Complete the statements to find 8 divided by 0.01

Recall that rational numbers are numbers that can be represented as the ratio, or fraction, of two integers. When expressed as d

Alenkasestr [34]

Answer:

1) $\frac{5}{-14}$ =0.3571428571 2) Terminating

Step-by-step explanation:

1st. To answer this question let's divide with long division algorithm 5 and -14.

-14)50

42

---

80

-70

--

100

98

---

20

14

----

60

-56

---

40

28

--

120

-112

----

80

70

--

100

-98

--

2

5 (dividend) : -14 (divisor) $\frac{5}{-14}$

Long division (check below)

With the same two numbers -14 and 5 we can write in a Long Division

5) -14

5) -14

10

40

So $\frac{-14}{5}=-2.8$

2nd. Both are terminating ones for they have finite quantities of numbers. -2.8 and 0.3571428571, as we can see that these are rational numbers for they can be written as a/b, and b≠ 0.

$\frac{5}{-14}$ form.

$\frac{-14}{5}$ form.

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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