Step-by-step explanation:
z = (x − μ) / σ
z = (65 − 69) / 3.5
z = -1.14
D=event that chip selected is defective
d=event that chip selected is NOT defective
Four possible scenarios for the first two selections:
P(DDD)=15/100*14/99*13/98=13/4620
P(DdD)=15/100*85/99*14/98=17/924
P(dDD)=85/100*15/99*14/99=17/924
P(ddD)=85/100*84/99*15/98=17/154
Probability of third selection being defective is the sum of all cases,
P(XXD)=P(DDD)+P(DdD)+P(dDD)+P(ddD)
=3/20
I do not see anything wrong with it
Answer:
z<1.5
Step-by-step explanation:
9+10=?
9 = 3x3
10= 5x2
3 (from 3x3) x 5 (from 5x2) = 15
3 (from 3x3) x 2 (from 5x2) = 6
15+6=21
Therefore 9+10=21
