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
Let

be the length of the rectangle and

be the width. In the problem it is given that

. It is also given that the area

. Substituting in the length in terms of width, we have

. Using the zero product property,

. Solving these we get the width

. However, it doesn't make sense for the width to be negative, so the width must be

. From that we can tell the length

.
Answer:
i think the answer is C
Step-by-step explanation:
Answer:
-3/12
Step-by-step explanation: