MNPL "z" argument possibilities
Does "z" the argument of MNPL function could be a vector of numeric ?
MNPL(z = 1.00) # 0.5K
MNPL(z = 2.39) # 0.6K
MNPL(z = c(1.00, 2.39)) Does this should return something ?
Does "z" the argument of MNPL function could be a vector of numeric ?
MNPL(z = 1.00) # 0.5K
MNPL(z = 2.39) # 0.6K
MNPL(z = c(1.00, 2.39)) Does this should return something ?