-343,000 x 1/3 = -114,333.333333333219
Answer:
- 273 mL of 5%
- 117 mL of 15%
Step-by-step explanation:
Let q represent the quantity of 15% dressing used. Then the amount of 5% dressing is (390 -q). The amount of vinegar in the mix is ...
0.15q + 0.05(390 -q) = 0.08(390)
0.10q = 31.2 -19.5 = 11.7 . . . . . . subtract 0.05(390) and simplify
q = 117 . . . . . . . . . . . . . . . . . . multiply by 10
390-q = 273
The chef should use 273 mL of the first brand (5% vinegar) and 117 mL of the second brand (15% vinegar).
__
<em>Additional comment</em>
You may have noticed that the value of q is (0.08 -0.05)/(0.10 -0.05)×390. The fraction of the mix that is the highest contributor is the ratio of the difference between the mix value and least contributor, divided by the difference between the contributors: (8-5)/(15-5) = 3/10, the fraction that is 15% vinegar. This is the generic solution to mixture problems.
(14 meter/sec) x (1 mile/1609.344 meter) x (3,600 sec/hour) =
(14 x 3600) / (1609.344) = <em>31.32 mile/hour </em>(rounded)
Volume is length x width x height aka (L•W•H)
The function "choose k from n", nCk, is defined as
nCk = n!/(k!*(n-k)!) . . . . . where "!" indicates the factorial
a) No position sensitivity.
The number of possibilities is the number of ways you can choose 5 players from a roster of 12.
12C5 = 12*11*10*9*8/(5*4*3*2*1) = 792
You can put 792 different teams on the floor.
b) 1 of 2 centers, 2 of 5 guards, 2 of 5 forwards.
The number of possibilities is the product of the number of ways, for each position, you can choose the required number of players from those capable of playing the position.
(2C1)*(5C2)*(5C2) = 2*10*10 = 200
You can put 200 different teams on the floor.