First we must calculate the current power the curling iron can handle:
![P=I* V](https://img.qammunity.org/2023/formulas/physics/college/za6cm07vedv9xql54z0v3jn87fv7d4rj1r.png)
where
I = current
V = tension
Then:
![\begin{gathered} 600=I*120 \\ I=5A \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/85czutlzxtsx6urkf8rchvogeuv34w61l7.png)
For the hair dryer we get:
![\begin{gathered} P=I* V \\ 1200=I*120 \\ I=10A \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/be6w8s0xckkim306ghqzj7izwlhj8s9j6d.png)
Lets suppose all appliances are working in parallel together. The total current must be:
![\begin{gathered} I=I_1+I_2 \\ I=5+10 \\ I=15A \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/uwnom00wa9jv2aqudc1xzmxoaaplczxeo4.png)
As we can see the appliances are drawning all current from the outlet fused.
Answer: the current that can be safely drawn from the outlet fused is zero.