A. Multiple password changes and verifications
You won’t need a password for most online stores. The rest of the answers are all required.
Answer:
Explanation:
March 1
Cash 1700
Share Capital 1700
To record the issuance of shares
March 3
Equipment 1400
Cash 1400
To record the purchase of equipment on cash
March 5
Rent Expense 470
Cash 470
To record the rent expense
March 7
No entry neither cash nor serves are provided.
March 12
Purchase 117
Cash 117
To record the purchases
March 15
Cash 670
Income 670
To record the services Income.
March 19
Advance 570
Payable 570
To record the advance cash receipt for services not yet provided thus advance is our liability.
March 25
Payable 228 570/25*10
Income 228
To record the services income against advance given.
March 30
Utilities Expense 82
Cash 82
To record the monthly utilities expense
March 31
Dividend Pay 85
Cash 85
To record the payment of dividend.
2)
<u>Share Capital Account</u>
Opening = 0
Cash 1700
<u>Closing=1700</u> <u> </u>
<u>Cash Account</u>
Opening = 0
share capital 1700 Equipment 1400
Rent 470
Purchase 117
Income 670
Adv. Pay 570
Utilities 82
Dividend 85
<u>Closing balance- 786</u>
<u>Equipment</u>
Opening = 0
Cash 1400
<u>Closing-1400</u>
<u>Rent Expense</u>
Opening = 0
Cash 470
<u>Closing-470</u>
<u>Purchase Expense</u>
Opening = 0
Cash 117
<u>Closing-117</u>
<u>Income Account</u>
Opening = 0
Cash 670
Payable 228
<u>Closing- 898</u> <u> </u>
<u>Payable Account</u>
Opening = 0
Cash 570
Income 282
<u>Closing-342</u> <u> </u>
<u>Utilities Expense</u>
Opening = 0
Cash 82
<u>Closing-82</u>
<u>Dividend</u>
Opening = 0
Cash 85
<u>Closing-85</u>
3) Trail Balance
Head Of Account Debit Credir
Cash 786
Share Capital 1700
Equipment 1400
Rent Expense 470
Purchases 117
Income 898
Payable 342
Utilities 82
Dividend 85
Total 2940 2940
Answer:
Option C. $0.11
Option D. $0.95
Explanation:
As we know that the Transfer Price is set at either selling price for an outside market or variable cost plus opportunity cost if the product sold is to internal market present within the organization (Inter group or inter division sales).
However, the division can still charge upper limit price to the division which is $1 market price of the product.
Upper limit = $1
As it is given that the selling of the additional units will be among divisions which means its inter division market. Hence the lower limit will be used here.
Lower Limit = Variable cost + opportunity cost
Here
Variable cost is $10 cents
And
Opportunity cost will be zero here as the division will be using its excess capacity to sell to the other division, so there is no opportunity cost.
So, by putting values, we have:
Lower Limit = $0.1 - $0 = $0.1
Upper limit = $1
Thus the transfer price set for each bell can be between $1 and $0.1. So the $0.11 and $0.95 falls between these range and both are correct options here.
Answer:
Scrap
Explanation:
The scrap material is that material that is not usable for the or the services are no longer available and these products are not used so far for the production process. It is totally and completely discarded and used as a by product production process
Hence, the correct option is scrap
And all other options are wrong and incorrect
Answer:
Answer for the question:
There are ten polluting firms, Firm1,. . . ,Firm10. Each firm emits 100 pounds of pollution prior to any regulations (so there are currently 1,000 pounds being emitted). Each firm has constant marginal abatement costs, but the costs vary across firms. Conveniently, the firms’ names indicate their marginal abatement costs. Firm1’s marginal abatement costs are constant at $1 per pound, Firm2’s marginal abatement costs are constant at $2 per pound,. . . , and Firm10’s marginal abatement costs are $10 per pound.
a. Suppose the regulator wants to achieve a 25% reduction in pollution (250 pounds). What is the cost effective allocation of emis- sions across the ten firms?
b. What are the total abatement costs for society to achieve a 250 pound reduction in emissions?
c. The marginal damage of pollution in this city is given by MD= 4-1/250 X, where X is the total reduction in pollution. What is the optimal level of pollution?
is given in the attachment.
Explanation:
