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BlackZzzverrR [31]
3 years ago
10

A number cube with faces labeled from 1 to 6 will be rolled once.

Mathematics
1 answer:
kramer3 years ago
3 0

Answer:

\Omega=\{1,2,3,4,5,6\}

A=\{1,2,3,4\}

Step-by-step explanation:

<u>Sample Space</u>

The sample space of a random experience is a set of all the possible outcomes of that experience. It's usually denoted by the letter \Omega.

We have a number cube with all faces labeled from 1 to 6. That cube is to be rolled once. The visible number shown in the cube is recorded as the outcome. The possible outcomes are listed as the sample space below:

\Omega=\{1,2,3,4,5,6\}

Now we are required to give the outcomes for the event of rolling a number less than 5. Let's call A to such event. The set of possible outcomes for A has all the numbers from 1 to 4 as follows

A=\{1,2,3,4\}

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You have 4 one-dollar bills and 4 five-dollar bills in your piggy bank. You select two bills at random, with replacement. What i
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The probability that one would have selected 2 five-dollar bills is; 1/4.

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<img src="https://tex.z-dn.net/?f=%284%20%2B%206i%29%20%7B%7D%5E%7B2%7D%20" id="TexFormula1" title="(4 + 6i) {}^{2} " alt="(4 +
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Answer:

36i^{2}+48i+16

Step-by-step explanation:

(4+6i)^{2} is essentially (4+6i)(4+6i) (you are squaring (4+6i), so multiplying it by itself), so to simplify, you would need to distribute.

A good way to go about this is using the F.O.I.L. method, multiplying the First numbers in the parentheses, then the Outers, then Inners, and finally, Lasts.

(4+6i)(4+6i) >> (4+6i)(4+6i) >> (4+6i)(4+6i) >> (4+6i)(4+6i)

((4*4)+(4*6i)+(6i*4)(6i*6i))

After doing so, you would end up with 16+24i+24i+36i^{2}, and after combining like terms, 16+48i+36i^{2}.

To write the expression in standard form, though, just order the terms from highest to lowest power!

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