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Gelneren [198K]
3 years ago
5

5x+3y>_ 9 ^ This sign is supposed to be greater than Or equal to. Sorry.

Mathematics
1 answer:
Andre45 [30]3 years ago
5 0
It’s saying the equation is going to be equal to.

HOPE THIS HELPS
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A new car sells for $18,000 but it will depreciate over time. After just one year,
WITCHER [35]

Answer:

a.) 18000 x (0.88)^x

x=number of years

b.) value after 10 yrs is represented by

18000 x (0.88)^10 = about $5013

Step-by-step explanation:

8 0
3 years ago
Give the value of the digit 8 in each number.
Anna71 [15]

Answer:

1) Hundreadth

2) Ten

3) Thousandths

4) Hundred

6 0
3 years ago
Read 2 more answers
-x-2y-4z=-6<br> X-5y-2z=-2<br> 2x-3y+2z=2
Svetlanka [38]

Answer:

\sqrt{2} or (1/2)

Step-by-step explanation:

7 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
At the market, 8 apples cost $4, 4. How much do 9 apples cost?
ella [17]

Answer: i need more information 8 apples cost $4, 4???????


Step-by-step explanation:


7 0
3 years ago
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