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I have the following formula:
=SE(1,5*Q2>F8<2*Q2;"RUIM";"BOM")
In Q2 have 44 and in F8 have 70 and my result gives BOM. Where is the error?
1 answer
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You will need to use the function E along with function SE;
SE the value of F8 (70) is greater than 66 E less than 88, so it’s "BAD," otherwise, "GOOD".
The formula will be:
=SE(E(F8>1,5*Q2; F8<2*Q2);"RUIM";"BOM")
Detail
When working with range it should be considered whether to include the reference values themselves - the extremes
Behold:
70 > 66 = VERDADEIRO
66 > 66 = FALSO
If in your spreadsheet the value of F8 is 66, then you will not enter the concept of "BAD";
Range formula (including ends)
=SE(E(F8>=1,5*Q2; F8<=2*Q2);"RUIM";"BOM")
wiki
"In mathematics, an interval (real) is a set that contains each real number between two indicated extremes, whether or not it contains the extremes themselves."
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Staff worked out with the following formula:
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=SE(E((1,5*$Q$2)<F8;(2*$Q$2)>F8);"RUIM";"B0M")
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OK, thank you all for your help. Thanks.
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@Gomesandré Help the site by voting and/or accepting the answers that helped you.
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The values for TRUE and FALSE are not reversed?
– Ismael
Ismael, reversing there is all BAD, or is not doing the logical test......
– Gomes André
Why do
1,5*Q2>F8<2*Q2? This makes it very unreadable. Putting the values you defined, it gets66 > 70 < 88, that is, it gives false and so gives"BOM".– PedroMVM
As you have two conditions, you intend to use
EorOU? Explaining what Peter said, 66 > 70 = FALSE but, 70 < 88 = TRUE - Consulting the truth tableFALSO E VERDADEIRO = FALSO– Ismael
Pedro, the criterion that is needed is: if the value of F8 is between 66 and 88 I attribute the BAD concept. If it is outside this range the concept is GOOD What would then be the given formula?????
– Gomes André
Try to change
1,5*Q2>F8<2*Q2forE(F8>=(1,5*Q2);F8<=(2*Q2))– danieltakeshi