The Lagrangian for this function and the given constraints is

which has partial derivatives (set equal to 0) satisfying

This is a fairly standard linear system. Solving yields Lagrange multipliers of

and

, and at the same time we find only one critical point at

.
Check the Hessian for

, given by


is positive definite, since

for any vector

, which means

attains a minimum value of

at

. There is no maximum over the given constraints.
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:
6(3x - 5) + 2(3x - 5) - 8(3x - 5) = 0
Step-by-step explanation:
The answer is b. 1/27
3^-3=1/27 and 3^4=81. 1/27×81=3. 3×3=9. 3^-5=1/243. 1/243×9= 1/27. The answer is b. 1/27.