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lesantik [10]
3 years ago
11

Please help asap . i will give brainliest -elena:)

Mathematics
1 answer:
sweet [91]3 years ago
3 0

Answer:

84

Step-by-step explanation:

14x6

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Musya8 [376]

Answer:

ok

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
What are linear equations? Explain In full depth please​
jasenka [17]

Answer:

, a linear equation is an equation that may be put in the form where are the variables, and are the coefficients, which are often real numbers. The coefficients may be considered as parameters of the equation, and may be arbitrary expressions, provided they do not contain any of the variables.

Step-by-step explanation:

y= 2x+1 is a linear equation

4 0
3 years ago
when 200 apples are sold at Rs 1.50 each there will be rupees hundred loss how many apples should be sold for rupees 150 to gain
Lera25 [3.4K]

Answer:

Step-by-step explanation:

Sales price of 200 apples = 300

At this price, there was loss of Rs. 100

Means the cost of the apples was

300 + 100 = Rs. 400

To earn Rs. 100, they should have been sold for Rs. 500

4 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
Explain why 0.04/3.6 has the same answer as 4/360
lisov135 [29]
Because when you divide you get the same answer or something to do with percents
7 0
3 years ago
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