First, calculate for the total operating cost of the park through the equation,
TC = TV + TF
where TC is the total cost,
TV is the total variable cost which is equal to the product of the variable cost per visitor and number of visitor, and
TF is the total fixed cost.
Substituting the known values,
TC = ($15)(1,750,000) + $60,000,000 = $86,250,000
Then, the total revenue is the product of the cost of ticket and the number of visitors.
TR = ($50/visitor)(1,750,000 visitors) = $87,500,000
Subtracting the two values will give us an answer of $1,250,000.
ANSWER: $1,250,000
Answer:
Expected dividend yield = 10.0%
Expected capital gains yield = 5.0%
Explanation:
D0 = $1.50 (Given)
E(D1) = D0 * (1 + g) = $1.50 * (1.05) = $1.575
E(P0) = $15.75 (Given)
E(P1) = $15.75 * (1.05)1 = $16.5375
Expected dividend yield = E(D1) / E(P0)
= $1.575 / $15.75 = 0.100 = 10.0%
Expected capital gains yield = (E(P1) - E(P0)) / E(P0)
($16.5375 - $15.75) / $15.75 = 0.050 = 5.0%
The answer will be B. It increased
Answer:

Explanation:
You need to assume that the total <em>expenses</em> were equal to the<em> cost of the supplies</em>, i.e. there were not other expenses but the<em> $1,500 for supplies to sell.</em>
The total income or revenue was <em>$3,700</em>.
The <em>percentage of the expenses to the revenue</em> is:

Answer:
They should operate Mine 1 for 1 hour and Mine 2 for 3 hours to meet the contractual obligations and minimize cost.
Explanation:
The formulation of the linear programming is:
Objective function:

Restrictions:
- High-grade ore: 
- Medium-grade ore: 
- Low-grade ore: 
- No negative hours: 
We start graphing the restrictions in a M1-M2 plane.
In the figure attached, we have the feasible region, where all the restrictions are validated, and the four points of intersection of 2 restrictions.
In one of this four points lies the minimum cost.
Graphically, we can graph the cost function over this feasible region, with different cost levels. When the line cost intersects one of the four points with the lowest level of cost, this is the optimum combination.
(NOTE: it is best to start with a low guessing of the cost and going up until it reaches one point in the feasible region).
The solution is for the point (M1=1, M2=3), with a cost of C=$680.
The cost function graph is attached.