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Ainat [17]
3 years ago
9

find five consecutive even integers if the sum of the first and fifth is 2 less than 3 times the fourth.

Mathematics
1 answer:
gulaghasi [49]3 years ago
3 0
2a,2a+2,2a+4,2a+6,2a+8-\ \ five\ consecutive\ even\ integers\\\\
2a+2a+8+2=3(2a+6)\\\\
2a+2a+10=6a+18\\\\
4a+10=6a+18\ \ \ |Subtract\ 4a\\\\
10=2a+18\ \ \ |Subtract\ 18\\\\
10-18=2a\\\\
-8=2a\ \ \ |Divide\ by\ 2\\\\
a=-4\\\\
\boxed{Numbers:\
-4,-2,0,2,4}
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Using probability concepts, it is found that:

a) \frac{4}{13} probability of drawing a card below a 6.

b) \frac{4}{9} odds of drawing a card below a 6.

c) We should expect to draw a card below 6 about 4 times out of 13 attempts, which as an odd, it also 4 times for every 9 times we draw a card above 6, which is the third option.

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  • A probability is the <u>number of desired outcomes divided by the number of total outcomes</u>.

Item a:

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Then:

p = \frac{20}{52} = \frac{4}{13}

\frac{4}{13} probability of drawing a card below a 6.

Item b:

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\frac{4}{13 - 4} = \frac{4}{9}

\frac{4}{9} odds of drawing a card below a 6.

Item c:

  • The law of large numbers states that with a <u>large number of trials, the percentage of each outcome is close to it's theoretical probability.</u>
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A similar problem is given at brainly.com/question/24233657

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