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evablogger [386]
3 years ago
8

A six-sided number cube is rolled twice.

Mathematics
1 answer:
Marina CMI [18]3 years ago
4 0

Answer: A and D

Step-by-step explanation:

Dependent events mean that one event depends on the outcome of the previous event.

A) If the first roll is a 3, and we want to sum more than 7 with the second roll, then the possible outcomes 5 and 6. Now if the first roll is 4, then the possible outcomes for the second event will be 4, 5 and 6. So you can see how the outcome space of the second event changes depending on the first event, so the events are dependent.

B) Here we do not have dependence, each event only depends on it's own outcome.

C) Again, both events only depend on it's own outcome, so the events are not dependent.

D) This is similar as the case for A, these events are dependent because is not the same if the outcome of the first event is 3, than if the outcome is 5 (the outcome needed for the second roll changes depending on the outcome of the first event)

E) Same as B and C, the events are independent.

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Answer:

The answer is below

Step-by-step explanation:

The system of equations:

8x-9y+13z=11

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The equations can be represented in matrix form as:

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Therefore:

\left[\begin{array}{ccc}8&-9&13\\-8&-5&5\\3&4&-8\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}11\\15\\-10\end{array}\right]\\\\\\\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}8&-9&13\\-8&-5&5\\3&4&-8\end{array}\right] ^{-1}\left[\begin{array}{c}11\\15\\-10\end{array}\right]\\

\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{19} &-\frac{1}{19}&\frac{1}{19}\\-\frac{49}{380}&-\frac{103}{380}&-\frac{36}{95} \\-\frac{17}{380}&-\frac{69}{380}&-\frac{28}{95} \end{array}\right] \left[\begin{array}{c}11\\15\\-10\end{array}\right]\\\\\\\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}-0.74\\-1.69\\0.13\end{array}\right] \\

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