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saul85 [17]
3 years ago
13

The graph of a rational function has vertical asymptote at x=-3 and x=3 and a horizontal asymptote At y=1.write a function that

has those attributes.
Mathematics
1 answer:
liubo4ka [24]3 years ago
4 0

Answer:

1/(x^2-9)   + 1.

Step-by-step explanation:

It will be a fractional function with the denominator = 0 when x = 3 or -3.

Therefore the  denominator could be x^2 - 9.

At high and low values of x the function  will have an horizontal  asymptote at y = 0 so to get an asymptote at x = 1 the function could be

1/ x^2-9)   + 1.

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Answer:

113.1 =VOLUME , 4/3 X 3.14 (3) ^3 = 113.1

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3 years ago
ANSWER CORRET AND ILL MARK U BRAINLIEST
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Answer:

two is the last one, three us the first one, and four is the second one, and six is the definition of parallelogram

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
Which expression is equivalent to 3x+10−x+12<br> a.24x<br> b.4x + 22<br> c.26x<br> d.2x + 22
jok3333 [9.3K]
The answer is:  [D]:  2x + 22  .
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8 0
3 years ago
A football coach is trying to decide: When a team is ahead late in the game, which strategy is better?
PSYCHO15rus [73]

Answer:

You are more likely to win by playing regular defense.

Step-by-step explanation:

Assume out of 100 reviewed games, there were 50 regular defense games and 50 prevent defense games. And out of 50 regular defense games, 38 were win, 12 were lose. And out of 50 prevent defense game, 29 were win, 21 were lose.

Probability to win the game by playing regular defense is:

P(win | regular) = 38/50 = 0.76

Probability to win the game by playing prevent defense is:

P(win | prevent) = 29/50 = 0.58

Since the probability of winning by regular defense game is more than prevent defense game (0.76 > 0.58), you are more likely to win by playing regular defense.

5 0
3 years ago
Read 2 more answers
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