Answer:
The number of shares of stock issued in the stock dividend is 5,312.20 shares.
Explanation:
This can be determined as follows:
Number of shares before stock dividend = Number of shares reported on January 1, 2019 - Number of shares purchased for its treasury on March 24, 2019 + Number of treasury shares were sold on August 19, 2019 = 268,000 - 3,000 + 610 = 265,610
Number of dividend shares = Number of shares before stock dividend * Rate of stock dividend issued = 265,610 * 2% = 5,312.20
Therefore, the number of shares of stock issued in the stock dividend is 5,312.20 shares.
Answer:
REVENUES
Explanation:
Revenue, often referred to as sales, is the income received from normal business operations and includes discounts and deductions for returned merchandise. It is the top line or gross income on a company's income statement from which all charges, costs, and expenses are subtracted to arrive at net income.
$3.00 probably because if it,s $3.00 it should be very elastic
Answer:
$1,282.80
Explanation:
The PMT formula is used for this question. The attachment is shown below:
The NPER shows the time period
Given that,
Present value = $300,000 - $30000 = $270,000
Future value = $0
Rate of interest = 4% ÷ 12 months = 0.33%
NPER = 30 years × 12 months = 360 months
The formula is shown below:
= PMT(Rate;NPER;-PV;FV;type)
The present value come in negative
So, after solving this, the answer is $1,282.80
Answer:
a) $3
b) $2
c) 1449
Explanation:
Given:
The cost for a carton of milk = $3
Selling price for a carton of milk = $5
Salvage value = $0 [since When the milk expires, it is thrown out ]3
Mean of historical monthly demand = 1,500
Standard deviation = 200
Now,
a) cost of overstocking = Cost for a carton of milk - Salvage value
= $3 - $0
= $3
cost of under-stocking = Selling price - cost for a carton of milk
= $5 - $3
= $2
b) critical ratio =
or
critical ratio =
or
critical ratio = 0.4
c) optimal quantity of milk cartons = Mean + ( z × standard deviation )
here, z is the z-score for the critical ration of 0.4
we know
z-score(0.4) = -0.253
thus,
optimal quantity of milk cartons = 1,500 + ( -0.253 × 200 )
= 1500 - 50.6
= 1449.4 ≈ 1449 units