Answer:
133 fishes
Step-by-step explanation:
Units of food A = 400 units
Units of food B = 400 units
Fish Bass required 2 units of A and 4 units of B.
Fish Trout requires 5 units of A and 2 units of B.
i. For food A,
total units of food A required = 2 + 5
= 7 units
number of bass and trout that would consume food A = 2 x 
= 114.3
number of bass and trout that would consume food A = 114
ii. For food B,
total units of food B required = 4 + 2
= 6 units
number of bass and trout that would consume food B = 2 x 
= 133.3
number of bass and trout that would consume food B = 133
Thus, the maximum number of fish that the lake can support is 133.
The dog eats 7 cans plus d cans?
this any help?!!?
Answer:
13
Step-by-step explanation:
1 1/3 = 4/3
18 ÷ 4/3
= 13
For this problem, 0.25 in = 50 m. So
11 in = (11 in) • (50/0.25 m/in) = 2200 m
Alternatively, if 0.25 in = 50 m, that means 1 in = 200 m. So 11 in = 11 (200 m) = 2200 m.