Nick attempted 20 free throws and made 12 of them.
Recall that the experimental probability is given by
![P=(number\: of\: favorable\: outcomes)/(total\: number\: of\: outcomes)](https://img.qammunity.org/2023/formulas/mathematics/college/k1nuy2jjicakcirhq9w26c0dd4p1q8mph7.png)
For the given case,
Number of favorable outcomes = 12 (since he made 12 successful throws)
Total number of outcomes = 20 (since he attempted a total of 20 throws)
![P(make\: free\: throw)=(12)/(20)=(3)/(5)=0.6](https://img.qammunity.org/2023/formulas/mathematics/college/3h91yet9srjtkufvrpnbx1queeyahxy6s3.png)
Therefore, there is 3/5 probability that Nick will make a free throw on his next try.