Use the Midpoint Rule with n = 10 to approximate the length of c(t) = (5 + sin(4t), 6 + sin(7t)) for 0 ≤ t ≤ 2π. (Round your ans
wer to two decimal places.)
1 answer:
Answer:
34.43
Step-by-step explanation:
A differential of length in terms of t will be ...
dL(t) = √(x'(t)^2 +y'(t)^2)
where ...
x'(t) = 4cos(4t)
y'(t) = 7cos(7t)
The length of c(t) will be the integral of this differential on the interval [0, 2π].
Dividing that interval into 10 equal pieces means each one has a width of (2π)/10 = π/5. The midpoint of pieces numbered 1 to 10 will be ...
(π/5)(n -1/2), so the area of the piece will be ...
sub-interval area ≈ (π/5)·dL((π/5)(n -1/2))
It is convenient to let a spreadsheet or graphing calculator do the function evaluation and summing of areas.
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The attachment shows the curve c(t) whose length we are estimating (red), and the differential length function (blue) we are integrating. We use the function p(n) to compute the midpoint of the sub-interval. The sum of sub-interval areas is shown as 34.43.
The length of the curve is estimated to be 34.43.
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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%.