A. 18.60, because you multiply 18.6 by .05 to get .93, and then subtract .93 from 18.6 to get 17.67.
Answer:
Option B. 5P5 × 20P15
Step-by-step explanation:
It is very important to remember that the second grade students are sitting in the front row, therefore, it is only necessary to organize 15 first grade students in 20 seats.
Permutations allow you to calculate the number of ways in which m objects can be arranged in n positions.
The permutation of m in n is written as:
nPm
Where n is the number of elements and m are chosen.
The way in which the 5 second grade students can be organized in the 5 seats is from the first row is:
5P5
Then, the number of ways in which 15 first-year students can be organized into 20 seats is:
20P15
Then, the number of ways to organize all students on the bus is the product of both permutations
5P5 * 20P15
Answer:
- P(≥1 working) = 0.9936
- She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.
Step-by-step explanation:
1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...
... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936
2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.
If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.
This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.
_____
My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)
The answer is B. 5P5 x 20P15.
This is because to make the second grade students to sit in the first row we have 5 seats and 5 students so we will permute the 5 students to those 5 seats. So 5P5.
Now we are left with 20 seats and 15 first grade students so we can simply permute those 15 students into those 20 seats. So 20P15.
Finally using the counting rule principle we will multiply both of these so 5P5 x 20P15.
