So I’m in IB and we have to come up with a creative personal project ... any ideas?
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Answer:
Date Account title Debit Credit
12/31/2019 Lease Receivable $175,934
Cost of Goods sold $120,000
Sales Revenue $175,934
Inventory $120,000
Date Account title Debit Credit
12/31/2019 Cash $40,800
Deposit Liability $40,800
The rental amount is constant and is made on the first day of the lease period so this is an annuity due.
As the collectability is probable, you need to find the present value of this lease:
= 40,800 * Present value of annuity due factor, 5 year, 8%
= 40,800 * 4.3121
= $175,933.68
= $175,934
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 :
the average number of customers awaiting repairs = 0.30
(b) Determine system utilization.
The system utilization is determined as follows:
(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 -
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