in any given year a factory has a 10% probability of having a serious accident. about every how many years might the factory exp
ect to have a accident
1 answer:
Years for one serious accident is 10 year
<u>Given:</u>
Probability of having a serious accident = 10% every year
P(serious accident) = 0.1
Number of years = n
<u>Computation:</u>
Years for one serious accident = 1 / Probability of having a serious accident
Years for one serious accident = 1 / P(serious accident)
Years for one serious accident = 1 / 0.1
Years for one serious accident = 10 year
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1. 21
2. -4
3. 20
4. 1
5. 4
6. -6
Step-by-step explanation:
8x -3 = 2(x-1/2)
⇒ 8x -3= 2x -2*(1/2)
⇒ 8x -3 = 2x -1
⇒ 8x -2x= -1+ 3
⇒ 6x = 2
⇒ x= 2/6
⇒ x= 1/3
The final answer is x= 1/3~
32 is closer to 33 than it is to 40
Answer:
x < -11/6.
Step-by-step explanation:
−12x + 13 > 35
-12x > 35 - 13
-12x > 22
Divide both sides by -12 and invert the inequality sign:
x < -22/12
x < -11/6.
The answer is B through ASA (Angle-Side-Angle) similarity criterion.
