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olganol [36]
3 years ago
14

If 6x = 54, then x = ___. (Only input whole number) Numerical Answers Expected! First Person GETS BRAINLEIST

Mathematics
2 answers:
Flura [38]3 years ago
7 0

6x=54\\ \\ x=\frac{54}{6} \\ \\ x=9

Hope that helps!

Shtirlitz [24]3 years ago
7 0
X = 9. this is because you just have to do 54 ÷ 6. that leaves you with 9. Hope this helps x
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The school auditorium has 80 rows of seats. There are 50 seats in each row. What is the total number of seats in the chhool audi
Snezhnost [94]
The number of rows = 80

number of seats in 1 row = 50

total number of seats is the product of the number of rows and number of seats per row = 80*50 = 4000
6 0
3 years ago
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Find the image (0,0) after two reflections first across y=3 and then across x-axis
Alina [70]
Refrection of point (0, 0) across y = 3 gives point (0, 6)
Refrection of point (0, -6) across the x-axis gives point (0, -6)
3 0
3 years ago
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Help please help me please
allsm [11]

Answer:

A. 2 (0)=0

B. 2 (2)=4

Step-by-step explanation:

You have to multiply the X numbers by 2 for example 2x , x=2 , 2x2=4

8 0
2 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
30 POINTS FOR ANSWER!!<br> 2 numbers sum 120 and product 400<br> INSTANT ANSWER NEEDED
OLga [1]
I think the problem is
2X+20=400
You just have to answer I think X and X equals that two numbers I’m not 100% sure but I’m trying to help
4 0
3 years ago
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