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n200080 [17]
3 years ago
10

Assuming that Borland retires shares it reacquires, record the appropriate journal entry for each of the following transactions:

Business
1 answer:
KiRa [710]3 years ago
5 0

Answer:

The first transaction is that 10 million shares are being reacquired at 32.50 per share so we need to find out how much cash is spent to buy these shares.

32.5*10 million = $325 million

We will debit treasury stock and credit cash because the company is buying shares from the market and paying cash

The second transaction is reacquiring 10 million shares at 36 per share so we need to find how much cash is spent

10 million *36= $360 million

We will debit treasury stock and credit cash because the company is buying shares from the market and paying cash

In the third transaction 1 million shares are being sold for 42, so need to figure out how much cash the company gets from the transaction

42* 1 million = 42 million

We will debit cash and credit common stock as the company is issuing shares to the market and getting cash for it

In the fourth transaction 1 million shares are being sold for 36, so need to figure out how much cash the company gets from the transaction

36* 1 million = 36 million

We will debit cash and credit common stock as the company is issuing shares to the market and getting cash for it

Journal entries

                                                            Debit                          Credit

Treasury stock                                      325 million

Cash                                                                                        325 million

Treasury stock                                      360 million                  

Cash                                                                                        360 million

Cash                                                        42 million

Common stock                                                                          42 million

Cash                                                        36 million

Common stock                                                                           36 million                

Explanation:        

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Partially completed units in ending work in process are 100 percent complete with regard to their direct materials costs if the
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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
3 years ago
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