Answer:
interest revenue 13,095.1
Explanation:
We will calcualte the loan interest:
first year
machine value x interest rate = interest revenue first year
250,000 x 10% = 25,000 interest revenue for the first year
cuota - interest = amortization
144,049 - 25,000 = 119.049 amortization
carrying value
250,000 - 119,049 = 130.951
second year
carrying value x interest
130,951 x 10% = 13,095.1 interest revenue for the second year
amortization
144,049 - 13,095.1 = 130.953,9
For the second year, the interest revenue will be of 13,095.1
B. No, a mortgage loan originator is prohibited from working for two companies as a MLO at the same time.
C. No, working for two lending companies at the same time may cause a conflict of interests.
D. Yes, however, she must apply for a dual license which allows her to work for both companies
Answer:
I BELIVE IS
Explanation:
No, a MLO is prohibited from working for two companies at the same time. HOPE IT HELPS
B) Discontinuous change
C) Resource scarcity
D) Buyer dependence
Answer:
C) Resource scarcity
Explanation:
According to my experience in the field of supply chain management, it can be said that the term that best illustrates this scenario is resource scarcity. This term refers to when the amount of an available resource becomes extremely limited, usually causing it to cost more. In this situation leather being banned has caused it to become extremely rare and limited in that location, thus causing Bagfur's business to get dropped since they deal exclusively in leather products.
Answer:
500 runs
Explanation:
In this question, we are asked to calculate the optimal number of production runs the company should make each year.
Please check attachment for complete solution and step by step explanation
The optimal number of production runs per year for a company that manufactures silverware is determined by minimizing the total cost per year, taking into account the fixed cost per run, the cost per unit, and the cost of storing a unit for a full year. This is achieved when the incremental cost of producing and storing one more set of silverware equals the incremental revenue from selling one more set. The calculation involves differentiating the total cost function with respect to the quantity produced in a single run, and solving this derivative equal to zero.
This question is about determining the optimal number of product runs per year for a company that makes silverware. The optimal number of product runs should minimize the total cost which includes production costs and storage costs. To find this optimal number of product runs, we need to take into consideration, the fixed cost per run, the cost per unit of silverware, and the cost of storing a set for a full year.
Let's define Q as the quantity of silverware sets produced in a single run, C as the cost per run excluding the cost per unit of silverware, V as the variable cost per unit of silverware, and S as the storage cost per set of silverware for a full year. The total cost for a year can then be expressed as:
TC = C * 2500/Q + V + S * Q
Note that the first term of the equation, C * 2500/Q, represents the fixed costs per set of silverware, and the last term, S * Q, represents the total storage cost for the units produced in a single run. Given the values for C ($200), V ($5), and S ($4), the task is to find the value of Q that minimizes TC. You can accomplish this by taking the derivative of TC with respect to Q, setting it equal to zero, and solving for Q. This is a calculus operation beyond the scope of this response, but the concept is that the optimal number of production runs per year is achieved when the incremental cost of producing and storing one more set of silverware is equal to the incremental revenue from selling one more set.
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