28a⁴b: 2 · 2 · 7 · a · a · a · a · b
16ab: 2 · 2 · 2 · 2 · a · b
GCF(28a⁴b, 16ab) = 4ab
The answer is A.
I've attached the complete question.
Answer:
Only participant 1 is not cheating while the rest are cheating.
Because only participant 1 has a z-score that falls within the 95% confidence interval.
Step-by-step explanation:
We are given;
Mean; μ = 3.3
Standard deviation; s = 1
Participant 1: X = 4
Participant 2: X = 6
Participant 3: X = 7
Participant 4: X = 0
Z - score for participant 1:
z = (x - μ)/s
z = (4 - 3.3)/1
z = 0.7
Z-score for participant 2;
z = (6 - 3.3)/1
z = 2.7
Z-score for participant 3;
z = (7 - 3.3)/1
z = 3.7
Z-score for participant 4;
z = (0 - 3.3)/1
z = -3.3
Now from tables, the z-score value for confidence interval of 95% is between -1.96 and 1.96
Now, from all the participants z-score only participant 1 has a z-score that falls within the 95% confidence interval.
Thus, only participant 1 is not cheating while the rest are cheating.
n - 4 = 16 Given
+ 4 + 4 Cancel out 4 to isolate n variable
n = 20
Prove.
20 - 4 = 16
16 = 16
True.
