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rjkz [21]
3 years ago
5

Use​ l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove

r 7 x cubed EndFraction ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Over 7 x cubed EndFraction equals nothing ​(Type an exact​ answer.)
Mathematics
1 answer:
steposvetlana [31]3 years ago
8 0

Answer:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}

Step-by-step explanation:

The limit is:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}

by replacing, and applying repeatedly you obtain:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}

hence, the limit of the function is -1/14

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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) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

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) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0
3 years ago
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