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grigory [225]
3 years ago
10

The Allen Marble Company has a target current ratio of 2.0 but has experienced some difficulties financing its expanding sales i

n the past few months. At present the firm has current assets of ​$2.5 million and a current ratio of 2.5. If Allen expands its receivables and inventories using its​ short-term line of​ credit, how much additional​ short-term funding can it borrow before its current ratio standard is​ reached?
Business
1 answer:
kap26 [50]3 years ago
6 0

Answer:

The additional short-term funding that can be borrowed is 500,000

Explanation:

First the amount of current liabilities must be known:

Current Ratio = Current Asset / Current Liabilities  

2.5 = 2,500,000 / X  

X = 2,500,000 / 2.5

X= 1,000,000

To know how much to expand receivables and inventories in the formula of the current ratio, to the amount of current assets and current liabilities must add an amount such that the result is 2.0.  

(2,500,000 + x) / (1,000,000 + x) = 2.0

(2,500,000 + x) = 2.0 * (1,000,000 + x)

 2,500,000 + x = (2.0* 1,000,000) + (2.0 x)

 2,500,000 + x = 2,000,000 + 2.0 x

 2,500,000 - 2,000,000 = 2.0 x – x

   500,000 = x

 So the maximum that should be expand inventory and receivables is $500,000.

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Answer:

Indenture

Deferred call provision

Explanation:

Indenture is defined as the contract that describes the terms of a borrowing arrangement between a firm that sells a bond issue and the investors who purchase the bonds.

A call provision is defined as the right that the issuer of a security has to call or redeem the security at certain times and under specific conditions.

The call provision in which the issuer is prevented from calling a portion or the entire issue for several years during the early years of the bond issue is called deferred call provision.

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3 years ago
Much of our communication is nonverbal, which accounts for at least: select one:
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The answer is D. 45 to 55%
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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.

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Answer:

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The codification '718-50-25' is named as 'Recognition' and the given point is mentioned in point 'b' of '718-50-25-1'.  

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Hope this helps!

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