Find and graph the feasible region for the following constraints: x + y < 5. 2x<span> + y > 4 ... y = 10/3. x = 30/3 - 10/3 = 20/3. Intersects at (20/3, 10/3). -x + </span>2y<span> = 0. x - </span>2y = 0.
Cost less salvage value = 970,000 - 4500 = 965,500
Capacity of machine = 1,000,000 units.
units consumed at the end of second year = 200,000 + 300,000 = 500,000 units.
Capacity remaining = 1,000,000 - 500,000 = 500,000 units
Book value at end of second year = (500,000/1,000,000)*965,500 + 4500
= $487,250
27 is the answer, I can explain how I got that if you need.
Answer:
In 1981, the Australian humpback whale population was 350
Po = Initial population = 350
rate of increase = 14% annually
P(t) = Po*(1.14)^t
P(t) = 350*(1.14)^t
Where
t = number of years that have passed since 1981
Year 2000
2000 - 1981 = 19 years
P(19) = 350*(1.14)^19
P(19) = 350*12.055
P(19) = 4219.49
P(19) ≈ 4219
Year 2018
2018 - 1981 = 37 years
P(37) = 350*(1.14)^37
P(37) = 350*127.4909
P(37) = 44621.84
P(37) ≈ 44622
There would be about 44622 humpback whales in the year 2018
Answer:
Hello I can I get some answers choics?
Step-by-step explanation:
