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lozanna [386]
3 years ago
10

A real option enables the investor to buy an option for a small initial investment, hold it until a decision point arrives, and

then exercise or abandon the option. true or false?
Business
1 answer:
Degger [83]3 years ago
6 0

The statement,"A real option enables the investor to buy an option for a small initial investment, hold it until a decision point arrives, and then exercise or abandon the option." is False .

<u>Explanation: </u>

A real option is to give corporate investment options to a company's executives. It is called "actual" because it usually refers to projects that involve a tangible asset rather than a financial product. Physical assets such as equipment, capital assets and the products are tangible assets.

The decision to extend or delay or wait or to leave a proposal may be real options. Real options require decisions or preferences that give people discretion and possible benefits when making financial decisions.

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Tell me about one of your most passionate beliefs and explain how it has an impact on the way you think, behave, attend events,
WITCHER [35]

Answer:

One of my most passionate belief is "Fail Better". This is the belief which I developed since childhood, my family kept on saying this me when I was just 10 years old. Since then, It has been deeply rooted and firmly suited in my mind. It has influenced me from my childhood, I never felt hesitated in taking risks, trying new ideas, things and adventures in my whole life. This has definitely impacted the way I think, behave and communicate with others. This belief was then further been transformed and translated into "Fall seven times, Get up eight". I have followed these rules very strongly in my whole life.  I have started many small businesses when I was in college, almost I failed in them but I learnt so many things which none could have taught me.

5 0
3 years ago
Ivanhoe Company purchased a new machine on October 1, 2017, at a cost of $77,980. The company estimated that the machine has a s
ad-work [718]

Answer:

Annual depreciation= $10,160 a year

Explanation:

Giving the following information:

Ivanhoe Company purchased a new machine on October 1, 2017, for $77,980. The company estimated that the machine has a salvage value of $6,860. The machine is expected to be used for 72,900 working hours during its 7-year life.

Annual depreciation= (original cost - salvage value)/estimated life (years)

Annual depreciation= (77,980 - 6,860)/7= $10,160 a year

6 0
3 years ago
Zephron Music purchased inventory for $4,400 and also paid a $260 freight bill. Zephron Music returned 25​% of the goods to the
asambeis [7]

Answer:

Cost of the inventory kept by Zephron Music is $3495

Explanation:

<u><em>Zephron Music purchased inventory for $4,400 and also paid a $260 freight bill</em></u>

Inventory $4660 (debit)

Trade Payable $ 4400 (credit)

Bank $260 (credit)

Recognise an Asset - Inventory and De-recognise asset - Bank

<u><em>Zephron Music returned 25​% of the goods to the seller, took a 1​% purchase discount</em></u>

Trade Payable $1212

Inventory $1165 (credit)

Discount Received $47 (credit)

Therefore Inventory Balance = $4660-1165 = $3495

6 0
3 years ago
Suppose that output (Y ) in an economy is given by the following aggregate production function: Yt = Kt + Nt where Kt is capital
shusha [124]

Answer:

Check the explanation

Explanation:

Yt = Kt + Nt

Taking output per worker, we divide by Nt

Yt/Nt = Kt/Nt + 1

yt = kt + 1

where yt is output per worker and kt is capital per worker.

a) With population being constant, savings rate s and depreciation rate δ.

ΔKt = It - δKt

dividing by Nt, we get

ΔKt/Nt = It/Nt - δKt/Nt ..... [1]

for kt = Kt/Nt, taking derivative

d(kt)/dt = d(Kt/Nt)/dt ... since Nt is a constant, we have

d(kt)/dt = d(Kt/Nt)/dt = (dKt/dt)/Nt = ΔKt/Nt = It/Nt - δKt/Nt = it - δkt

thus, Capital accumulation Δkt = i – δkt

In steady state, Δkt = 0

That is I – δkt = 0

S = I means that I = s.yt

Thus, s.yt – δkt = 0

Then kt* = s/δ(yt) = s(kt+1)/(δ )

kt*= skt/(δ) + s/(δ)

kt* - skt*/(δ) = s/(δ)

kt*(1- s/(δ) = s/(δ)

kt*((δ - s)/(δ) = s/(δ)

kt*(δ-s)) = s

kt* = s/(δ -s)

