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.
Answer:
5n
Step-by-step explanation:
Summarizing what we know:
Casey has n nickels, and Meghan 4 times as many: 4n
Then the total number of nickets these two people have is n + 4n, or 5n.
A is 30% more than B. So, A = B + 30% of B = B + 30/100 *B = B + 3/10 *B = 13/10 *B. B is 60% less than C. So, B = C - 60% of C = C - 60/100 *C = C - 3/5 *C = 2/5 *C.
