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In-s [12.5K]
3 years ago
9

Juan found a supplier that costs 20% less than all other suppliers who provide the same quality this is an example of

Business
1 answer:
grigory [225]3 years ago
5 0

Answer:

hello the answer is A

Explanation:

hope this helps:))))

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In business writing, the main idea of a written work should be located
Nadya [2.5K]

Answer:

c

Explanation:

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6 0
2 years ago
Read 2 more answers
Some industries’ competition is much more intense than others. Retail grocery stores such as Kroger, Safeway, and Albertson’s in
Lemur [1.5K]

Answer:

rivalry among existing competitors

Explanation:

The Porters' 5 forces is used to analyse the competitiveness among firms in an industry.

Porter's 5 forces include :

  • Competition in the industry : the higher the number of companies in the industry, the lower the power an individual firm possesses. For example, if an industry increases it price, a consumer can easily shift to the consumption of substitutes
  •  Potential of new entrants into the industry : If there are low barriers to entry in an industry, firms in the industry experience greater competition  
  • Power of suppliers : the higher the number of suppliers in the industry, the higher the bargaining power of firms in the industry and the greater the power they possess
  •  Power of customers : the larger the number of customers, the greater the power firms possess
  • Threat of substitute product : if there are little or no substitutes for the goods produced by companies, the greater the power the firms possess
6 0
2 years ago
The degree of operating leverage can be measured by​ ________. A. multiplying the contribution margin by sales revenue B. dividi
Varvara68 [4.7K]

Answer:

Option C is the answer

Explanation:

The degree of operating leverage is measured by dividing the contribution margin by operating income.

The degree of operating leverage (DOL) is the ratio of contribution margin to operating income. It measures how much the operating income of a company will change in response to a change in sales. A Companies that have higher proportion of fixed costs to variable cost will have greater levels of operating leverage.

4 0
3 years ago
Machinery purchased for $150,000 by Tom Brady Co. in 2010 was originally estimated to have a life of 12 years with a salvage val
Phoenix [80]

Answer:

$7,312.50

Explanation:

The computation of the depreciation expense for 2017 is shown below:

Book Value is

= Cost - Accumulated Depreciation

= $150,000 - {[($150,000 - $24,000) ÷ 12 ] × 7y}

= $150,000 - [($126,000 ÷ 12 ) × 7]

= $150,000 - ($10,500 × 7)

= $150,000 - $73,500

= $76,500

Now the depreciation expense for 2017 :

= ($76,500 - $18,000) ÷ (15 - 7) years

= $58,500 ÷ 8 years

= $7,312.50

5 0
2 years ago
Agan Interior Design provides home and office decorating assistance to its customers. In normal operation, an average of 2.5 cus
Phantasy [73]

Answer:

A) Single-server single-phase model (M/M/1).

\lambda=2.5 \,customers/hour\\\\\mu=6\,customers/hour

B) The goal is not met, as the average time waiting for service is 5.56 minutes.

C) The new mean service rate is 7.5 customers/hour.

In this case, the average time waiting for service is 4 minutes, so the goal is met.

Explanation:

A) This situation can be modeled as a single-server single-phase model (M/M/1).

The mean arrival rate is 2.5 customers per hour.

\lambda=2.5 \,customer/h

The mean service rate is 6 customers per hour, calculated as:

\mu=\frac{60\, min/h}{10 \,min/customer}=6\, customer/h

B) The average waiting time for a customer can be expressed as:

W_q=\frac{\lambda}{\mu}\frac{1}{\mu-\lambda}  =\frac{2.5}{6}\frac{1}{6-2.5} =0.417*0.222=0.093\,hours\\\\W_q=0.093\,hours*(60min/h)=5.56 \,min

The average waiting time is 5.56 minutes, so it is more than the goal of 5 minutes.

C) If the average time spent per customer to 8 minutes, the mean service rate becomes

\mu=\frac{60\, min/h}{8 \,min/customer}=7.5\, customer/h

An the average waiting time for the service now becomes:

W_q=\frac{\lambda}{\mu}\frac{1}{\mu-\lambda}  =\frac{2.5}{7.5}\frac{1}{7.5-2.5} =0.333*0.2=0.067\,hours\\\\W_q=0.067\,hours*(60min/h)=4 \,min

The average time is now 4 minutes, so the goal is achieved.

6 0
3 years ago
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