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german
3 years ago
9

May someone help me with this question please anyone please

Mathematics
1 answer:
djverab [1.8K]3 years ago
4 0
5;\ 5;\ 5;\ 5;\ 7;\ 9\\\\\overline{x}=\dfrac{5+5+5+5+7+9}{6}=\dfrac{36}{6}=6\\\\\delta^2=\dfrac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+(x_3-\overline{x})^2+...+(x_n-\overline{x})^2}{n}\\\\\delta^2=\dfrac{(5-6)^2+(5-6)^2+(5-6)^2+(5-6)^2+}{6}
\dfrac{+(7-6)^2+(9-6)^2}{6}\\\\\delta^2=\dfrac{(-1)^2+(-1)^2+(-1)^2+(-1)^2+1^2+3^2}{6}=\dfrac{14}{6}=\dfrac{7}{3}

standard\ deviation:\\\\\sqrt{\delta^2}=\sqrt{\dfrac{7}{3}}=\dfrac{\sqrt7}{\sqrt3}=\dfrac{\sqrt7\cdot\sqrt3}{\sqrt3\cdot\sqrt3}=\boxed{\dfrac{\sqrt{21}}{3}}\approx\boxed{1.53}
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Complete Question

The complete question is shown on the first uploaded image

Answer:

First Question

Option A is correct

Second  Question

Option C is correct

Third   Question

     D =  A^{-1}  *  B^{-1} *  C^{-1}

Fourth   Question

  So substituting for D in  (ABC) D =  I

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This proof that  ABC is invertible

Step-by-step explanation:

From the question we are told that

   A , B and  C are invertible which means that A^{-1} , B^{-1}, C^{-1} exist

Now

 From the question

          (ABC) D =  I

Where I is an identity matrix

   Now when we multiply both sides by  A^{-1}  we have

          A^{-1}  A BCD =  A^{-1} * I

          IBCD =  A^{-1}

Now when we multiply both sides by  B^{-1}  we have  

         B^{-1 } *I BCD =  A^{-1}  *  B^{-1}

         I CD =  A^{-1}  *  B^{-1}

Now when we multiply both sides by  C^{-1}  we have  

          C^{-1} * I CD =  A^{-1}  *  B^{-1} *  C^{-1}

              I D =  A^{-1}  *  B^{-1} *  C^{-1}

                 D =  A^{-1}  *  B^{-1} *  C^{-1}

So substituting for D in the above equation

                 (ABC) *  A^{-1}  *  B^{-1} *  C^{-1} =  I

                 I =  I

This proof that  ABC is invertible

 

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