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3 years ago

#29 is the only one I need

Mathematics

1 answer:

const2013 [10]3 years ago

7 0

So the neat number is all the numbers added together I'm pretty sure or it's either all the numbers added together times the amount of numbers
There are

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In a small chess tournament, 20 matches were played. Find out how many people were involved if it is known that each participant

Nina [5.8K]

Answer:

<em>5</em><em> players participated in the tournament.</em>

Step-by-step explanation:

In a small chess tournament, 20 matches were played.

Let us assume that n number of players participated in the tournament

As in each game 2 players play, so the number of ways they can play is,

$=\ ^nC_2$

As they played 2 games with every other participant in the tournament.

So the total number of games is,

$=\ 2\times ^nC_2$

But it is given to be 20, so

$\Rightarrow \ 2\times ^nC_2=20$

$\Rightarrow \ ^nC_2=10$

$\Rightarrow \dfrac{n!}{2!(n-2)!}=10$

$\Rightarrow \dfrac{n(n-1)}{2}=10$

$\Rightarrow {n(n-1)=20$

As $5\times 4=20$ , so we get n=5.

Therefore, 5 players participated in the tournament.

5 0

2 years ago

Read 2 more answers

I Need Help In #1-5

Citrus2011 [14]

If you are having trouble with ratios you can look at them as fractions. The first number being the numerator and the second number being the denominator or the other way around. When you are asked to find an equivalent ratio, you just have to find an equivalent fraction.

7 0

3 years ago

An equation in slope-intercept form is ______

Amanda [17]

an equation in slope intercept form is

y=mx+b

8 0

3 years ago

Read 2 more answers

You need to invest $1000 in a bank account and are give two options. The first option is to earn $50 every month you leave the m

garri49 [273]

Answer:

<h3><u>Option 1</u></h3>

Earn $50 every month.

Let x = number of months the money is left in the account
Let y = the amount in the account
Initial amount = $1,000

$\implies y = 50x + 1000$

This is a <u>linear function</u>.

<h3><u>Option 2</u></h3>

Earn 3% interest each month.

(Assuming the interest earned each month is <u>compounding interest</u>.)

Let x = number of months the money is left in the account
Let y = the amount in the account
Initial amount = $1,000

$\implies y = 1000(1.03)^x$

This is an <u>exponential function</u>.

<h3><u>Table of values</u></h3>

<u />

$\large \begin{array}{| c | l | l |}\cline{1-3} & \multicolumn{2}{|c|}{\sf Account\:Balance} \\ \cline{1-3} & \sf Option\:1 & \sf Option\:2 \\\sf Month & \sf \$50\:per\:mth & \sf 3\%\:per\:mth \\\cline{1-3} 0 & \$1000 & \$1000 \\\cline{1-3} 1 & \$1050 & \$1030 \\\cline{1-3} 2 & \$1100 & \$1060.90 \\\cline{1-3} 3 & \$1150 & \$1092.73 \\\cline{1-3} 4 & \$1200 & \$1125.51 \\\cline{1-3} 5 & \$1250 & \$1159.27 \\\cline{1-3} 6 & \$1300 & \$1194.05 \\\cline{1-3} 7 & \$1350 & \$1229.87 \\\cline{1-3}\end{array}$

From the table of values, it appears that <u>Account Option 1</u> is the best choice, as the accumulative growth of this account is higher than the other account option.

However, there will be a point in time when Account Option 2 starts accruing more than Account Option 2 each month. To find this, graph the two functions and find the <u>point of intersection</u>.

From the attached graph, Account Option 1 accrues more until month 32. From month 33, Account Option 2 accrues more in the account.

<h3><u>Conclusion</u></h3>

If the money is going to be invested for less than 33 months then Account Option 1 is the better choice. However, if the money is going to be invested for 33 months or more, then Account Option 2 is the better choice.

3 0

1 year ago

¿Se pueden reducir términos semejantes en la expresión? y +3x +4y+x+3y+2x. ¿Por qué?

ludmilkaskok [199]

Answer:

8y + 6x

Step-by-step explanation:

Combine like terms

y + 3x +4y + x + 3y + 2x

= (y + 4y + 3y) + ( 3x + x + 2x)

= 8y + 6x

6 0

2 years ago