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V125BC [204]
3 years ago
11

Need help with number 10 please

Mathematics
1 answer:
Naddika [18.5K]3 years ago
3 0

Answer:

  Pilar is correct. Both systems have solution (-2, 4).

Step-by-step explanation:

Graphing, or any other means of solving these systems of equations reveals both systems have the same solution: (x, y) = (-2, 4).

_____

By Cramer's rule, the solution to the first set is ...

  x = (2(6)-4(2))/(2(5)-4(3)) = 4/-2 = -2

  y = (2(5)-6(3))/-2 = -8/-2 = 4

(x, y) = (-2, 4) is the solution to the first system

__

Since the first equation of the second system is the same as the first equation of the first system, we know the above solution satisfies that equation. We only need to check to see if the same point satisfies the second equation of the second system:

  11(-2) +8(4) = -22 +32 = 10 . . . . . yes,

(-2, 4) is the solution of the second system

Pilar is correct that both systems have the same solution.

_____

Cramer's Rule tells you the solution to

  • ax +by = c
  • dx +ey = g

is ...

  • x = (bg -ec)/(bd -ea)
  • y = (cd -ga)/(bd -ea) . . . . same denominator as for x

I like to use Cramer's Rule when the solution is needed without a lot of steps. Here, we could use the Elimination method to eliminate the y-variable and solve for x. Then more steps are needed to find y. The writing out and explanation are tedious, so it is much simpler just to use a formula that gives the answer.

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The probability that the larger of the two rolls was equal to 5 is; 2/9

<h3>Probability of possible outcomes</h3>

The total number of possible outcomes when the dice is rolled twice is 36 outcomes.

Now, for the larger of the two rolls to equal 5, the total number of favourable  outcomes are 8 which are;

(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5,4)

Thus, probability that the larger of the two rolls was equal to 5 is;

P(larger of 2 rolls equals 5) = 8/36 = 2/9

Read more about probability of possible outcomes at; brainly.com/question/19916581

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Sphinxa [80]

The valid probability distributions are the ones in options C and D.

<h3>Which of the following are valid probability distributions?</h3>

For discrete random variables with probabilities p₁, p₂, ..., pₙ, there are two rules:

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So, for the first rule we can discard the first option, where we have negative probabilities.

To check the other 4 options, just add the probabilities and see if the addition gives 1.

The options that add up to 1 are C and D, so these two are the correct options.

D: 1/5 + 1/10 + 1/10 + 1/10 + 1/5 + 1/10 + 1/10 + 1/10 = 1

C: 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1

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