Atrip to europe for a class of n students has a fixed cost of x dollars. if 5 students decide not to go, then how much more will it cost each student to take the trip?
a) x/5
b) x/n-5
c) (x//5)
d) (x//n-5)
e) (x/n-/n)

Answer would be E) (x/n-5) - (x/n).

$x is fixed and the cost is shared with the whole class.

Whole Class = n

So, if 5 students decides to remove themselves from the trip ...

Whole class - 5 students = n - 5

If $x is shared with (n-2) students, each students has to pay ...

$(x/n-2)

But the question wants to know how much MORE each student has to pay ...

Adjusted cost - Original cost = $(x/n-2) - $(x/n)

Why $(x/n-2) - $(x/n) ?
$(x/n-2) will incur more cost for each student than $(x/n) as it is divided by a smaller denominator.

Imagine if lesser students went for the trip but has to pay the same amount for their trip. Wouldn't each of them have to pay more?

N students, fixed cost = x dollars.

Originally each student to contribute =   x/n

If 5 , decide not to go, the number left = (n - 5)

Cost =   x/(n - 5)

The extra cost is  x/(n - 5) -   x/n

Option E.

It is 2 out of 8 she will pick green and 3 out of 7 she will pick green but red and green would be 5 out of 5 so 50% chance

You need to show the box plots