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PIT_PIT [208]
3 years ago
13

Verify that the points are the vertices of a parallelogram, and find its area. A(1, 1, 3), B(2, −5, 6), C(4, −2, −1), D(3, 4, −4

)
Mathematics
1 answer:
almond37 [142]3 years ago
6 0

Answer:

Area = 71.3 sq unit

Step-by-step explanation:

a) We will use properties of parallelogram to verify the given vertices.

Property: Have a pair of parallel opposite sides

Hence, vectors AB AD BD BC CD AC. A pair should be parallel:

vectors

AB = A - B = (1, 1 , 3) - ( 2 , -5 , 6 ) = (-1 , 6 , -3)

AD = A - D = (1, 1 , 3) - (3, 4 , -4) = (-2 , -3 , 7)

BC = B - C = (2 , -5 , 6) - (4, -2 , -1) = (-2 ,-3 , 7)

CD = C - D = (4, -2 , -1) - (3, 4 , -4) = (1 , -6 , 3)

Hence we can see  that AB //CD and AD // BC as their unit vector co-efficents are identical/scalar multiple of each other.

b)

Equation of line AB = (1,1,3) + t*(-1 , 6 , -3 )

Point E = (1-t , 1+6t , 3 -3t) ... denotes an position of point E on line AB.

Choose a point on line CD, lets suppose C = ( 4 , -2 , -1 )

Vector EC = ( 1-t , 1+6t , 3-3t ) - (4 , -2 , -1 ) = (-3 -t , 3 +6t , 4 -3t)

For EC to be perpendicular to AB then dot product of AB . EC = 0

Hence,

EC . AB = -1 * (-3-t) -2*(3+6t) -3*(4-3t) = 0

3 + t -6 -12t -12+9t = 0

-2t-3 = 0

t = -3/2

Vector EC = (-3 -t , 3 +6t , 4 -3t) = (-1.5 , -6 , 8.5)

Area = magnitude (AB) * magnitude (EC)

Area = sqrt ((-1)^2 + 6^2 + (-3)^2) * sqrt ((-1.5)^2 + 6^2 + (8.5)^2)

Area = sqrt (46) * sqrt (442) / 2

Area = 71.3 unit^2

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The perimeter of a rectangle garden is 54yd. One side has a length of 15 yd. Find the area of the garden.
Lelechka [254]

Answer:

=》Area = 180 sq.yd

Step-by-step explanation:

perimeter = 2l + 2b

where, we have l = 15

  • so (2×15) + 2b = 54
  • 30 + 2b = 54
  • 2b = 24
  • b = 12

now, Area = l × b = 12 × 15 = 180 sq.yd

5 0
2 years ago
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The Municipal Transit Authority wants to know if, on weekdays, more passengers ride the northbound blue line train towards the c
Kitty [74]

Answer:

The 90% confidence interval  -49.8

The null hypothesis is  H_o : \mu_1 = \mu_2

The alternative hypothesis H_a  :  \mu_1 <  \mu_2

The distribution test statistics is t = -3.222

The rejection region is  p-value < \alpha

The decision rule is reject the null hypothesis

The conclusion is

      There is sufficient evidence to conclude that there are more passengers riding the 8:30  train

The p-value  is  p-value  =0.000951

Step-by-step explanation:

From the question we are told that

    The first sample size n_1 = 30

    The first sample mean is  \= x _1 = 323

    The first standard deviation is s_1 = 41

    The second sample size is n_2 = 45

    The second sample mean is  \=x_2 = 356

    The second standard deviation is s_2 = 45

given that the confidence level is 90% then the level of significance is mathematically represented as

          \alpha  = (100 -90)\%

         \alpha  = 0.10

Generally the critical value of \frac{\alpha }{2} obtained from the normal distribution table is  

   Z_{\frac{\alpha }{2} } = 1.645

Generally the pooled variance is mathematically represented as

        s^2 = \frac{(n_1 - 1)s_1^2  + (n_2 -1)s_2^2 }{n_1 + n_2 -2}

      s^2 = \frac{(30 -1)(41^2) + (45-1)45^2}{30+45 -2}

     s^2 = 1888.34

Generally the standard error is mathematically represented as

     SE =  \sqrt{\frac{s^2}{n_1} + \frac{s^2}{n_2}  }

=>  SE =  \sqrt{\frac{1888.34}{30} + \frac{1888.34}{45}  }

=>   SE =  10.24

Generally the margin of error is mathematically evaluated as  

      E =  Z_{\frac{\alpha }{2} } * SE

       E =  1.645* 10.24

       E = 16.85

Generally the 90% confidence interval is mathematically represented as

     \=x_1 -\=x_2 -E < \mu_1 -\mu_2 < \=x_1 -\=x_2 +E

     323 -356 -16.84

     -49.8

The null hypothesis is  H_o : \mu_1 = \mu_2

The alternative hypothesis H_a  :  \mu_1 <  \mu_2

Generally the test statistics is mathematically represented as

     t =  \frac{\= x_1 - \=x_2 }{SE}

=>   t = \frac{323-356}{10.24}

=>   t = -3.222

Generally the degree of freedom is mathematically represented as

     df =  n_1+n_2 -2

      df = 30 + 45 -2

     df = 73

The p-value is obtained from the student t distribution table at degree of freedom of 73 at 0.05 level of significance

    The value is  p-value  =0.000951

Here the level of significance is  \alpha =  5\%  =  0.05

Given that the p-value < \alpha then we  reject the null hypothesis

Then the conclusion is  

  There is sufficient evidence to conclude that there are more passengers riding the 8:30  train

               

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Answer:

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4 0
3 years ago
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