Answer: The amount of cash provided by operating activities is $179,600.
Explanation:
<u>Statement of cash flows for the company</u>
Net income $210,600
Add Depreciation expense 27,000
Add Loss on sale of equipment 1,800
Increase in Accounts receivable (16,900)
Increase in Inventory (41,700)
Decrease in Prepaid expenses 5,000
Decrease in Accounts payable (6,200)
Cash flows fron operating activities $179,600
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
Answer:
ENCOUNTER
Explanation:
Seth appears to be taken aback by the number of files on his desk and his coworker’s comments about his boss. This reaction depicts the Encounter stage of the socialization process.
