MATHEMATICS COLLEGE

In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experience that 1%, 3%, and 2% of the products made by each machine, respectively, are defective. A finished product is randomly selected and found to be non-defective, what is the probability that it was made by machine B1?

Answers

Answer 1

Answer:

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) = $(P(B|A)P(A))/(P(B)) = (P(B|A)P(A))/(P(B|A)P(A) + P(B|a)P(a))$

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) = $(P(N|B1)P(B1))/(P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3))$ = $((0.297)(0.3))/((0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5))$ = 0.1138

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

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Answers

i took personal finance, sorry if i,m wrong but i think its C

From 1985 to 2007, the number B B of federally insured banks could be approximated by B ( t ) = − 329.4 t + 13747 B(t)=-329.4t+13747 where t is the year and t=0 corresponds to 1985. How many federally insured banks were there in 1990?

Answers

The number of federally insured banks in the year 1990 was 12,100.

Given data:

To find the number of federally insured banks in the year 1990, you need to substitute the value of t = 1990 into the given function B(t):

B(t) = -329.4t + 13747

Since t corresponds to the year and t = 0 corresponds to 1985, we need to calculate for t = 1990 - 1985.

t = 5.

Substitute t = 5 into the function:

B(5) = -329.4 * 5 + 13747

Now, perform the calculations:

B(5) = -1647 + 13747

B(5) = 12100

Hence, there were approximately 12,100 federally insured banks in the year 1990.

To learn more about linear equations, refer:

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Final answer:

Using the given equation and by substituting t with 5 (for the year 1990, five years after 1985) in that equation, we find that there were 13363 federally insured banks in 1990.

Explanation:

The question is asking how many federally insured banks there were in 1990, given that the number of banks in any given year from 1985 to 2007 is given by the equation

B(t) = -329.4t + 13747

. In this equation,

is the year with t=0 corresponding to 1985. To find the answer, we should substitute in the appropriate value of t, which would be 5 because 1990 is five years after 1985. Thus, the equation becomes B(5) = -329.4*5 + 13747. This simplifies to 15010 - 1647, which equals

13363

. Therefore, there were 13363 federally insured banks in 1990.

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What is 04.151515... into a fraction

Answers

Answer:

$\displaystyle 4.\overline{15} = (137)/(33)$ .

Step-by-step explanation:

Start by separating this decimal number into its integer part and its fraction part:

$4.151515\cdots = 4 + 0.151515\cdots$

The most challenging task here is to express $0.151515\cdots$ as a proper fraction. Once that fraction is found, expressing the original number $4.151515\cdots$ will be as simple as rewriting a mixed number as an improper fraction.

Let $x = 0.151515\cdots$ . $(x + 4)$ would then represent the original number.

Note that the repeating digits appear in groups of two. Therefore, if the digits in $x$ are shifted to the left by two places, the repeating part will continue to match:

$\begin{aligned}x = 0.&151515\cdots && \n 100\, x = 15.& 151515\cdots \end{aligned}$ .

Note, that this "shifting" is as simple as multiplying the initial number by $100$ (same as $10$ raised to the power of the number of digits that needs to be shifted.)