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Anastaziya [24]
3 years ago
12

Describing Tasks for Stationary Engineers Click this link to view O*NET’s Tasks section for Stationary Engineers. Note that comm

on tasks are listed toward the top, and less common tasks are listed toward the bottom. According to O*NET, what are common tasks performed by Stationary Engineers? Check all that apply. observing and interpreting readings on gauges, meters, and charts managing financial resources to order supplies and equipment writing computer programs to control equipment testing boiler water quality or arranging for testing creating safety regulations after experimentation operating or tending stationary engines, boilers, and auxiliary equipment
Engineering
2 answers:
Natali5045456 [20]3 years ago
8 0

Answer:

A, D, F

Explanation:

took on edge.

11111nata11111 [884]3 years ago
8 0

Answer:

The answer is A,D,F

A:observing and interpreting readings on gauges, meters, and charts

D:testing boiler water quality or arranging for testing

F:operating or tending stationary engines, boilers, and auxiliary equipment

Explanation:

I got all right on edge.

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Answer:

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Explanation:

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3 years ago
A sheet of steel 1.5 mm thick has nitrogen (N2) atmospheres on both sides at 1200°C and is permitted to achieve steady-state dif
Gelneren [198K]

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Explanation:

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Vehicles begin to arrive at a parking lot at 8:10 am at a constant rate of 6 veh/min until 8:25 am. There is no arrival from 8:2
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3 years ago
You are recording a friend's 3-minute song with 24 bits per sample at 96 kHz sampling rate for a 5.1 surround sound system (6 ch
djyliett [7]

Answer:

save space = 1.9626 Gibibits

Explanation:

given data

recording song time = 3 minute = 180 seconds

song per sample =  24 bits per sample

frequency = 96 kHz

surround sound system = 5.1

solution

first we get here required space for 24 bits at  96 kHz that is

required space = 24 × 96 × 10³ × 6 × 180

required space = 2.48832 × 10^{9}  bits

as 1 Gib bit = 2^{30} bits

so required space = 2.48832 × 10^{9}  bits ÷  2^{30} bits

required space = 2.3174 Gibibits   ...............1

and

space required to save record 16 bit at 44.1 kHz

space required = 16 × 44.1 × 10³ × 3 × 180

space required = 0.381024 × 10^{9}  bits

space required = 0.381024 × 10^{9}  bits  ÷  2^{30} bits  

space required = 0.3548 Gibibits    ...........2

so

we get here that save space in 16 bit at 44.1 kHz

save space = 2.3174 Gibibit - 0.3548 Gibibits

save space = 1.9626 Gibibits

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