I think that this is a combination problem. From the given, the 8 students are taken 3 at a time. This can be solved through using the formula of combination which is C(n,r) = n!/(n-r)!r!. In this case, n is 8 while r is 3. Hence, upon substitution of the values, we have
C(8,3) = 8!/(8-3)!3!
C(8,3) = 56
There are 56 3-person teams that can be formed from the 8 students.
The system of linear equations represents the situation is;
x + y = 125
x + y = 1255x + 8y = 775
<h3>Simultaneous equation</h3>
Simultaneous equation is an equation in two unknown values are being solved for at the same time.
let
- number of quick washes = x
- number of premium washes = y
x + y = 125
5x + 8y = 775
From equation (1)
x = 125 - y
5x + 8y = 775
5(125 - y) + 8y = 775
625 - 5y + 8y = 775
- 5y + 8y = 775 - 625
3y = 150
y = 150/3
y = 50
x + y = 125
x + 50 = 125
x = 125 - 50
x = 75
Therefore, the number of quick washes and premium washes Monica’s school band had is 75 and 50 respectively.
Learn more about simultaneous equation:
brainly.com/question/16863577
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Amount of money you will have = 42.54(1+6/100)^5=56.9
The hypotenuse would be square root of 41
