# How to find the percentage error?

i have got the practical value 0.396

the actual value is 0.379

please tell me the answer also explain me the method to find

case 2:

if the practical value is 0.389

and actual value is 0.379

what is the percentage error?

You sure this is a health question?

Its a simple equation:

measured - actual

------------------------- x 100

acutal

or go to this website to do it through a computer: http://www.marshu.com/articles/percent-e...

i'm not really sure but i think it's:

bigger value minus the smaller value and then divide them to the bigger value then multiply by 100. =D

for that i guess it's,

Case 1:

(0.396 - 0.379) / 0.396 = 0.043 x 100 = 4.3%

Case 2:

(.389 - .379) / .389 = 0.0257 times 100 = 2.57%

[i hope this helps, although i'm not quite sure]

Steamyst,I am surprised at your outburst.and now to address the question at hand.Put your question in the correct category you lazy bastard.

the actual value is 0.379

please tell me the answer also explain me the method to find

case 2:

if the practical value is 0.389

and actual value is 0.379

what is the percentage error?

**Answers:**Put your ? question in the correct category. You lazy bastard.You sure this is a health question?

Its a simple equation:

measured - actual

------------------------- x 100

acutal

or go to this website to do it through a computer: http://www.marshu.com/articles/percent-e...

i'm not really sure but i think it's:

bigger value minus the smaller value and then divide them to the bigger value then multiply by 100. =D

for that i guess it's,

Case 1:

(0.396 - 0.379) / 0.396 = 0.043 x 100 = 4.3%

Case 2:

(.389 - .379) / .389 = 0.0257 times 100 = 2.57%

[i hope this helps, although i'm not quite sure]

Steamyst,I am surprised at your outburst.and now to address the question at hand.Put your question in the correct category you lazy bastard.

The health and medicine information post by website user , AnyQA.com not guarantee correctness , is for informational purposes only and is not a substitute for medical advice or treatment for any medical conditions.**More Related Questions and Answers ...**

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