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In a JUnit report, if we set the "Health report amplification factor" to 2 and there's a 1% failure, what's the health score?
Suggested Answer: A
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You would take 100% then subtract (1%*2), thus 98%.
12 comments · newest first
MA
Mauricio
11 months ago
How can a 1% failure lead to a 75% score? Sounds fishy!
SH
Shannan
11 months ago
Definitely not 80%, that's way off.
AJ
Aja
12 months ago
Wait, is it really that simple?
BU
Buck
12 months ago
I think it should be 98%.
SU
Suzan
1 year ago
The health score is calculated as 100% - (failure rate * amplification factor).
TA
Tandra
1 year ago
I feel like I saw a question like this in our last mock exam. I think the health score should be 80%, but I can't recall the exact reasoning behind it.
RO
Rosamond
1 year ago
I'm a bit confused about the amplification factor. Does it mean we multiply the failure rate by 2? If so, then wouldn't that change the score significantly?
GL
Glennis
1 year ago
I remember practicing a similar question where we had to adjust scores based on failure rates. I think the answer might be 98% since we reduce the failure percentage first.
FR
Frederick
1 year ago
I think the health score is calculated by taking the failure rate and adjusting it with the amplification factor, but I'm not entirely sure how that works.
MA
Matthew
1 year ago
I feel pretty confident about this one. I think the key is to focus on the specific computers mentioned in the table and their capabilities in relation to Microsoft Defender for Cloud.
BR
Brett
1 year ago
I'm drawing a blank on the exact sequence. I know these concepts, but putting them in the right order is tripping me up. Gotta think this through carefully.
SU
Sue
1 year ago
I keep thinking FY11 was strong too, but I can't quite remember if it matched FY12 or not.