Answer:
0.6421
Step-by-step explanation:
In this case we have 3 trials and we have 2 options for each one. The driver has or hasn't been under alcohol influence. The probability that the driver has is 0.29 and the probabiility that the driver hasn't is 1 - 0.29 = 0.71
each trial is independent because we are assuming that the population of drivers in between 21 and 25 years old is very big.
The probability that one of them was under alcohol influence can be found by finding the probability that non of them was under alcohol influence because:
1 = p(x = 0) + p(x ≥ 1)
p(x ≥ 1) = 1 - p(0)
The probability that none of them was under alcohol influence is going to be:
0.71×0.71×0.71 = 0.3579
The probability of finding at least one driver that has been under alcohol influence is:
0.6421
Answer:
Step-by-step explanation:
1. .....................................................................................................
A. This is a functional relationship as the number of workers depend on the number of job sites.
B. This is a functional relationship as the amount of money is dependent on the number of withdrawals
C. This is a functional relationship as the number of vitamins depend on the number of monkeys
D. This is NOT a functional relationship as the output is a fixed value. The performance and scoring may repeat.
2. .....................................................................................................
<h3>Given</h3>
<h3>To find </h3>
<h3>Solution</h3>
- A/B =
- (62x - 100)/(2x - 3) =
- (31*2x - 31*3 - 7)/(2x - 3) =
- (31(2x - 3) -7)/(2x - 3) =
- 31 - 7/(2x - 3)
Correct option is C.
I believe the answer would be 4.2.
Add all the points together and divide by 10
-10.2 - 12.85 which would be -23.05
the 10.2 is negative because it is below zero
Step-by-step explanation:
1. C = the graphing is slowly increasing, then he walks across the top so it is flat
2. D = it started out slow and gradually got higher, which is what the story said
3. H = he had to stop walking, and there is a part of no movement in the graph H
4.