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 correct answer is C. 4π / 3
Hope this helps!
Answer:
The answer for your question is 2.
Step-by-step explanation:
This can be expressed as exponential growth of the form:
f=ir^t, f=final amount, i=initial amount, r=common ratio (rate), t=time
In this case we are given the point (2, 6500) and i=5000 so we can solve for the rate...
6500=5000r^2 divide both sides by 5000
1.3=r^2 take the square root of both sides and note that we know r>0
r=1.3^(1/2) so our equation becomes:
f=5000(1.3)^(1/2)^t and knowing that (a^b)^c=a^(b*c) we can say:
f=5000(1.3)^(t/2) so for t=18
f=5000(1.3)^9
f≈53022 (to nearest whole bacteria)
Answer:
c
Step-by-step explanation:
i think c but not sure
