )

P ( 64 < X < 66 )

From normal distribution application ;

Z = ( X - µ ) / ( σ / √(n)), plugging in the values, Z = ( 64 - 65 ) / ( 4 / √(12)) = Z = -0.866Z = ( 66 - 65 ) / ( 4 / √(12)) = Z = 0.866

P ( -0.87 < Z < 0.87 )

P ( 64 < X < 66 ) = P ( Z < 0.87 ) - P ( Z < -0.87 )P ( 64 < X < 66 ) = 0.8068 - 0.1932P ( 64 < X < 66 ) = 0.6135

c) From the values gotten for (a) and (b), it is indicative that the probability in part (b) is much higher because the standard deviation is smaller for the x distribution.