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Vsevolod [243]
3 years ago
8

Is rolling a 3 standard doe and then rolling a sum greater than 4 when counting the next roll an independent event?

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
5 0

Answer:

False

Step-by-step explanation:

In this question, you roll 3 from the standard die and then add it with the next roll to see if the sums are greater than 4.

From this explanation you can see that you use the result from your first roll for the second event, so we can conclude that the event is dependent.  

Imagine if we change the result of the first roll into 5, without adding the second roll we can know that the sum will be greater than 4. The first event result will influence the second event, so it is a dependent event.

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Solve for s. 1/2 s = 7/2. A 1/7. B 2/7. C 3. D 7.
fenix001 [56]

Simple equation and only requires one step to solve. Remember, a number and a variable juxtaposed means multiplications.

\frac{1}{2} s = \frac{7}{2} \\ \\\frac{(\frac{1}{2}s)}{\frac{1}{2}} = \frac{\frac{7}{2}}{\frac{1}{2}} \\ \\\\s = \frac{\frac{7}{2}}{\frac{1}{2}} \\ \\\\s = \frac{7}{2} * \frac{2}{1} \\ \\\\s = 7

One step? I know. It seems more because I was slowing down the calculations for you so could understand everything happening. For someone who can do this in their sleep it would, for them, take one step.

4 0
3 years ago
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Svetradugi [14.3K]

Answer:

There is really no answer to this because this is the furtherest you can go to simplify this expression. I can help solve it if you give me more information, or you tell me what the value of a and b is so I can solve it.

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kodGreya [7K]
You first need to convert the fractions so they are both eighths.

1 3/4 = 1 6/8

Then you simply need to add:
1 6/7 + 3 3/8 = 4 9/8 = 5 1/8

Your answer is 5 1/8.
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(7,2) is the answer.
There you go.
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The area of the figure is 84

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