SOLUTION
The clerk sold three piece of troupe of ribbon to three customers
The lenght receive by each customer is
![\begin{gathered} 5(7)/(27)yards\text{ } \\ 1(6)/(7)yards \\ \text{and } \\ 14(1)/(9)yards \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/55wthl3y8y2ywft2mn3kogvzh45y421red.png)
The total lenght of the type of Ribbon is gotten by suming the lenght of each yards received by the customers
Hence we have
![\begin{gathered} \text{The total lenght is } \\ (5(7)/(27)+1(6)/(7)+14(1)/(9))yards \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7jhclff130lhc50pus1db7hnix2xeamb95.png)
Then we add the whole numbers together and the fraction
![(5+1+14)((7)/(27)+(6)/(7)+(1)/(9))](https://img.qammunity.org/2023/formulas/mathematics/college/z3xo1xesj4riaej2lrhz3z78fs82z85m12.png)
![\begin{gathered} 20((49+162+21)/(189)) \\ \text{where LCM=189} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bk3q29v9x48gji589bea5bqnoqrpytborg.png)
Then we have
![\begin{gathered} 20(232)/(189) \\ 20(1(43)/(189)) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9dkg2kzukx642g9i9ru1vaa30tr0kj18yo.png)
Then by adding the whole numbers we obtain
![21(43)/(189)](https://img.qammunity.org/2023/formulas/mathematics/college/voud9pzalhnepgwc0f12pqwyk09ce0ybfd.png)
Therefore the lenght of type of ri