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joja [24]
3 years ago
9

Al sumar 2x+5y-4z+9 y 2x-4y+6 se obtiene?

Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
8 0

Answer: (Responder)

4x + y + 15 – 4z

Step-by-step explanation: (Explicación paso a paso)

(2x + 5y – 4z) + (9 + 2x – 4y + 6)

Group Like Terms (Agrupar términos similares)

2x + 2x + 5y – 4y – 4z + 9 + 6

Add Similar Elements (Agregar elementos similares)

4x + 5y – 4y – 4z + 9 + 6

4x + y – 4z + 9 + 6

Add The Numbers (Suma los números)

4x + y + 15 – 4z

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sa (surface area) = a^{2} + 2a \sqrt{x} \frac{a^2}{4} + h^2\\

Step-by-step explanation:

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Find the slope of a line parallel to each given line Y equals -1÷3 X +3
g100num [7]

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-1/3x+3

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8 0
3 years ago
Read 2 more answers
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
9.99x10 ^-2 + 1.11 × 10^-2 i dont have the same scientific calculator as the one from school. Would like to verify my math is co
Rufina [12.5K]
The answer should not depend on which machine or which pencil you use to
find it.  If you work a problem two different ways and get two different answers,
then at least one of them is wrong, and there's a pretty good chance that both
of them are.

(9.99 of anything) + (1.11 of the same thing) = 11.1 of them

9.99 (x 10^-2) + 1.11 (x 10^-2) = <em>11.1 (x 10^-2)</em> .

Can we do any more with that ?

10^-2  =  1 / 10^2  =  1 / 100 .

11.1 x 10^-2  =  11.1 / 100  =  <em>0.111</em>
3 0
3 years ago
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