Determine the values of the variables in isosceles trapezoid CHLE below.
1 answer:
Answer:
The values of variables x and m are 11 and 17
Step-by-step explanation:
The question has missing details as the diagram of the trapezoid isn't attached.
(See attachment).
Given that trapezoid CHLE is isosceles then the angles at the base area equal (4x)
And
The angles at the top are also equal
8m = 11x + 15
At this point, the four angles in the trapezoid are 8m, 11x + 15, 4x and 4x..
The sum of interior= 360
So,
11x + 15 + 8m + 4x + 4x = 360
Collect like terms
11x + 4x + 4x + 8m = 360 - 15
19x + 8m = 345
Substitute 11x + 15 for 8m
19x + 11x + 15 = 345
30x + 15 = 345
30x = 345 - 15
30x = 330
Divide through by 30
30x/30 = 330/30
x = 11
Recall that 8m = 11x + 15;
8m = 11(11) + 15
8m = 121 + 15
8m = 136
Divide through by 8
8m/8 = 136/8
m = 17
Hence, the values of variables x and m are 11 and 17
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Answer:
The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.
Step-by-step explanation:
Using Bayes' Theorem
P(A|B) =
where
P(B|A) is probability of event B given event A
P(B|a) is probability of event B not given event A
and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.
For this problem,
Let P(B1) = Probability of machine B1 = 0.3
P(B2) = Probability of machine B2 = 0.2
P(B3) = Probability of machine B3 = 0.5
Let P(D) = Probability of a defective product
P(N) = Probability of a Non-defective product
P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003
P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006
P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010
Likewise,
P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997
P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994
P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990
For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)
P(B1|N) = =
= 0.1138
Hence the probability that a non-defective product is produced by machine B1 is 11.38%.