How much is $23,000 worth after 10 years at 6% annual interest?
1 answer:
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Step-by-step explanation:
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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.