One example is “engagement” consider to be “business activities”
Answer:
It is a violation of NASD rules against guaranteeing a customer against loss.
Explanation:
In this case the RR is guaranteeing the customer against loss. The customer initially bought the shares for $20 the new price is $10. The RR now coming in to buy the shares above market value is a way to guarantee the customer against loss, and its a NASD violation.
Answer:
The monetary value is $24,201.23
Explanation:
Giving the following information:
Cash flows:
Year 1= $6,800
Year 2= 6,800
Year 3= 6,800
Year 4= $15,000.
The discount rate is 15 percent.
We need to discount each cash flow to the present value:
PV= FV/(1+i)^n
Year 1= 6,800/1.15= 5,913.04
Year 2= 6,800/1.15^2= 5,141.78
Year 3= 6,800/1.15^3= 4,471.11
Year 4= 15,000/ 1.15^4= 8,576.30
Total= $24,201.23
Answer:
A) R(x) = 120x - 0.5x^2
B) P(x) = - 0.75x^2 + 120x - 2500
C) 80
D) 2300
E) 80
Explanation:
Given the following :
Price of suit 'x' :
p = 120 - 0.5x
Cost of producing 'x' suits :
C(x)=2500 + 0.25 x^2
A) calculate total revenue 'R(x)'
Total Revenue = price × total quantity sold, If total quantity sold = 'x'
R(x) = (120 - 0.5x) * x
R(x) = 120x - 0.5x^2
B) Total profit, 'p(x)'
Profit = Total revenue - Cost of production
P(x) = R(x) - C(x)
P(x) = (120x - 0.5x^2) - (2500 + 0.25x^2)
P(x) = 120x - 0.5x^2 - 2500 - 0.25x^2
P(x) = - 0.5x^2 - 0.25x^2 + 120x - 2500
P(x) = - 0.75x^2 + 120x - 2500
C) To maximize profit
Find the marginal profit 'p' (x)'
First derivative of p(x)
d/dx (p(x)) = - 2(0.75)x + 120
P'(x) = - 1.5x + 120
-1.5x + 120 = 0
-1.5x = - 120
x = 120 / 1.5
x = 80
D) maximum profit
P(x) = - 0.75x^2 + 120x - 2500
P(80) = - 0.75(80)^2 + 120(80) - 2500
= -0.75(6400) + 9600 - 2500
= -4800 + 9600 - 2500
= 2300
E) price per suit in other to maximize profit
P = 120 - 0.5x
P = 120 - 0.5(80)
P = 120 - 40
P = $80
