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olasank [31]
3 years ago
10

First, determine if the example is EXLUSIVE (M.E.) or

Mathematics
1 answer:
attashe74 [19]3 years ago
7 0

Answer:

Sorry I had to go to dinner here you go hope this helps! :D

Step-by-step explanation:

So, before we start, there are two things we have to go over.

First off, we have "or"

When using or between two probabilitess, you are basically adding them together.

For instance, you first question asks what the probability is of rolling a 5 or a number greater than 3(assuming this is a 6 sided dice).

So, rolling a 5 is a change of 1/6. Since 1 of the 6 dice iwll be 5. The other probability is a 1/2, since 3 of the 6 dice are over 3.

Now lets add these two together because of the "or":

1/2+1/6=3/6+1/6=4/6

So the probabily of rolling a 5 OR rolling a number greater than 3 is 4/6.

This is inclusive, since rolling a 5 is possible and greater than 3.

Now "and" is a bit different and more complex than or is.

Here is an example:

What is the change of rolling a even number or rolling a number above 4?

Well, rolling a even number has a probability of 1/2. Rolling a number above 4 is 1/3.

To find the "and", we multiply the two probabilities.

So 1/3*1/2=1/6

So the probability of rolling a even number and a number above 4 is 1/6.

Now, lets get back to the answers.

So what is the probability of rolling a 2 or a 6?

There is a 1/6 probability to get a 2, and a 1/6 probability to get a 6.

Add it together since they use or:

1/6+1/6=2/6

So the probability of rolling a 2 or a 6 is 2/6, or 1/3.

This is exlusive since you cant roll both a 6 and a 2 at the same time.

Next we have what the probability is of drawing a 6 or a king.

There are 13 different cards in a deck.

There is a 4/52 chance of drawing a 6(spade/heart/etc), and a 4/52 chance of drawing a king. This simplyfies to 1/13 for each.

This is:

1/13+1/13=2/13

So the probability of drawing a 6 or a king is 2/13

This is a exclusive event since a king and 6 cant be drawn at the same time.

Finally, we have:

What is the probability of drawing a queen or a spade?

Again, there is a 4/52 chance of drawing a queen. A spade, however, is different.

13 of the 52 cards can be spades, which simplies to 1/4.

So we have 1/13+1/4=4/52+13/52=17/52

So there is a 17/52 probability of drawing a queen or spade.

This is a inclusive event since it is possible to draw both a spade and queen at the same time.

Hope this answers all your questions and helps you understand what "and" "or" are in probability.

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Step-by-step explanation:

<em>First</em><em> </em><em>things</em><em> </em><em>first</em><em>,</em><em> </em><em>we</em><em> </em><em>need</em><em> </em><em>to</em><em> </em><em>solve</em><em> </em><em>for</em><em> </em><em>the</em><em> </em><em>side</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>octagon</em><em>. </em><em>We</em><em> </em><em>can</em><em> </em><em>do</em><em> </em><em>so</em><em> </em><em>by</em><em> </em><em>using</em><em> </em><em><u>Pythagorean</u></em><em><u> </u></em><em><u>Theorem</u></em><em><u>:</u></em><em> </em><em>a</em><em>²</em><em> </em><em>+</em><em> </em><em>b</em><em>²</em><em> </em><em>=</em><em> </em><em>c</em><em>²</em><em> </em><em>where</em><em> </em><em>c</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em><u>hypotenuse</u></em><em> </em><em>and</em><em> </em><em>a</em><em> </em><em>and</em><em> </em><em>b</em><em> </em><em>are</em><em> </em><em>the</em><em> </em><em>other</em><em> </em><em>two</em><em> </em><em>sides</em><em>. </em>

12² + b² = 13²

144 + b² = 169

b² = 25

b = 5

<em>Now</em><em> </em><em>that</em><em> </em><em>we</em><em> </em><em>know</em><em> </em><em>half</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>side</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>octagon</em><em> </em><em>is</em><em> </em><em>5</em><em>,</em><em> </em><em>we</em><em> </em><em>know</em><em> </em><em>that</em><em> </em><em>one</em><em> </em><em>side</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>octagon</em><em> </em><em>is</em><em> </em><em>1</em><em>0</em><em>.</em><em> </em>

<em>a</em><em>.</em><em> </em><em>To</em><em> </em><em>find</em><em> </em><em>perimeter</em><em> </em><em>we</em><em> </em><em>just</em><em> </em><em>add</em><em> </em><em>every</em><em> </em><em>side</em><em> </em><em>together</em><em> </em><em>(</em><em>or</em><em> </em><em>since</em><em> </em><em>every</em><em> </em><em>side</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>size</em><em>,</em><em> </em><em>we</em><em> </em><em>multiply</em><em> </em><em>by</em><em> </em><em>8</em><em>)</em>

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3 years ago
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Answer:

0.5455

Step-by-step explanation:

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