IF function
That was it. Thanks guys!!!!!
"Ron Coderre" wrote:
Try this:
=IF(I19="yes",(D15/52)*0.6,0)
In quotes, Excel interprets "yes" as simple text. Without the quotes, it
looks for a function or range with that name.
Does that help?
***********
Regards,
Ron
XL2002, WinXP
"cwilliams" wrote:
I am trying to find a value of a cell. I keep recieving error messages #Name?
If I19 is yes then I want it to calculate a value using (D15/52)*60%
If I19 is no then I want the cell to show"0"
IF(I19=yes,(((D15/52)*0.6)),"0")
Any suggestions?
Thanks!
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