Answer:
The correct answer is B
Explanation:
The net cost of goods is computed as if the paid in the discounting period:
Net Cost of goods = Inventory cost - (Inventory cost × Discounting percentage)
where
Inventory cost is $9,000
Discounting percentage is 2%
Putting the values above:
Net Cost of goods = $9,000 - ($9,000 × 2%)
Net Cost of goods = $9,000 - $180
Net Cost of goods = $8,280
Therefore, the amount of $8,280 will be paid by the company if paid within the discounting period and avail the discount of $180.
In order to compute for the effective annual rate, the
working equation would be [( 1 + i/n)^n] – 1. The i
corresponds to the nominal rate while n is the number of compounding periods
per year which in this case is 12. The answer would be 5.116%.
Answer and Explanation:
The computation of the incremental net income is shown below:
<u>Particulars Sell Process Further Incremental Net income
</u>
Sales $20,000.00 $50,000.00 $30,000.00
(10,000 units × $2) (10,000 × $5)
Less:
Additional
Processing cost $18,000.00 $18,000.00
Total $20,000.00 $32,000.00 $12,000.00
Answer:
a-Dec-31. Dr Utility expense 485
Cr Utility bills payable 485
b-Jan-11. Dr Utility bills payable 485
Cr Cash 485
c-Dec-31. Dr Salary expense 3990
Cr Salary payable 3990
d-Dec-31. Dr bank 51600
Cr Loan payable 51600
e-Dec-31 Dr Interest expense 215
Cr interest payable 215
f-Dec-31 Dr Account receivable 340
Cr Service revenue account 340
g-Dec-31. Dr Cash 6840
Cr Advance Rent 6840
Explanation:
a-Utility expense incurred for the m/o Dec will be paid in Jan.
c- Salaries of 3990 will be paid on Jan of 4 days.
e-Interest expense for the m/o Dec will be (51600*5%=2580/12=215.
f-The service fee is receivable which will be paid on Jan.
g- Advance rent is received from client.
Answer:
maximum income is $900
Explanation:
given data
oil change = $20
per day = 40 customer
increase = $ 2
dailer customers = 2
owner charge = $ 2
to find out
income from the business
solution
we know current income is 40 × 20
current income = $800
we consider here price increase x and income as function y
so y = (20 +2x) × ( 40 - 2x) ........1
y = −4x² + 40x + 800
take derivative and put dy/dx = 0 for maximum
dy/dx = -8x + 40
0 = -8x + 40
x=5
so here from 1
y = (20 +2x) × ( 40 - 2x)
y = (20 +2(5)) × ( 40 - 2(5))
y = 30 × 30
y = 900
so maximum income is $900
