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jolli1 [7]
3 years ago
10

I HAVE 2 MIN PLS HELP

Mathematics
1 answer:
Maksim231197 [3]3 years ago
5 0

Answer:

C (ignore this it asked for 20 characters)

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I am having trouble with this equation 3 ( x-5 ) =9<br><br>can anyone help pls
Murljashka [212]
First you distributive property:
3x-15=9
Add 15
3x=24
Divide
x=8
4 0
3 years ago
Read 2 more answers
In a widget factory, machines A, B, and C manufacture, respectively, 20, 30, and 50 percent of the total. Of their output 6, 3,
steposvetlana [31]

Answer:

38.71% probability it was manufactured by machine A.

29.03% probability it was manufactured by machine B.

32.26% probability it was manufactured by machine C.

Step-by-step explanation:

We have these following probabilities:

A 20% probability that the chip was fabricated by machine A.

A 30% probability that the chip was fabricated by machine B.

A 50% probability that the chip was fabricated by machine C.

A 6% probability that a chip fabricated by machine A was defective.

A 3% probability that a chip fabricated by machine B was defective.

A 2% probability that a chip fabricated by machine C was defective.

The question can be formulated as:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

What are the probabilities that it was manufactured by machines A, B, and C?

Machine A

What is the probability that the widget was manufactures by machine A, given that it is defective?

P(B) is the probability that it was manufactures by machine A. So P(B) = 0.20

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine A. So P(A/B) = 0.06

P(A) is the probability that a widget is defective. This is the sum of 6% of 20%, 3% of 30% and 2% of 50%. So

P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031

So

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.2*0.06}{0.031} = 0.3871

38.71% probability it was manufactured by machine A.

Machine B

What is the probability that the widget was manufactures by machine B, given that it is defective?

P(B) is the probability that it was manufactures by machine B. So P(B) = 0.30

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine B. So P(A/B) = 0.03

P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031

So

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.03}{0.031} = 0.2903

29.03% probability it was manufactured by machine B.

Machine C

What is the probability that the widget was manufactures by machine C, given that it is defective?

P(B) is the probability that it was manufactures by machine C. So P(B) = 0.50

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine C. So P(A/B) = 0.02

P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031

So

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.02}{0.031} = 0.3226

32.26% probability it was manufactured by machine C.

6 0
3 years ago
Help me please !!!<br><br> math 20-3
Lesechka [4]

Answer:

Number 25:   B or C idk

Number 26:   D  1.50M

Tell me if im right i tried sorry if not

Step-by-step explanation:

7 0
3 years ago
Draw the image of AABC under the translation (x,y) → (x-6,y-7).
arsen [322]

Answer:

Step-by-step explanation:

8 0
2 years ago
PLEASE HELP!!!!!! ILL GIVE BRAINLIEST *EXTRA 40 POINTS** !! DONT SKIP :((
Rom4ik [11]
X= -10
Y=7
Y=1/5X+9
Y=1/5 •(-10)+9
Y= -10/5 + 9
Y= -2+9
7=7

Y= -1/2 • (-10) +2
Y= 10/2 +2
Y=5+2
7=7

Yes
4 0
2 years ago
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