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makkiz [27]
2 years ago
13

In how many ways can 3

Mathematics
1 answer:
statuscvo [17]2 years ago
3 0
<h3>Answer:  2300</h3>

Work Shown:

_{n} C _{r} = \frac{n!}{r!(n-r)!}\\\\_{25} C _{3}  = \frac{25!}{3!*(25-3)!}\\\\_{25} C _{3}  = \frac{25!}{3!*22!}\\\\_{25} C _{3}  = \frac{25*24*23*22!}{3!*22!}\\\\ _{25} C _{3}  = \frac{25*24*23}{3!}\\\\ _{25} C _{3}  = \frac{25*24*23}{3*2*1}\\\\ _{25} C _{3}  = \frac{13800}{6}\\\\ _{25} C _{3}  = 2300\\\\

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Which inequality correctly compares Two-fifths, Six-sevenths, Five-eighths, and 1?
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Answer:

<u>2/5 < 5/8 < 6/7 < 1 </u>

   <u>OR</u>    

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Step-by-step explanation:

It is required to compare Two-fifths, Six-sevenths, Five-eighths, and 1

Two-fifths = 2/5

Six-sevenths = 6/7

Five-eighths = 5/8

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We need to make the numbers in order from the least to the greatest or from the greatest to the least

The easy method is convert the rational numbers to decimal numbers

So,

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1 = 1

So, the numbers form the least to the greatest are:

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So,

2/5 , 5/8 , 6/7 , 1

The inequality correctly compares the numbers are:

<u>2/5 < 5/8 < 6/7 < 1</u>

Or can be written from the greatest to the least as:

<u>1 > 6/7 > 5/8 > 2/5 </u>

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3 years ago
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