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frutty [35]
3 years ago
9

Twenty students ride on your bus

Mathematics
1 answer:
kotegsom [21]3 years ago
3 0

Answer: 8

Step-by-step explanation:

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Two planes intersect at a...<br> A. Plane<br> B. Power<br> C. Point<br> D. Line
Anna71 [15]

Answer:

D. Line

Step-by-step explanation:

7 0
3 years ago
A company is designing a new cylindrical water bottle. The volume of the bottle will be 220 cmcubed. The height of the water bot
FrozenT [24]

Answer:

Radius = 2.90 cm

Step-by-step explanation:

Volume \: of \: Cylinder \:  = \pi {r}^{2} h

220 = 3.14 × {r}^{2} × 8.3

220 = 26.062 × {r}^{2}

220/26.062 = {r}^{2}

{r}^{2} = 8.44 ( approx )

r = √8.44

r = 2.90 cm. (approx)

Hope this helps .....

Please mark as Brainliest!!!!

4 0
4 years ago
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

7 0
3 years ago
A rectangle length is 3 times its width. The area is 27 square inches
Nikolay [14]
The length would be 9 and the width would be 3
3 0
3 years ago
Divide: 407:10<br> Guys I need help I don’t know how to do divide SOMEONE HELP ME QUICKLY!!!
katen-ka-za [31]

Answer:

40

7 Remainder

7 0
3 years ago
Read 2 more answers
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