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GuDViN [60]
3 years ago
6

Izzy Division of Marine Boats Corporation had the following results last year (in thousands). Sales $4,700,000​ Operating income

$700,000​ Total assets $3,900,000​ Current liabilities $300,000​ Management's target rate of return is 13% and the weighted average cost of capital is 12%. What is the Izzy Division's Residual Income (RI)?
Business
1 answer:
ASHA 777 [7]3 years ago
4 0

Answer:

Izzy Division's Residual Income is $193,000

Explanation:

given data

Sales = $4,700,000​

Operating income =  $700,000​

Total assets = $3,900,000​

Current liabilities = $300,000​

target rate of return = 13%

average cost of capital = 12%

to find out

Izzy Division's Residual Income

solution

first we find here Minimum required income that is express as

Minimum required income = Total assets × Target rate of return   ................1

Minimum required income = $ 3,900,000​ × 13%

Minimum required income = $ 507,000

and

now we find residual income that is express as

Residual income = Operating income - Minimum required income    ..........2

Residual income = $700000 - $507000

Residual income = $193,000

Izzy Division's Residual Income is $193,000

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During its first and second years of operations, Rogers Company, a corporation using a periodic inventory system, made undiscove
elena-s [515]

Answer:

Net Income understated by $20,000

Explanation:

In the first year, closing inventory was overstated by $80,000. The implications of the above would be,

Net Income for the first year would be overstated by $80,000

In the Second year,

Opening Stock would be overstated by $80,000

Due to this, cost of production stands overstated by $80,000.

Now, given in the question that closing stock for second year is overstated by $60,000 i.e profits are overstated by $60,000.

This means, the net effect on profits would be, $80,000 less $60,000 i.e $20,000 understated profits for the second year.  

4 0
3 years ago
Problem 10A specialty coffeehouse sells Colombian coffee at a fairly steady rate of 280 pounds annually. The beans are purchased
SOVA2 [1]

Answer:

The computations are shown below:

Explanation:

a. The computation of the economic order quantity is shown below:

= \sqrt{\frac{2\times \text{Annual demand}\times \text{Ordering cost}}{\text{Carrying cost}}}

= \sqrt{\frac{2\times \text{280}\times \text{\$45}}{\text{\$0.48}}}

= 229 units

The carrying cost is come from

= $2.40 × 20%

b. Time between placement of orders is

= Economic order quantity ÷Annual demand

= 229 ÷ 280

= 0.8179 years

So,

= 0.8179 × 365 days

= 298.53 days

We assume 365 days in a year

c. The average annual cost of ordering cost and carrying cost equals to

= Holding cost + ordering cost

= (Economic order quantity ÷ 2 × Holding cost)  + (Annual demand ÷ Economic order quantity × ordering cost)

= (229 units ÷ 2 × $0.48) + (280 ÷ 229 units × $45)

= $54.96 + $55.02

= $109.98

d)   Now the reorder level is

= Demand × lead time + safety stock

where, Demand equal to

= Expected demand ÷ total number of weeks in a year

= 280 pounds ÷ 52 weeks

= 5.38461

So, the reorder point would be  

=  5.38461 × 3 + $0

= 16.15 pounds

7 0
3 years ago
In general, how would your best friend describe you as a risk taker?
earnstyle [38]

Answer:

answer is B.............

6 0
2 years ago
Lenox Company has the following financial data: 2020 2019 Assets Current Assets: Cash and Cash Equivalents $ 2,000 $ 1,900 Accou
Kobotan [32]

Answer:

25.55 days

Explanation:

Days Sales in Receivable = Accounts Receivable ÷ (Sales / 365)

therefore

2020 = $2,800 ÷ ($ 40,000 / 365) = 25.55 days

4 0
3 years ago
Repair calls are handled by one repairman at a photocopy shop. Repair time, including travel time, is exponentially distributed,
wolverine [178]

Answer:

the average number of customers awaiting repairs = 0.30

the system utilization = 42

the amount of time that the repairman is not out on a call is  = 4.64 hours

the probability of two or more customers in the system = 0.1764

Explanation:

Given that :

Repair time, including travel time =  mean of 1.6 hours per call.

Requests for copier repairs = mean rate of 2.1 per eight-hour day

i.e mean rate R = 2.1/day

Time = 8 hours

thus; mean rate μ = 8 hours/ 1.6 hours = 5

(a)

Let the average number of customers awaiting repairs be I_i :

I_i = \dfrac{R^2}{\mu (\mu-R)}

I_i = \dfrac{2.1^2}{5 (5-2.1)}

I_i = \dfrac{4.41}{5 (2.9)}

I_i = \dfrac{4.41}{14.5}

\mathbf{I_i = 0.30}

the average number of customers awaiting repairs = 0.30

(b) Determine system utilization.

The system utilization is determined as follows:

\delta = \dfrac{R}{\mu}

\delta = \dfrac{2.1}{5}

{\delta = 0.42}

\mathbf{\delta = 42}

(c) The amount of time during an eight-hour day that the repairman is not out on a call is calculated as :

Percentage of Idle time = 1 - \delta

Percentage of Idle time = 1 - 0.42

Percentage of Idle time = 0.58

However during an 8 hour day; The amount of time that the repairman is not out on a call is = 0.58 × 8 = 4.64 hours

(d)

the probability of two or more customers in the system by assuming Poisson Distribution is:

P(N ≥ 2) = 1 - (P₀+ P₁)

where;

P₀ = 0.58

P₁ = 0.58  × 0.42 = 0.2436

P(N ≥ 2) = 1 - ( 0.58 + 0.2436)

P(N ≥ 2) = 1 - 0.8236

P(N ≥ 2) = 0.1764

Thus; the probability of two or more customers in the system is 0.1764

7 0
4 years ago
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