Answer:
...
Step-by-step explanation:
t would be graph: B (0,4)
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.
The third one is correct so it would be C.
Just substitute X and Y into the question.
Easy, take your problem,
-5(a-6)+2a
Multiply the 5.
-5a+30+2a
Then just add apples and oranges.
-5a+2a= -3a.
Making your equation -3a+30.
