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3 years ago

One less than the quotient of a number and negative five

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

6 0

<h2>Answer:</h2><h3>-5x-1</h3><h2>Step-by-step explanation:</h2>

Let's say the unknown number is x

<u>Quotient = is the answer to a multiplication question</u>

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

We have to multiple -5 by x:

_---------------__-------------__-----------____-----------------------------------------

Lastly: we have to minus 1 from -5x

<h3>= -5x-1</h3><h3>````````````````````````````````````````````````````````````````````````````````````````````````````````````````</h3><h3 /><h3>If you learned from my answer please rate it below from a scale of 1 to 5 using the stars to let me know how well you learned from my answer. </h3><h3 /><h3>Make sure to also press the like button (Thanks button) if you really liked my answer / explanation </h3><h3 /><h3>If it was incorrect or you didn't learn from it please review my answer accordingly (report if incorrect) </h3><h3 /><h3>````````````````````````````````````````````````````````````````````````````````````````````````````````````````</h3><h3 /><h3>I only want to spread knowledgeable and truthful information let me know if it is incorrect or not helpful so I / Brainly can take it down. </h3><h3 /><h3>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ </h3><h3 /><h3>Lastly, If your the owner of this question please mark me Brainliest if you really learned from my answer I really do appreciate it. </h3><h3 />

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How many 5-member chess teams can be chosen from 15 interested players? Consider only the members selected, not their board posi

ryzh [129]

<u>Answer</u>:

3003 number of 5-member chess teams can be chosen from 15 interested players.

<u>Step-by-step explanation:</u>

Given:

Number of the interested players = 15

To Find:

Number of 5-member chess teams that can be chosen = ?

Solution:

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula $nCr = \frac{n!}{r!(n - r)!}$

where

n represents the total number of items,

r represents the number of items being chosen at a time.

Now we have n = 15 and r = 5

Substituting the values,

$15C_5 = \frac{15!}{5!(15- 5)!}$

$15C_5 = \frac{15!}{5!(10)!}$

$15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 }{5 \times 4 \times 3 \times 2 \times 1(10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}$