Answer:
The minimum number of different tanks needed to safely house all the fish is:
Step-by-step explanation:
To identify the minimum number of different tanks, we're gonna concentrate in a fish species, in this case can be the A: as you see in the table, the A species can live with all the fish excepting the F and G, by their side, the F and G can't live together , by this reason, this three species must live in a different tank, in the next form:
- Tank 1: <em>A</em>
- Tank 2: <em>F</em>
- Tank 3: <em>G</em>
Now the B species, it can live with A, F and G, but for this example we can put in the tank 1 (the tank of the A species). The C especies can live with A, F and G, but how we have A and B together, we're gonna put the C especies in the tank 3 (the tank of the G especies). The D species can live with A and G, we're gonna put in the tank 1 because can live with B species too. The E species can live with A and F, we're gonna put in the tank 2 (the tank of the F species) because the E species can't live with D that is in the in the tank 1. Al last, the H species just can live with A, E, F, and H species, by this reason, the only tank that can be put is the tank 2. In this form, the order is the next:
- Tank 1: <em>A, B, D</em>.
- Tank 2: <em>F, E, H</em>.
- Tank 3: <em>G, C</em>.
And t<u>he owner of the pet store must buy three different tanks to display these tropical fish</u>.
Answer:
Step-by-step explanation:
-x/4 - 6 ≥ -8.
What was the end goal of this problem?
Answer:
m=-3
x1,y1=1,-5
we have,
y-y1=m(x-x1)
y-(-5)=-3(x-1)
y+5=-3x+3
3x+y+2=0 is the required eqn
Answer:
0.1,0.5,0.8,0.9
Step-by-step explanation:
Mode 17,
Mean is 19.3
Median 18.