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agasfer [191]
2 years ago
15

Solve the system of equations. A. (3,2) B. (-2,0) C. (0, -2) D. (-4,2)

Mathematics
2 answers:
SVETLANKA909090 [29]2 years ago
5 0

Answer:

d. (-4,2)

Step-by-step explanation:

GREYUIT [131]2 years ago
5 0
  • Y = 1/2X + 4 ( Equation – 1)

  • Y = –5/4X – 3 ( Equation – 2)

Here, Y = 1/2X + 4

Or, –5/4X – 3 = 1/2X + 4 (From equation 1 and 2)

Or, –5/4X = 1/2X + 4 + 3

Or, –5/4X = 1/2X + 7

Or, 1/2X + 7 = –5/4X

Or, 7 = –5/4 X – 1/2 X

Or, –5/4X – 2/4 X = 7

Or, –7/4 X = 7

Or, –7 X = 7 × 4

Or, –7X = 28

Or, X = 28/–7

Or, X = – 4

Here, in the fourth option, we can see that X= –4, Hence Option <u>D— (-4,2)</u> Is correct.

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if you roll a fair 6-sided die 9 times, what is the probability that at least 2 of the rolls come up as a 3 or a 4?
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Using the binomial distribution, it is found that there is a 0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For each die, there are only two possible outcomes, either a 3 or a 4 is rolled, or it is not. The result of a roll is independent of any other roll, hence, the <em>binomial distribution</em> is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
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  • p is the probability of a success on a single trial.

In this problem:

  • There are 9 rolls, hence n = 9.
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The desired probability is:

P(X \geq 2) = 1 - P(X < 2)

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0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For more on the binomial distribution, you can check brainly.com/question/24863377

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