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Minchanka [31]
3 years ago
15

Suppose you have a 6-sided number cube and a coin. Event A is that you roll the number cube and land on 1, and event B is that y

ou flip the coin and land on tails. How could you describe events A and B?
Choose all that apply.

-independent events
-dependent events
-mutually exclusive events
-not mutually exclusive events
Mathematics
2 answers:
ratelena [41]3 years ago
6 0

Answer:  The correct option is (A) Independent events.

Step-by-step explanation:  Given that we have a 6-sided Suppose you have a 6-sided number cube and a coin.

Event A and event B are defined as follows:

Event A : the event of rolling 1 on the number cube,

Event B: the event of landing on a tail while flipping the coin.

We are to select the correct description of the events A and B from the given options.

<u><em>Independent Events:</em></u><em> </em> Two events are said to be independent if the happening of one of them does not affect the happening of the other.

<u><em>Dependent events:</em></u> Two events are said to be dependent if the happening of one of them affect the happening of the other.

<u><em>Mutually exclusive events:</em></u>  Two events are mutually exclusive or disjoint if they cannot both occur at the same time.

<u><em>Not-mutually exclusive:</em></u>  Two events are said to be mutually non exclusive events if both the events have at least one common outcome between them.

Now, looking at the definition of events A and B, we can say that the happening of any one of them does not affect the happening of the other event.

Because, the rolling of 1 on a number cube and landing a tail on a coin are independent of each other.

So, the events A and B are INDEPENDENT events.

Option (A) is CORRECT.

Anton [14]3 years ago
5 0
They are independent events because the outcome of one does not affect the outcome of the other
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Below

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x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

Step-by-step explanation:

5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}

Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}

\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x

\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

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If you find this, it was an accident
Anna007 [38]
Alright, lols yes yes
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