What key would you use if 10 students chose cartoons
2 answers:
Answer:
Each Box = 2 students
Step-by-step explanation:
Consider the provided diagram,
We need to define what key would you use if 10 students chose cartoons.
It is given that
Key: Each Box = 3 students
But with the help of this key we can not divide 10 students in group.
So we need to change the key with the number which is multiple of 10.
Hence Key: Each Box = 2 students
For better understanding refer the figure 1.
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Answer:
52.56% probability that eight or more of the flights will arrive on time.
Step-by-step explanation:
For each flight, there are only two possible outcomes. Either it is on time, or it is not. The probability of a flight being on time is independent from other flights. So we use the binomial probability distribution to solve this question.]
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
At a certain airport, 75% of the flights arrive on time.
This means that
A sample of 10 flights is studied.
This means that
Find the probability that eight or more of the flights will arrive on time.
In which
52.56% probability that eight or more of the flights will arrive on time.
Answer:
1. The Venn diagrams are attached
2. When the statistics students number = 10·x + 3, we have;
The number of students that study
a. Algebra = 128/11
b. Statistic = 213/11
When the statistics students number = 2·x + 3, we have;
The number of students that study
a. Algebra = 16
b. Statistic = 15
Step-by-step explanation:
The parameters given are;
Total number of students = 30
Number of students that study algebra n(A) = x + 10
Number of students that study statistics n(B) = 10·x + 3
Number of student that study both algebra and statistics n(A∩B) = 4
Number of student that study only algebra n(A\B) = 2·x
Number of students that study neither algebra or statistics n(A∪B)' = 3
Therefore;
The number of students that study either algebra or statistics = n(A∪B)
From set theory we have;
n(A∪B) = n(A) + n(B) - n(A∩B)
n(A∪B) = 30 - 3 = 27
Therefore, we have;
n(A∪B) = x + 10 + 10·x + 3 - 4 = 27
11·x+13 = 27 + 4 = 31
11·x = 18
x = 18/11
The number of students that study
a. Algebra
n(A) = 18/11 + 10 = 128/11
b. Statistic
n(B) = 213/11
Hence, we have;
n(A - B) = n(A) - n(A∩B) = 128/11 - 4 = 84/11
Similarly, we have;
n(B - A) = n(B) - n(A∩B) = 213/11 - 4 = 169/11
However, assuming n(B) = (2·x + 3), we have;
n(A∪B) = n(A) + n(B) - n(A∩B)
n(A∪B) = 30 - 3 = 27
Therefore, we have;
n(A∪B) = x + 10 + 2·x + 3 - 4 = 27
2·x+3 + x + 10= 27 + 4 = 31
3·x = 18
x = 6
Therefore, the number of students that study
a. Algebra
n(A) = 16
b. Statistics
n(B) = 15
Hence, we have;
n(A - B) = n(A) - n(A∩B) = 16 - 4 = 12
Similarly, we have;
n(B - A) = n(B) - n(A∩B) = 15 - 4 = 11
The Venn diagrams can be presented as follows;
