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navik [9.2K]
3 years ago
11

Find the value of x. B 6 х A 2. 3 x = [?]

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

B^6 × A^2.3 = 0

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

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Emma's final grade in her math class is the average (mean) of her 5 chapter tests she took throughout the year. If she received
NikAS [45]

Answer:

Step-by-step explanation:

The mean (or average) of a set of numbers can be found with the following formula:

Mean = \frac{A_{1} + A_{2} + ... + A_{N}}{N}

For this problem, we have five numbers, so we add them together and divide by 5:

Mean = \frac{92 + 85 + 78 + 97 + 88}{5}

Mean = \frac{440}{5}

Mean = 88

8 0
3 years ago
Read 2 more answers
Compare 8.25 and 8.250
GenaCL600 [577]
The answer is = because of you take away the 0 in the decimal 8.250 it's the same decimal as 8.25 and in decimals the zeros farthest to the right don't matter.
8.25 = 8.250
3 0
3 years ago
What is the probability of having 3 children that are all boys?
olganol [36]
So you have a 50% chance to have a boy vs a girl (1/2). If you want 3 children, that would be 1/3*1/2= 1/6 or 16.67% chance
5 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
The expression 3(x-9) is equivalent to<br> 3(x)+9.<br> 3(x)+3(9).<br> 3(x) – 9.<br> 3(x) – 3(9).
sukhopar [10]

Answer:

3(x) – 3(9)

Step-by-step explanation:

3(x-9)

Distribute the 3 to each term in the parentheses

3*x -3*9

3x -27

3 0
2 years ago
Read 2 more answers
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