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skelet666 [1.2K]
3 years ago
11

What is 100 divided by 10?

Mathematics
2 answers:
hammer [34]3 years ago
6 0
Why are you asking these questions????????

100/10 is 10. Don't believe me? Multiply 10 by 10

Or a simpler way. When multiplying by 10, add a 0 next to the number being multiplied by 10. When dividing, delete one 0. Or add a decimal when no 0s are available.
siniylev [52]3 years ago
3 0
100 divided by 10 is 10.
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PLS HELP!! WILL GIVE BRAINLIEST Chris claims that Path A is 2 miles longer than Path B. What calculation error
maria [59]

Answer:

Step-by-step explanation:

He forgot the scale. It is 2 inches longer on the graph, but each inch represents 1/4 mile. So it is 2 quarter miles, or 1/2 mile

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2 years ago
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Two Step Equations<br><br> 9=5- 9/3
Alex787 [66]
9=5 -y/3
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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
Please help me to solve this question. ​
tino4ka555 [31]

Answer:

A=266CM

Step-by-step explanation:

p=2l + 2w                                   l=2x+1             w=x+5

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66=4x+2 + 2x+10                      l=18+1                 w=14

66=6x+12                                   l=19

66-12=6x

54=6x                                      A=L x W

x=9                                         A=19 x 14

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3 0
3 years ago
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Harlamova29_29 [7]

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