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Valentin [98]
3 years ago
12

How many unique planes can be determined by four noncoplanar points?

Mathematics
1 answer:
marta [7]3 years ago
3 0

Answer:

Four unique planes

Step-by-step explanation:

Given that the points are non co-planar, triangular planes can be formed by the joining of three points

The points will therefore appear to be at the corners of a triangular pyramid or tetrahedron such that together the four points will form a three dimensional figure bounded by triangular planes

The number of triangular planes that can therefore be formed is given by the combination of four objects taking three at a time as follows;

₄C₃ = 4!/(3!×(4-3)! = 4

Which gives four possible unique planes.

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Answer:

where are you from me

Step-by-step explanation:

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4 0
3 years ago
Y = -x -1 y = -2x + 7 find x and y​
goldenfox [79]

Answer:

Step-by-step explanation:

Y = -x -1..........(1)

y = -2x + 7....(2)

Equating the two equations

-x -1 = -2x + 7

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x = 8

Putting x in (1)

y = -8 - 1

y = -9

3 0
3 years ago
Use the substitution method to solve the linear system.<br> 4×- 7y = 20<br> X- 3y = 10
Allushta [10]
Answer: x = -2, y = -4
(WORK SHOWN BELOW)

7 0
2 years ago
1. Let the test statistics Z have a standard normal distribution when H0 is true. Find the p-value for each of the following sit
Nonamiya [84]

Answer:

1 a  p -value  =   0.030054

1b   p -value  =   0.0029798

1c  p -value  = 0.0039768  

2a  p-value  =   0.00099966

2b  p-value  =  0.00999706

2c  p-value   = 0.0654412

Step-by-step explanation:

Considering question a

  The alternative hypothesis is H1:μ>μ0

   The test statistics is  z =1.88

Generally from the z-table  the  probability of   z =1.88 for a right tailed test is

    p -value  =  P(Z > 1.88) = 0.030054

Considering question b

  The alternative hypothesis is H1:μ<μ0

   The test statistics is  z=−2.75

Generally from the z-table  the  probability of   z=−2.75 for a left tailed test is

    p -value  =  P(Z < -2.75) = 0.0029798

Considering question c

  The alternative hypothesis is H1:μ≠μ0

   The test statistics is  z=2.88

Generally from the z-table  the  probability of  z=2.88 for a right  tailed test is

    p -value  = P(Z >2.88) =  0.0019884    

Generally the p-value for the two-tailed test is

    p -value  = 2 *  P(Z >2.88) =  2 * 0.0019884    

=> p -value  = 0.0039768  

Considering question 2a

    The alternative hypothesis is H1:μ>μ0

     The sample size is  n=16

     The  test statistic is  t =  3.733

Generally the degree of freedom is mathematically represented as

        df =  n - 1

=>     df =  16 - 1

=>     df =  15

Generally from the t distribution table  the probability of   t =  3.733 at a degree of freedom of  df =  15 for a right tailed test is  

       p-value  =  t_{3.733 ,  15} = 0.00099966

Considering question 2b

    The alternative hypothesis is H1:μ<μ0

     The degree of freedom is df=23

     The  test statistic is ,t= −2.500

Generally from the t distribution table  the probability of   t= −2.500 at a degree of freedom of  df=23 for a left  tailed test is  

       p-value  =  t_{-2.500 ,  23} = 0.00999706

Considering question 2c

    The alternative hypothesis is H1:μ≠μ0

     The sample size is  n= 7

     The  test statistic is ,t= −2.2500

Generally the degree of freedom is mathematically represented as

        df =  n - 1

=>     df =  7 - 1

=>     df =  6

Generally from the t distribution table  the probability of   t= −2.2500 at a degree of freedom of  df =  6 for a left   tailed test is  

       t_{-2.2500 , 6} = 0.03272060

Generally the p-value  for t= −2.2500 for a two tailed test is

     p-value  =  2 *  0.03272060 = 0.0654412

4 0
3 years ago
The high temperature in Northville one day was 14°F. The low temperature on the same day was –9°F. What was the difference betwe
strojnjashka [21]

Answer:

-15

Step-by-step explanation: you have to think about it like a chat which means you have to count and with that being said that’s how I got -15


6 0
3 years ago
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