Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.
For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 70% of fatalities involve an intoxicated driver, hence
.
- A sample of 15 fatalities is taken, hence
.
The probability is:

Hence







Then:

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.
A similar problem is given at brainly.com/question/24863377
If i did the math correctly one number is 50
19 has two factors: 1, 19
21 has four factors: 1, 3, 7, 21
23 has two factors: 1, 23
And we need a number that has more than four factors and is greater than 25.
Factors of 50- 1, 2, 5, 10, 25, 50
50has six factors and is greater than 25.
Answer:
The answer is 11 and 12
Step-by-step explanation:
you just find perfect squares