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Mariulka [41]
3 years ago
7

(-8) x 10 x 9 is equal to

Mathematics
2 answers:
Sergio [31]3 years ago
8 0

Answer:

-720

Step-by-step explanation:

labwork [276]3 years ago
6 0

Answer:

The answer is _ 720

Step-by-step explanation:

Thank you For your efforts

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A company produces a certain product, and each unit of this product may have 3 different types of defects. Let Di, D2,Ds represe
Stella [2.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) 0.88

b) 0.02

c) 0.01

d) 0.99

Step-by-step explanation:

Step one: State the given parameters

            P(D_{1} ) = 0.12                                   P(D_{2} ) = 0.07

           P(D_{3} ) = 0.05                                    P (D_{1} U D_{2} ) = 0.13

          P(D_{1}n D_{2}n D_{3}) = 0.01                        P(D_{1} U D_{3}) = 0.14

Step 2 : Obtain the probability that a unit does not have a type 1 defect

         P(\frac{}{D_{1} }) =  1 -P(D_{1} )

                    = 1 - 0.12

                    = 0.88  

Step 3 : Obtain the probability that a unit has both type 2 and 3 defect?

          The probability of the unit having both type 2 and type 3 defect is denoted as P(D_{2} n D_{3} )

   This is calculated as

                    P(D_{2}n D_{3}) =P(D_{2} ) + P(D_{3}) - P(D_{2} U D_{3})\\\\                          = 0.07 + 0,05 - 0.13

                    =   0.02

Therefore P(D_{2} n D_{3} ) = 0.02

Step 4 : Obtain the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect

                  Let P(\frac{}{D_{1}} n D_{2} n D_{3} ) denote the  probability that the unit has both a type 2 and type 3 ,but not a type 1 defect.

This can be calculated as follows :

                      P(\frac{}{D_{1}} n D_{2} n D_{3} ) = P(D_{2} n D_{3}) - P(D_{1} n D_{2}nD_{3})

                                               =   0.02 - 0.01

                                               =  0.01

Step 4 : Obtain the probability that a unit has at most two defects

               P(at most 2 defects)  = 1 - P(all three defects)

                                                  = 1- P(D_{1} n D_{2}nD_{3})

                                                  =  1 - 0.01

                                                  = 0.99

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Answer:

The line graph

Step-by-step explanation:

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