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vesna_86 [32]
3 years ago
8

RED YOU ARE VERY SUS

Mathematics
2 answers:
Allisa [31]3 years ago
7 0

Answer:

YES YES HE VENTED ITS NOT ME DONT WORRY

Marrrta [24]3 years ago
6 0

Answer:

red do be looking sus tho

Step-by-step explanation:

send him outta the airlock

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Open the picture and answer it pls
Nastasia [14]

Answer:

3.5

Step-by-step explanation:

11.5 - 2.5 = 9

9 - 5.5 = 3.5

and scalene triange means none of the 3 sides are equal...

3.5 does not equal 5.5 or 2.5

3 0
3 years ago
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How do I solve this?​
crimeas [40]

Answer:

The final answer is (2,-1)

(also idek what I did in the photo but I'm positive the answer is (2,-1)

7 0
3 years ago
When constructing an inscribed equilateral triangle, how many arcs will be drawn on the circle?
Lady bird [3.3K]
The answer is A(3). 3 arcs will be drawn
6 0
3 years ago
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Every time Luisa bakes a batch of brownies she uses 1/2 cup of chocolate if she has 3/2 cups of chocolate remaining how many bat
kakasveta [241]

Answer:

she can make 4 batches of brownies

Step-by-step explanation:

since she already made a batch of brownie with 1/2 cup and she has 3/2 cups left, that means she can make 3 more

3+1=4

hope this helps

3 0
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
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