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3 years ago

What are some common work contexts for Licensing Examiners and Inspectors? Select four options.

Engineering

2 answers:

Sonbull [250]3 years ago

7 0

Answer:

Telephone

Face-to-face discussions

Contact with others

Importance of being exact or accurate.

Explanation:

Akimi4 [234]3 years ago

6 0

According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:

Telephone
Face-to-face discussions
Contact with others
Importance of being exact or accurate.

O*NET is an acronym for occupational information network and it refers to a free resource center or online database that is updated from time to time with several occupational definitions, so as to help the following categories of people understand the current work situation in the United States of America:

Workforce development professionals

Students

Human resource (HR) managers

Job seekers

Business firms

On O*NET, work contexts are typically used to describe the physical and social elements that are common to a particular profession or occupational work. Also, the less common work contexts are listed toward the bottom while common work contexts are listed toward the top.

According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:

1. Telephone

2. Face-to-face discussions

3. Contact with others

4. Importance of being exact or accurate.

Read more on work contexts here: brainly.com/question/22826220

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Given V1 = 8 vpp, R1 = 6 kΩ, V2 = 5 vpp, R2 = 3 kΩ, V3 = 10 vpp, R3 = 3 kΩ, and Rf = 14 kΩ, find Vout. Vout = vpp. (Round your a

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Vout= 93.3V

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For this question, consider circuit in the attachment 1.

This is the circuit of an inverting amplifier. In an inverting amplifier

Vout/Vin= -Rf/Rin

To calculate the Vout, we must find Rin and Vin. For this we must solve the input circuit (attachment 2) using Thevinine theorem. Thevnine theorem states that all voltage sources in a circuit can be replaced by an equivalent voltage source Veq and and all resistances can be replaced by an equivalent resistance Req. To find out Req all voltage sources must be short circuited (attachment 3)

1/Req= 1/R1+1/R2+1/R3

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It is at an angle of 180° from Veq

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3 years ago