Answer:
The number of possible choices of my team and the opponents team is
Step-by-step explanation:
selecting the first team from n people we have 
possibility and choosing second team from the rest of n-1 people we have 
As { A, B} = {B , A}
Therefore, the total possibility is 
Since our choices are allowed to overlap, the second team is
Possibility of choosing both teams will be
![\frac{n(n-1)}{2} * \frac{n(n-1)}{2} \\\\= [\frac{n(n-1)}{2}] ^{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%20%20%2A%20%20%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%20%20%5C%5C%5C%5C%3D%20%5B%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%5D%20%5E%7B2%7D)
We now have the formula
1³ + 2³ + ........... + n³ =![[\frac{n(n+1)}{2}] ^{2}](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D%5D%20%5E%7B2%7D)
1³ + 2³ + ............ + (n-1)³ = ![[x^{2} \frac{n(n-1)}{2}] ^{2}](https://tex.z-dn.net/?f=%5Bx%5E%7B2%7D%20%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%5D%20%5E%7B2%7D)
=![\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] = [\frac{n(n-1)}{2}]^{3}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dn-1%5C%5CE%5C%5Ci%3D1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5B%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%5D%5E%7B3%7D)
Answer:
If there are N students, the total cost will be:
C(N) = $500 + $5*N.
And we want the amount that each student pays to be equal or less than $15.
then:
C(N)/N ≤ $15.
Then the inequality that represents this situation is:
($500 + $5*N)/N ≤ $15.
Now, let's solve this:
($500 + $5*N) ≤ $15*N
$500 ≤ $15*N - $5*N
$500 ≤ $10*N
$500/$10 ≤ N
50 ≤ N
So the minimum number of students needed is 50.
Answer:
300
Step-by-step explanation:
The total number of ratio units in 6:5 is 6+5=11, so each of those ratio units must stand for 550/11 = 50 detectives. Then the 6 ratio units of police detectives stand for 6·50 = 300 police detectives.
Could just be used as a variable.
3x+4=x+2
2x+4=2
2x=-2
-2/2
X=-1
Y=-1+2
Y= 1
Solution= (-1,1)
