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Setler79 [48]
3 years ago
13

How many times do you need to divide by ten to get from 8950.1 to 0.89501

Mathematics
1 answer:
Paul [167]3 years ago
8 0

8950.1 / 10000 = 0.89501...

1000 x 10 = 10000, so you’ll need to divide 8950.1 1000 times!

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Consider △RST and △RYX. If the triangles are similar, which must be true? A. RY/YS=RX/XT=XY/TS B.RY/RS=RX/RT=XY/TS C.RY/RS=RX/RT
trasher [3.6K]

Answer:

B.RY/RS=RX/RT=XY/TS

Step-by-step explanation:

If Triangles RST and RYX are similar, then the ratio of their corresponding sides is:

\dfrac{RS}{RY}= \dfrac{ST}{YX}=\dfrac{RT}{RX}\\$This could also be written by switching the numerator and denominators as:\\\dfrac{RY}{RS}= \dfrac{XY}{TS}=\dfrac{RX}{RT}

Out of the given options, Option B is true.

3 0
4 years ago
How many numbers are 10 units from 0 on the number line?​
Travka [436]

Answer:

2 Numbers; -10, and 10.

Step-by-step explanation:

5 0
3 years ago
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WILL MARK BRAINLIEST CAR!!!! HELP
Digiron [165]

Answer:

111°

Step-by-step explanation:

Let the centre of the circle be C

mRQ=157 (marked)

The angle at the centre of a circle standing on an arc is twice any angle at the circumference, standing on the same arc. So <SCR=2(SQR)=2(46)=92. mSR=<SCR=92

All the arc measure add up to 360 so:

mSQ+mRQ+mSR=360

mSQ+157+92=360

mSQ=360-249=111

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3 years ago
I need a answer asap
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Answer:

As the weight increases, the price increases.

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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

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0
3 years ago
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