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alekssr [168]
3 years ago
14

Which behavior would best describe someone who has good communication skills with customers ? a) Following up with some customer

s b) Talking to customers more than listening to them c) Repeating back what the customer says d) Interrupting customers frequently
SAT
2 answers:
s344n2d4d5 [400]3 years ago
6 0

Answer:

A)Following up with some customers

Explanation:

D)Is wrong because interrupting is a sign of bad communication

C)Repeating what people say is also a bad skill/habit

B)They may feel overwhelmed and feel you are talking to much

A) Correct

erastovalidia [21]3 years ago
3 0

Answer:

A)

it is good to follow up with your customers

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Answer:

Elected Official Is The Correct Answer!

Explanation:

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When the speaker in "any human to another" says that "sorrow never scorned to speak / to any who / were false or true," he means
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The speaker means that; <u>sorrow comes to every human being </u>

<h3>The poem " any human to another"</h3>

In the poem " any human to another" the summary of the poem is that every human have undergone one form of suffering and this unites every human on earth. therefore what the speaker says  means that sorrow comes to every human being.

Hence we can conclude that The speaker means that <u>sorrow comes to every human being </u>

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2 years ago
Newton’s law of cooling states that for a cooling substance with initial temperature t0, the temperature t(t) after t minutes ca
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The cooling rate of the substance is approximately 0.0732.

According to the statement, the Newton's law of cooling is defined by the following formula:

T(t) = T_{s} + (T_{o}-T_{s})\cdot e^{-k\cdot t} (1)

Where:

  • T_{s} - Final temperature, in degrees Celsius.
  • T_{o} - Initial temperature, in degrees Celsius.
  • t - Time, in minutes.
  • k - Cooling rate, in \frac{1}{min}.
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Please notice that substance reaches thermal equilibrium when T(t) = T_{s}, that is when temperature of the substance is equal to the temperature of surrounding air.

If we know that T_{o} = 80\,^{\circ}C, t = 15\,min, T_{s} = 50\,^{\circ}C and T(15) = 60\,^{\circ}C, then the cooling rate of the substance is:

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k = -\frac{1}{15}\cdot \ln \frac{1}{3}

k \approx 0.0732

The cooling rate of the substance is approximately 0.0732.

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Answer:

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