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Ahat [919]
3 years ago
13

In the context of a job advertisement, what does 'self-starting' mean?​

Business
1 answer:
andrey2020 [161]3 years ago
5 0

Answer:

A self-starter is a person who is motivated to set and achieve goals and takes initiative within the workplace. Self-starters are especially important in the workplace and highly sought after by employers because they can get their job done with supervision and work through issues and perform tasks on their own.

Explanation:

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Creative Sound Systems sold investments, land, and its own common stock for $37.0 million, $14.3 million, and $38.6 million, res
worty [1.4K]

Answer:

$18.3 million

Explanation:

Financing activities: It includes those activities which comes under the long term liabilities and shareholder equity balance. The issue of shares is an inflow of cash whereas redemption, dividend, and the purchase of treasury stock is an outflow of cash.

The computation of the amount reported as a net cash flows from financing activities is shown below:

Cash flow from Financing activities  

Issuance of common stock $38.6 million

Less: Purchase of treasury stock -$20.3 million

Net Cash flow from Financing activities           $18.3 million

8 0
3 years ago
What is the name of the Inca tot of stock market prices that averages 30 selected industrial stocks?
nydimaria [60]

Answer:

C

Explanation:

I'm smart boy that's y because y = u and u nedda pay attention in class blood

6 0
3 years ago
What is the argument in this formula? =AVERAGE(B11:H14) AVERAGE H11 B11:H14 B11
Whitepunk [10]

Answer:

The Answer is D. B11:H14

Explanation:

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8 0
3 years ago
If a department that applies process costing starts the reporting period with 40,000 physical units that were 80% complete with
andriy [413]

Answer:

True

Explanation:

<em>Equivalent units are notional whole units which represent incomplete work and are used to apportion cost between work progress and completed work. These units are determined as follows: </em>

Equivalent units = Degree of work done(%) × units of inventory

Equivalent units

<em>Direct material:</em>

Balance of work = 100-80 = 20%

Equivalent of work to be added = 20% × 40,000 = 8,000

<em>Labour</em>

Balance of work = 100-50= 50%

Equivalent of work to be added =50% × 40,000 = 20,000 units

                                    <em>Equivalent of work to be added(units)</em>

Material                               8000                                  

Labour                               20,000

8 0
2 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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