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777dan777 [17]
3 years ago
14

Wilson Dover Inc. The total value (debt plus equity) of Wilson Dover Inc. is $500 million and the face value of its 1-year coupo

n debt is $200 million. The volatility (σ) of Wilson Dover's total value is 0.60, and the risk-free rate is 5%. Assume that N(d1) = 0.9720 and N(d2) = 0.9050. Refer to the data for Wilson Dover Inc. What is the yield on Wilson Dover's debt? a. 7.05% b. 6.04% c. 6.70% d. 7.42% e. 6.36%
Business
1 answer:
Nutka1998 [239]3 years ago
6 0

Answer:

7.42%

Explanation:

Value = 500 million

Amount of debt = 200 million

Time = year

Volatility = 6%

Risk free rate = 0.05

Nd1 = 0.9720

Nd2 = 0.9050

We have to calculate the value =

500 - (500 x 0.9720 - 200 x e^-0.05 x 0.9050)

= 186.17 million

We now calculate the yield

(200/186.17)^1 - 1

= 0.0742

= 7.42%

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Answer:

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1. Identification of transactions.

2. Recording of identified transactions in a Journal

3. Posting into the general ledger

The general ledger provides details of all accounting activities by account enabling the bookkeeper to monitor financial positions and statuses by account.

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3 years ago
The former CEO of Sam's Club, a division with its own profit-and-loss responsibility, Rosalind Brewer, reported to Walmart's CEO
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B. strategic business unit

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3 0
4 years ago
Pressure with Two Liquids, Hg and Water. An open test tube at 293 K is filled at the bottom with 12.1 cm of Hg, and 5.6 cm of wa
Radda [10]

Answer:

1170839.28 dyn/cm^2

16.9816 psia

117.083928 kN/m^2

Explanation:

To calculate the absolute pressure in the bottom of tube we need to sum the atmosferic and gauge pressure.

P_{abs}=P_{atm}+P_G

And the gauge pressure is given by the contributions of columns of water (P_{w}) and mercury(P_{Hg}), we can calculate the contribution of each column as:

P= \rho g h (*)

where \rho is the respective density, g gravity and h is height.

So we have all the data required to use the above equations (P_{atm}, height and density of each column) we only need to be carefully with the units.

For simplicity we can to express all pressure contributions in mmHg ( P_{atm}, P_{w} and P_{Hg}). Note that the units "x" mmHg  means the pressure at the bottom of a column of mercury of "x" mm high. For example, in this case we have a column 12.1 cm of Hg, that is a column of 121 mmHg (passing from cm to mm only requires multiply by 10) pressure exerted by that column is 121 mmHg.

Now pressure of 5.6 cm (56 mm) of water would be 56 mm of water, but it is not the same that mmHg, since the density of water is lower, the pressure exerted by 1 mm of water is lower than the exerted by 1 mm of Hg. The conversion between mmHg and mm of water is given by the relation between the densities.

mmHg=\frac{\rho_w*mmH_2O}{\rho_{Hg}}

mmHg=\frac{0.998*mmH_2O}{13.55}=0.0737 mmH_2O

And pressure of water in mmHg is

0.0737*56=4.1246 mmHg

The absolute pressure is:

P_{abs}=P_{atm}+P_G= 756 + 121 + 4.1246  = 881.1246 mmHg = 88.11246cmHg

To pass to dyn/cm^2 units we need to use the equation (*)

P= \rho g h = 13.55 \frac{g}{cm^3} * 980.665 \frac{cm}{s^2} * 88.11246 cmHg = 1170839.28 \frac{g}{cm s^2} = 1170839.28 \frac{dyn}{cm^2}

Note: We need to use cm Hg for units coherence

Now passing from dyn/cm^2 to kN/m^2 (or kPa) we need to consider that 1 dyn is 10^{-8} kN and 1 cm^2 is 10^{-4} m^2.  

1170839.28 \frac{dyn}{cm^2} * \frac{10^{-8}kN}{1 dyn}*\frac{cm^2}{10^{-4}m^2}=117.083928kN/m^2

Now passing kN/m^2 to psia. We need to consider that 1 psia is 6.89476.

117.083928kN/m^2*\frac{1psia}{6.89476kN/m^2}=16.9816 psia

 

8 0
3 years ago
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