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3 years ago

Think for a moment about the potential problems with big data that the speaker mentioned:

Engineering

2 answers:

Natalka [10]3 years ago

5 0

Answer: 1. sadly yes, some people are treated unfairly for crimes people yet have commited. 2. no 3. yes

Explanation:

i did this last year

Nataliya [291]3 years ago

4 0

I think it’s the first option and 3rd one

Hope this is right and hale a nice day :)

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In this type of projection, the angles between the three axes are different:- A) Isometric B) Axonometric C) Trimetric D) Dimetn

nekit [7.7K]

Answer:

The correct answer is C) Trimetric

Explanation:

The most suitable answer is a trimetric projection because, in this type of projection, we see that the projection of the three angles between the axes are not equal. Therefore, to generate a trimetric projection of an object, it is necessary to have three separate scales.

7 0

3 years ago

Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an

ser-zykov [4K]

Answer:

volumetric flow rate = $0.0251 m^3/s$

Velocity in pipe section 1 = $6.513m/s$

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the volume will be mass/density

25/997 = $0.0251 m^3/s$

volumetric flow rate = $0.0251 m^3/s$

Average velocity calculations:

<em>Pipe section A:</em>

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s$

<em>Pipe section B:</em>

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s$

7 0

2 years ago

A 36 ft simply supported beam is loaded with concentrated loads 16 ft inwards from each support. On the left side, the dead load

lana66690 [7]

Answer:

1st part: Section W18X76 is adequate

2nd part: Section W21X62 is adequate

Explanation:

See the attached file for the calculation

8 0

2 years ago

(TCO 4) A system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without

steposvetlana [31]

Answer:

The frequency that the sampling system will generate in its output is 70 Hz

Explanation:

Given;

F = 190 Hz

Fs = 120 Hz

Output Frequency = F - nFs

When n = 1

Output Frequency = 190 - 120 = 70 Hz

Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz

5 0

3 years ago

What is the primary damage control telephone circuit for

user100 [1]

Answer:

hsyghcjqg9ug9duyssatayfjzurldh

6 0

2 years ago