capital per worker is given by kt*

b) with population growth rate of n,

d(kt)/dt = d(Kt/Nt)/dt =

= \frac{\frac{dKt}{dt}Nt - \frac{dNt}{dt}Kt}{N^{2}t}

= \frac{dKt/dt}{Nt} - \frac{dNt/dt}{Nt}.\frac{Kt}{Nt}

= ΔKt/Nt - n.kt

because (dNt/dt)/Nt = growth rate of population = n and Kt/Nt = kt (capital per worker)

so, d(kt)/dt = ΔKt/Nt - n.kt

Δkt = ΔKt/Nt - n.kt = It/Nt - δKt/Nt - n.kt ......(from [1])

Δkt = it - δkt - n.kt

at steady state Δkt = it - δkt - n.kt = 0

s.yt - (δ + n)kt = 0........... since it = s.yt

kt* = s.yt/(δ + n) =s(kt+1)/(δ + n)

kt*= skt/(δ + n) + s/(δ + n)

kt* - skt*/(δ + n) = s/(δ + n)

kt*(1- s/(δ + n)) = s/(δ + n)

kt*((δ + n - s)/(δ + n)) = s/(δ + n)

kt*(δ + n -s)) = s

kt* = s/(δ + n -s)

.... is the steady state level of capital per worker with population growth rate of n.

3. a) capital per worker. in steady state Δkt = 0 therefore, growth rate of kt is zero

b) output per worker, yt = kt + 1

g(yt) = g(kt) = 0

since capital per worker is not growing, output per worker also does not grow.

c)capital.

kt* = s/(δ + n -s)

Kt*/Nt = s/(δ + n -s)

Kt* = sNt/(δ + n -s)

taking derivative with respect to t.

d(Kt*)/dt = s/(δ + n -s). dNt/dt

(dNt/dt)/N =n (population growth rate)

so dNt/dt = n.Nt

d(Kt*)/dt = s/(δ + n -s).n.Nt

dividing by Kt*

(d(Kt*)/dt)/Kt* = s/(δ + n -s).n.Nt/Kt* = sn/(δ + n -s). (Nt/Kt)

\frac{sn}{\delta +n-s}.\frac{Nt}{Kt}

using K/N = k

\frac{s}{\delta +n-s}.\frac{n}{kt}

plugging the value of kt*

\frac{sn}{\delta +n-s}.\frac{(\delta + n -s)}{s}

n

thus, Capital K grows at rate n

d) Yt = Kt + Nt

dYt/dt = dKt/dt + dNt/dt = s/(δ + n -s).n.Nt + n.Nt

using d(Kt*)/dt = s/(δ + n -s).n.Nt from previous part and that (dNt/dt)/N =n

dYt/dt = n.Nt(s/(δ + n -s) + 1) = n.Nt(s+ δ + n -s)/(δ + n -s) = n.Nt((δ + n)/(δ + n -s)

dYt/dt = n.Nt((δ + n)/(δ + n -s)

dividing by Yt

g(Yt) = n.(δ + n)/(δ + n -s).Nt/Yt

since Yt/Nt = yt

g(Yt) = n.(δ + n)/(δ + n -s) (1/yt)

at kt* = s/(δ + n -s), yt* = kt* + 1

so yt* = s/(δ + n -s) + 1 = (s + δ + n -s)/(δ + n -s) = (δ + n)/(δ + n -s)

thus, g(Yt) = n.(δ + n)/(δ + n -s) (1/yt) =  n.(δ + n)/(δ + n -s) ((δ + n -s)/(δ + n)) = n

therefore, in steady state Yt grows at rate n.

5 0
3 years ago
9) Napier Co. provided the following information on selected transactions during 2018: Purchase of land by issuing bonds $1,000,
4vir4ik [10]

Answer:

($1,100,000)

Explanation:

Given that

Loans made to affiliated corporations = $1,400,000

Proceeds from sale of Equipment = $300,000

The computation of net cash provided (used) by investing activities is here below:-

Net cash provided(used) by investing activities = (Loans made to affiliated corporations) - Proceeds from sale of Equipment

= ($1,400,000) - $300,000

= ($1,100,000)

So, for computing the cash provided(used) by investing activities we simply applied the above formula.

7 0
4 years ago
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