In a factory that makes bolts, 20% of the bolts being made by a certain machine are defective. The factory owner wants to determine the probability that two bolts in a row will be defective. He decides to use random numbers as a simulation tool. The following list shows 100 two-digit numbers that were randomly generated to represent two bolts in a row. If he uses the digits 0 and 1 to represent defective bolts, how many times were two bolts in a row defective?

a.0

b.33

c.3

d.18

14.29%

step-by-step explanation:

hi do you need an answer for your question?

step-by-step explanation:

below

step-by-step explanation:

1. 5/1

2. 10/16

3. 90/2

4. 1/20

5. 3/1

6. 14/24

7. 1/4

8. 18/10

p(x)= 8 x³ - 36 x² + 54 x - 27

q(x) = 2 x -3

you can check your mistake and then you can choose the correct option.

dividend =   8 x³ - 36 x² + 54 x - 27

divisor = 2 x -3

if you want to find quotient, proceed as follows

q(x)= 0

2 x - 3=0

x =

now , i have used division method , to find quotient and remainder