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.
Euler's formula is given by:
C + V = A + 2
Where,
C: number of faces
V: number of vertices
A: number of edges.
Clearing A we have:
A = C + V-2
Substituting values:
A = 21 + 14-2
A = 33
Answer:
the missing number is:
C.33
R + w = 50
r = 6w + 1
now we sub ..
6w + 1 + w = 50
7w + 1 = 50
7w = 50 - 1
7w = 49
w = 49/7
w = 7 <== white beads
r = 6w + 1
r = 6(7) + 1
r = 42 + 1
r = 43 <=== red beads