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Alex787 [66]
3 years ago
6

Is 2,12 4,24 6,36 8,48 10,60 proportional

Mathematics
1 answer:
Elena L [17]3 years ago
4 0

Answer:

Yes

Step-by-step explanation:

12 divided by 2 = 6, 24 divided by 4 = 6, 36 divded by 6 = 6, 48 divded by 8 = 6, 60 divided by 10 = 6.  

All answers have 6 so yes it is proportional.

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Indicate in standard form the equation of the line passing through the given points. X(0, 6), Y(5, 6)
Olegator [25]
(0,6)(5,6)
notice how the y values are they same.....this means u have a horizontal line with a 0 slope.

y = 0x + 6...slope intercept form
0x + y = 6 <==== standard form
8 0
3 years ago
The equations of three lines are given below.
Yuliya22 [10]

Step-by-step explanation:

line 1 and 2 : parallel

as line 2 is actually

6x + 2y = 8

2y = -6x + 8

y = -3x + 4

so, they have the same slope (factor of x).

line 1 and 3 : neither

the slopes -3 and 3 are not parallel not perpendicular (90°).

line 2 and 3 : neither

as line 2 is parallel to line 1, it has the same relationship to line 3 as line 1.

6 0
1 year ago
Six friends will stay at a cabin in the woods this weekend. The distance to the cabin is 148.5 miles. Each person will drive one
ivolga24 [154]

Answer:

24.75/24¾

Step-by-step explanation:

148.5 ÷ 6 = 24¾ or 24.75

Answer: Each driver drove 24.75/24¾ miles

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
A garden has 8 white roses and 11 yellow roses
Kisachek [45]
8+11=19. So there’s 19 roses in total
4 0
3 years ago
Read 2 more answers
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