1x + 2 - 3 = 8 what is x?
2 answers:
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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;
Assume the first even integer is x.
The next even integer will be 2 digits after this digit. So the next even integer will be x+2.
The reciprocals of these two even integers can be written as
and 
respectively.
The sum of their reciprocals is 7/24. So we can set up the equation as:
Taking LCM on left hand side:
Cross Multiplying the denominators:
Using the quadratic formula to solve the above equation:
Since, the value of x can only be an integer, we discard the fractional value and keep x=6
So the first even integer is 6 and the next even integer is 8. The sum of reciprocals of 6 and 8 is 7/24
The next even integer will be 2 digits after this digit. So the next even integer will be x+2.
The reciprocals of these two even integers can be written as
The sum of their reciprocals is 7/24. So we can set up the equation as:
Taking LCM on left hand side:
Cross Multiplying the denominators:
Using the quadratic formula to solve the above equation:
Since, the value of x can only be an integer, we discard the fractional value and keep x=6
So the first even integer is 6 and the next even integer is 8. The sum of reciprocals of 6 and 8 is 7/24