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Sergeeva-Olga [200]
2 years ago
15

Mai has a dome paperweight that she can use as a magnifier. The paperweight is

Mathematics
1 answer:
Shkiper50 [21]2 years ago
6 0

Answer:

6 x 6 x 3 cm

Step-by-step explanation:

The base of the box needs to fit the base of the hemisphere.

The maximum width and length of the hemisphere is the diameter of the base of the hemisphere (which is circular)

Diameter of a circle = 2r

As the radius = 3 cm,  the diameter = 2 x 3 = 6 cm

Therefore, the width and length of the box should both be 6cm

The height of the box needs to accommodate the height of the hemisphere.  The height of the hemisphere is the radius, so the height of the box should be 3 cm.

Therefore, the dimensions of the box should be 6 x 6 x 3 cm

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The rule would be ~y=4x+3~

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3 years ago
Daniel wrote the number 45,358. How many times greater is the 5 in the thousands place than the 5 in the tens place
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1000 times cause its the value of the number
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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
Write the ratio of 8 apples to 4 oranges in 3 different ways.
uranmaximum [27]
8 : 4
16 : 8
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Simplest:
2 : 1
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2 years ago
Read 2 more answers
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The answer is the last one
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