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Aliun [14]
3 years ago
13

The table shows where sixth grade students at Sharonton

Mathematics
1 answer:
e-lub [12.9K]3 years ago
3 0
Did you add the picture of the table? I do not see it :(
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Hayleigh took a long multiple-choice, end-of-year math test. The ratio of the number of problems Hayleigh got incorrect to the n
timurjin [86]

Answer:

36

Step-by-step explanation:

If sh missed 8, then 8/2 is 4, 4x9 is 36

8 0
3 years ago
Which set of ordered pairs represents y as a function of x?
givi [52]
I think the answer is j
7 0
3 years ago
What is the following quotient? StartFraction StartRoot 96 EndRoot Over StartRoot 8 EndRoot
steposvetlana [31]

Answer:

D 12

Step-by-step explanation:

because your just solving 96/8

3 0
3 years ago
Anybody knows the answer?
vladimir1956 [14]

The top and bottom faces are 3 inches by 3 inches.

Area = 3 in. * 3 in. * 2 = 18 in.^2

The front, back, right, and left faces are 3 in. by 6 in.

Area = 3 in. * 6 in. * 4 = 72 in.^2

Total surface area = 18 in.^2 + 72 in.^2 = 90 in.^2

Answer: B. 90 square inches

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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