Income <span>will be the same under both variable and absorption costing</span> when there is zero beginning inventory and all inventory units produced are sold.
Variable costing fluctuates based on level of output while adsorption costing is when manufacturing costs are absorbed by the amount produced. When everything is sold and no inventory is being held, both will be zero since there is nothing to sell or have on hand.
It’s is definitely b and c
Answer:
Journal entries will be as follows;
Explanation:
1.The machine purchased is an asset so machinery a/c will be debited.
The cash used to purchase the machine is an outflow so it's credited on the cash a/c
2. Electricity wiring on the machine is part of the acquisition cost, hence we debit machinery account and the cash paid for that is credited on cash a/c
3. Cost of securing it in place is also an operating cost hence you debit machinery a/c and credit the cash used to pay for it in the cash a/c
<u>Journal entries</u>
1. Machinery account Dr 192,000
Cash account Cr 192,000
2.Machinery account Dr 8,000
Cash account Cr 8,000
3.Machinery account Dr 1,600
Cash account Cr 1,600
Answer:
a. c+b≤360....equation 1
b. 2 c+1.5 b≥500.... equation 2, where c and b are the number of cans and bottles of water respectively.
c. The number of water bottles to be sold have to be equal to or more than 142 to cover the cost of renting costumes.
Explanation:
a.
<em>Step 1: Determine maximum number of cans and bottles</em>
As indicated, the number of cans and bottles can not exceed a certain value. This means that the number of cans and bottles can be either equal to or less than that value. The maximum number of cans and bottles can be represented in the following expression;
c+b≤m
where;
c=unknown
b=unknown
m=360
replacing;
c+b≤360....equation 1
b.
<em>Step 2: Determine total amount needed to raise $500</em>
Since $500 dollars is the minimum amount needed, the sales have to be $500 and more. This can be expressed as;
(C×c)+(B×b)≥T
where;
T=total amount needed
C=price per can of lemonade
c=number of cans sold
B=price per bottle of water
b=number of bottles sold
In our case;
T=$500
C=$2
c=unknown
B=$1.50
b=unknown
replacing;
(2×c)+(1.5×b)≥500
2 c+1.5 b≥500.... equation 2
c.
<em>Step 3: Determine least number of bottles of water that must be sold</em>
The least number of bottles of water that must be sold to cover the cost of renting costumes can be solved using equation 2 above;
2 c+1.5 b≥500
where;
c=144
b=unknown
replacing;
(2×144)+1.5 b≥500
288+1.5 b≥500
1.5 b≥500-288
1.5 b≥212
b≥212/1.5=141.33=142
b≥142, meaning the number of water bottles to be sold have to be equal or more than 142 to cover the cost of renting costumes.
<span>Annualized consumption dropped immensely in November 2008. The 5 years prior to 2008 were some of our strongest yet, also, a GDP of $14 trillion is nothing to balk at. I am thoroughly surprised that the decline we experienced then came so soon after a long streak of winning.</span>