Answer:
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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
Answer:
Look up this on google area of circle with diameter of 8.
Step-by-step explanation:
It will be correct trust me
Answer:
Step-by-step explanation:
Multiply the first term by
, the second term by
, the third term by
.
<u>The first term becomes:</u>
<u>The second term becomes:</u>
<u>The third term becomes:</u>
<u>Their sum is:</u>
Correct choice is B