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PilotLPTM [1.2K]

3 years ago

9

Please help me with this, easy points.

1 answer:

m_a_m_a [10]3 years ago

4 0

Elimination is where you aline two equations with the varibles in the same place and the second equation substracts the first equation. Important to know that in the second equation all varibles and constant or y - intercepts are negative. Now with this just start with the first two equations, 5x + 2y -3z = 10 and 2x - 2y + 4z = 6.

5x + 2y - 3z = 10

- 2x + 2y - 4z = -6

3x + 4y - 7z = 4

Do the same thing with the last equation.

3x + 4y - 7z = 4

-x + y - 2z = -3

2x + 5y -9z = 1

Please tell me if I got it wrong, but from what I concluded there's no solution.

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第 5 个问题 a multiple choice exam has 10 questions. each question has 3 possible answers, of which one is correct. a student knows

tatuchka [14]

The chances that the student was merely guessing is 1/3.

Bayes Theorem determines the conditional probability of an event A given that event B has already occurred.

denoted by

$P(A/B)=\frac{P(A)*P(B/A)}{P(B)}$

let A be the event that the student knows the answer .

B be the event that the student does not knows the answer .

and

E be the event he gets answer correct .

According to the given question

$P(A)=\frac{4}{10} \\\\ P(B)=1-\frac{4}{10} =\frac{6}{10}$

Probability that the answer is correct ,given that he knows the answer is

$P(E/A)=1$

Probability that the answer is correct ,given that he guesses it is

$P(E/B)=\frac{1}{3}$ [as the MCQ has 3 options and only one is correct]

We need to find the probability that he guesses the answer given that it is correct.

Required probability $P(B/E)=\frac{P(B)*P(E/B)}{P(A)*P(E/A)+P(B)*P(E/B)}$

Substituting the values we get

$P(B/E)=\frac{\frac{6}{10} *\frac{1}{3} }{\frac{4}{10} *1+\frac{6}{10} *\frac{1}{3} }$

$=\frac{6}{30}*\frac{30}{18} \\ \\ =\frac{6}{18} \\ \\ =\frac{1}{3}$

Therefore , the chances that the student was merely guessing is 1/3.

Learn more about Probability here brainly.com/question/13140147

#SPJ4

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