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dlinn [17]
3 years ago
9

There are 18 apples and a number of oranges in a box. The probability of choosing an

Mathematics
1 answer:
-BARSIC- [3]3 years ago
3 0

Answer:

60 apples and oranges

Step-by-step explanation:

18/3/7 = 42

18/42 = 3/7

18+42 = 60

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t+1 = 0 // - 1

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I graphed but do not understand, Please explain. Figure shown.
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6 0
3 years ago
Which expression is equivalent to 2(c+7)-18x+4/5
mestny [16]

Answer:

2c - 18x + 74/5

Step-by-step explanation:

<u>Step 1:  Distribute </u>

2(c + 7) - 18x + 4/5

(2 * c) + (2 * 7) - 18x + 4/5

<em>2c + 14 - 18x + 4/5 </em>

<u>Step 2:  Combine like terms </u>

2c + 14 - 18x + 4/5

<em>2c - 18x + 74/5 </em>

Answer:  2c - 18x + 74/5

3 0
3 years ago
In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
diamong [38]

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) = \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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