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valkas [14]
2 years ago
9

Are the communication skills required for a team leader in a manufacturing plant different from those of a customer service exec

utive?
What would you do to improve your communication skills?

Discuss the importance of customer service and suggest ways to ensure good customer experience.
Engineering
1 answer:
Svetllana [295]2 years ago
5 0

Answer:

Explanation: Excellent customer service isn’t just down to your frontline staff, but a customer might be won or lost there. That’s why it’s so important every employee works together to resolve issues and create memorable moments – and empower your frontline agents to be as valuable to the customer as possible. It could be the difference between the customer coming back again, or going elsewhere.

The following tips are designed to help both customer service representatives, customer service management, and operations staff to work together to make experiences that matter. Developing customer service skills is important for the whole team to thrive – and to ensure customers keep coming back.

Customer service representative tips

Practice active listening

Learn to empathize with your customers

Use positive language

Improve your technical skills

Know your products and services

Look for common ground

Communicate clearly

Be solutions-focused

Admit mistakes

Be willing to learn

Customer service team management and operations tips

Provide first-class training

Set your standards high

Have a clear escalation pathway

Align your touchpoints

Create a culture of excellence

Be smart about automation

Use tools that boost speed and efficiency

Measure and analyze customer feedback

Use closed-loop feedback

Listen, understand and take action

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The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8
il63 [147K]

Answer:

a.)

US Sieve no.                         % finer (C₅ )

4                                                  100

10                                                95.61

20                                               82.98

40                                               61.50

60                                               42.08

100                                              20.19

200                                              6.3

Pan                                               0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no.             Mass of soil retained (C₂ )

4                                            0

10                                          18.5

20                                         53.2

40                                         90.5

60                                         81.8

100                                        92.2

200                                       58.5

Pan                                        26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×\frac{100}{w}

∴ we get

US Sieve no.               % retained (C₃ )               Cummulative % retained (C₄)

4                                            0                                           0

10                                          4.39                                      4.39

20                                         12.63                                     17.02

40                                         21.48                                     38.50

60                                         19.42                                     57.92

100                                        21.89                                     79.81

200                                       13.89                                     93.70

Pan                                        6.30                                      100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no.               Cummulative % retained (C₄)          % finer (C₅ )

4                                                     0                                          100

10                                                  4.39                                      95.61

20                                                 17.02                                     82.98

40                                                 38.50                                    61.50

60                                                 57.92                                    42.08

100                                                79.81                                     20.19

200                                                93.70                                   6.3

Pan                                                 100                                        0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = \frac{D60}{D10}

⇒ Cu = \frac{0.4}{0.12} = 3.33

d.)

Coefficient of Graduation = Cc = \frac{D30^{2}}{D10 . D60}

⇒ Cc = \frac{0.22^{2}}{(0.4) . (0.12)} = 1

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The effective inductance is 5 uH.

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