We should bring all of those numbers to the same denominator: here 45 is best, since it's a multiplication of all the denominators present (3,15,5,9).
So the numbers will be:
\(2\frac{30}{45}\), \(2\frac{18}{45}\), \(2\frac{27}{45}\), \(2\frac{20}{45}\)
I calculated this like this:
(on the example of \(2\frac{2}{3}\))
\(2\frac{2}{3}=2\frac{2*15}{3*15} =2\frac{30}{45}\)
I multiphied by 15 because 45/3=13
so now we can order those numbers from the smallest to the largest:
\(2\frac{18}{45}\) \(2\frac{20}{45}\), \(2\frac{27}{45}\), \(2\frac{30}{45}\),
which correspond to
\(2\frac{6}{15}\) \(2\frac{4}{5}\), \(2\frac{3}{5}\), \(2\frac{2}{3}\),
