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Fiesta28 [93]
3 years ago
10

Solve for x x-15.99 - 5806 A) -42.07 B) - 74.05 42.07 D) 74.05

Mathematics
1 answer:
Anna11 [10]3 years ago
5 0

Answer:

D) 74.05

Step-by-step explanation:

x = 15.99 + 58.06

x = 74.05

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Solve the given equation <br> 7ex – 9 = 10
maria [59]

Answer: x = 10.18

Step-by-step explanation:

7ex – 9 = 10

(x-9)*ln7 = ln 10

x-9 = 2.30/1.95

x = 10.18

3 0
3 years ago
Evaluate the expression 5h-1/4+ 4.5 (h=3)
Aliun [14]

Answer: I THINK IT IS:

19.25

Step-by-step explanation:

IT IS:

5h-1/4+4.5=5h=17/4

h=3=h=3

6 0
2 years ago
Simplify -7(-4p+1) + 2p
mariarad [96]

Step-by-step explanation:

-7(-4p+1) + 2p

-7 x -4 = 28p

-7 x 1 = -7

28p + 2p = 30p

so

30p-7

8 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
What is the estimate of 11,942 ​
ki77a [65]

Answer:

10000 (1sf)

12000(2sf)

11900(3sf)

11940(4sf)

Step-by-step explanation:

I would go for 12000 (2sf) if it the degree of accuracy isn't specified in the question

6 0
3 years ago
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