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Rudik [331]
3 years ago
13

What is the sign of the product (–5)(–3)(–8)(–6)? Positive, because the products (–5)(–3) and (–8)(–6) are negative, and the pro

duct of two negative numbers is positive Positive because the products (–5)(–3) and (–8)(–6) are positive, and the product of two positive numbers is positive Negative because the products (–5)(–3) and (–8)(–6) are negative, and the product of two negative numbers is negative Negative, because the products (–5)(--3) and (–8)(–6) are positive, and the product of two positive numbers is negative
Mathematics
2 answers:
chubhunter [2.5K]3 years ago
6 0

Answer:

The product (–5)(–3)(–8)(–6) should be Positive.

Step-by-step explanation:

The product of 2 negative numbers equal a positive product.

N * N = P

-5*-3=15

-8*-6=48

The product of 2 positive numbers is positive.

P * P = P

15 * 48 = 720

Your answer should be: Positive because the products (–5)(–3) and (–8)(–6) are positive, and the product of two positive numbers is positive.

Fudgin [204]3 years ago
5 0

Answer:

its A because Positive, because the products (–5)(–3) and (–8)(–6) are negative, and the product of two negative numbers is positive

the numbers are negative not positive which two negative numbers make positive

i know i am late its been a week on this question just wanted to say because he said the answer is B which was wrong

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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

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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

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3 years ago
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Answer:

Step-by-step explanation:

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3 years ago
Building A has 7,500 ft.² of office space for 320 employees. Building B has 9500 ft.² of office space for 317 employees. Which b
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Answer:

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6 0
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