The required probability is 
<u>Solution:</u>
Given, a shipment of 11 printers contains 2 that are defective.
We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.
Now, we know that, 
Probability for first draw to be non-defective 
(total printers = 11; total defective printers = 2)
Probability for second draw to be non defective 
(printers after first slot = 10; total defective printers = 2)
Then, total probability 
Answer:
Basically, you have to think: more bricks : more time
more workers : less time
2,400 bricks 6 workers takes 18 hours
You now have to solve for 4,500 blocks and 10 workers
4,500 / 2,400 = 1.875 (times greater)
10 / 6 = 1.666666666 (times less)
So, we get 18 hours, multiply it by 1.875 and divide it by 1.666666666
which equals 20.25 hours
So, it seems you were correct on your third try.
Step-by-step explanation:
B, because 24/4=6 :))))))
(5/8) / (3/8) =
5/8 * 8/3 =
5/3 =
1 2/3....a = 1, b = 2, c = 3
Step-by-step explanation:
5x4-3x2+2x-4/x-3
=2x+5×4-3×2-4/x-3
=2x+32/x-3
