Answer:
Step-by-step explanation:
given that the Chocolate House specializes in hand-dipped chocolates for special occasions. Three employees do all of the product packaging
Clerk I II III total
Pack 0.33 0.23 0.44 1
Defective 0.02 0.025 0.015
Pack&def 0.0066 0.00575 0.0066 0.01895
a) probability that a randomly selected box of chocolates was packed by Clerk 2 and does not contain any defective chocolate
= P(II clerk) -P(II clerk and defective) = 
b) the probability that a randomly selected box contains defective chocolate=P(I and def)+P(ii and def)+P(iiiand def)
=0.01895
c) Suppose a randomly selected box of chocolates is defective. The probability that it was packaged by Clerk 3
=P(clerk 3 and def)/P(defective)
=
Answer:
1-c
2-b
3-a
Step-by-step explanation:
1. 7y=14
y=2
-7y=-14
y=2
so the equation has one solution
2. 0=0
so the equation has infinite solution
3. 0=2 false
so the equation has no solution
Try this:
1) note that weight of pure antifreeze before mixing and after mixing is the same. So, if 'x' is weight of pure antifreeze in 50% solution, it is possible to make up equation before mixing: 0.5x+0.2*90.
2) there are 0.2*90=18 gal. of pure antifreeze in the 20% solution. If 'x' gal. is the weight of pure antifreeze in 50% sol. and 18 gal. is the weight of pure antifreeze in 20% sol., it is possible to make up an equation after mixing: 0.4(x+18).
3) using the both parts: 0.5x+0.2*90=0.4(x+18) ⇒ x=54 gal. of <u>pure</u> weight.
4) to find 50% solution of 54 gal. pure weight just 54:0.5=108 gal.
Answer: 108 gal.