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Allushta [10]
3 years ago
11

Find the equation for the plane that contains the line x=−1+3t , y=1+2t, z=2+4t and is perpendicular to the plane containing the

two lines
L1: x=2t+1,y=-t-1,z=6t+5

L2:x=t+1,y=t-1,z=-3t+5
Mathematics
1 answer:
Ivan3 years ago
8 0

Let L be the line given by the vector equation

(-1,1,2) + \lambda(3,2,4) \ , \lambda \in \mathbb{R}.

First, we use the director vectors of the lines L1 and L2 to get the

vector equation of the plane containing them, which we denote by \Pi_1. This is,

\\\\\Pi_1  : (1,-1,5) + \alpha (2,-1,6) + \beta (1,1,-3) \ , \alpha, \beta \in \mathbb{R}\\\\\\

We observe that \vec{N} = (2,-1,6)\times(1,1,-3) = (-3,12,3) \ne (0,0,0). Therefore, the vector equation of \Pi_1 defines a plane and \vec{N} is a normal vector to \Pi_1.

 

Finally, the vector equation for the wanted plane, which we denote by \Pi, is

\Pi : (-1,1,2) + r(3,2,4) + s(-3,12,3), r,s \in \mathbb{R} \ .

Thus, if s = 0, then L \subset \Pi and since \vec{N} is parallel to \Pi, then it is perpendicular to \Pi_1.

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3 years ago
Pilates is a popular set of exercises for the treatment of individuals with lower back pain. The method has six basic principles
Bas_tet [7]

Answer:

a) We reject H₀     We find difference between the mean of the two groups

b) we accept H₀ : There is no statistical difference  ( in the scale of pain) when the difference in samples means is 1

Step-by-step explanation:

Group 1: Only with educational material

Sample size     n₁  =  42

Sample mean   x₁  =  3,2

Sample standard deviation  s₁  =  2,3

Group 2: With exercises with the treatment of individuals

Sample size    n₂   =  42

Sample mean    x₂  = 5,2  

Sample standard deviation   s₂  = 2,3

Test Hypothesis

Null Hypothesis                     H₀           x₂  -  x₁  = 0  or    x₁  =  x₂

Alternative Hypothesis         Hₐ           x₂  -  x ₁ > 0  or    x₂  >  x₁

Significance level  α  = 0,01

Sample sizes are n₁  =  n₂  >  30

We use z-test

for α  = 0,01     z(c)  ≈  2,32

To calculate z(s)

z(s)  =  (  x₂  -  x₁ ) / √ (2,3)²/42  + (2,3)²/42

z(s)  = ( 5,2  -  3,2 )/ √0,2519

z(s)  =  2 / 0,5

z(s) = 4

Comparing  z(s)  and z(c)

z(s) > z(c)

z(s) is in the rejection region. We reject H₀ the true average pain level for the control condition exceeds that for the treatment condition.

b) Does the true average pain level for the control condition exceed that for the treatment condition by more than 1.

In this case

z( s) =  x₂ - x₁ / 0,5

z(s) =  1 /0,5

z(s) = 2

Comparing z(s) and z(c)  now

z(s) < z(c)        2 < 2,32

And z(s) is in the acceptance region ( for the same significance level) and we should accept H₀  equivalent to say that the two groups statistical have the same mean

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