Well it is a toy manfacturing company and the intermediate good would be a toy plane
Answer:
Do = $2.00
D1= Do(1+g)1 = $2(1+0.2)1 = $2.40
D2= Do(1+g)2 = $2(1+0.2)2 = $2.88
D3= Do(1+g)3 = $2(1+0.2)3 = $3.456
D4= Do(1+g)4 = $2(1+0.2)4 = $4.1472
D5= Do(1+g)5 = $2(1+0.2)5 = $4.97664
PHASE 1
V1 = D1/1+ke + D2/(1+ke)2 + D3/(1+ke)3 +D4/(1+ke)4 + D5/(1+ke)5
V1 = 2.40/(1+0.15) + 2.88/(1+0.15)2 + 3.456/(1+0.15)3 + 4.1472/(1+0.15)4 + 4.97664/(1+0.15)5
V1 = $2.0870 + $2.1777 + $2.2723 + $2.3712 + $2.4742
V1 = $11.3824
PHASE 2
V2 = DN(1+g)/ (Ke-g )(1+k e)n
V2 = $4.97664(1+0.02)/(0.15-0.02)(1+0.02)5
V2 = $5.0762/0.1435
V2 = $35.3742
Po = V1 + V2
Po = $11.3824 + $35.3742
Po = $46.76
Explanation: This is a typical question on valuation of shares with two growth rate regimes. In the first phase, the value of the share would be obtained by capitalizing the dividend for each year by the cost of equity of the company. The dividend for year 1 to year 5 was obtained by subjecting the current dividend paid(Do) to growth rate. The growth rate In the first regime was 20%.
In the second phase, the value of shares would be calculated by taking cognizance of the second growth rate of 2%. In this phase, the last dividend paid in year 5 would be discounted at the appropriate discount rate after it has been adjusted for growth.
Answer:
Interest will be $855 x 10 years= $8,550
Explanation:
Interest
6÷100=0.06
0.06x14,250=$855
$855x10=$8,550.
How much to have paid back
At the end of 10years $8,550 would have been paid as interest
Total sum will be $14,250+$8,550=$22,800 to be paid back.
Answer:
The correct answer is Assign costs of work process.
Explanation:
Among the main changes to be able to allocate costs, Julio must take the costs of work in process in a single account, instead of directly to different department accounts. This will ensure better control of the information, avoiding mistakes in the planning process.
