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3 years ago

The tables show some input and output values:

2 answers:

5 0

The answer is 2.Only B because, functions can't have input values of the same number. Table B has <span>7, 3 & 7,2 which both have 7 as their input value. </span>

quester [9]3 years ago

5 0

Answer:

I think @Toko2chanoqoy56 was trying to say functions can't have the same output, otherwise what he is saying makes absolutley no sense.

Step-by-step explanation:

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If DM=35, what is the value of r? Where DG is r+5 and GM is 3r-14

AlekseyPX

R=17.5
DG=22.5
GM=38.5

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2 years ago

The regression line for the given data is = 5.044x + 56.11. determine the residual of a data point for which x = 2 and y = 66.

iren [92.7K]

Do you have answer choices ?

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3 years ago

Find the first, fourth, and eighth terms of the sequence. a(n)=-2 x 2^n-1

leonid [27]

Answer:

2, 16 and 256.

Step-by-step explanation:

Just substitute for n:-

first term ( when n = 1) = 2 * 2^(1-1)

= 2 * 1 = 2

4th term = 2 * 2^(4-1) = 16

8th term = 2* 2^(8-1) = 256

8 0

3 years ago

Solve for x

Anon25 [30]

Answer:

3. x = y - 3

Step-by-step explanation:

You want to move all numbers on the same side of x to the other side with y. Whenever doing this, we have to reverse the operation. In this problem, we have + 3 with x. So we will need to subtract 3 from both sides.

x + 3 = y

x = y - 3

6 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