with the code int foobar(int*n){*n=*n+1; return*n;}when called int k=6; printf("k=%d\n",k) what get printed? * foobar (k)=6,*foobar (k)=7,k=*foobar (k)=6, k=7​

While using pert (program evaluation review technique) and cpm (critical path method) techniques, activities that are not on the critical path are more inflexible than those activities on the critical path. select one: true false