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liraira [26]
3 years ago
9

PDC Bank is working on creating an AI application that enables customers to send SMS to the AI application to allow banking acti

vities from their registered ID. Jane, the project engineer, has taken bank customer data from the last few years from the server and is using it to train the ML to recognize and authenticate actual users and to ensure unauthorized users are barred from entering the application. Suppose the AI application has been compromised, and the reason has been identified as compromised data being used to improve the ML accuracy. What kind of attack is the PDC Bank application subjected to?
Engineering
1 answer:
bogdanovich [222]3 years ago
5 0

The kind of attack that the PDC Bank application is subjected to due to compromised data is: Adversarial artificial intelligence

Adversarial artificial intelligence is a form of attack on a neural network where a wrong algorithm is entered into the system to deceive the end user.

The aim of this machine learning method is to deceive the end user.

This is what happens to the PDC Bank when its data was compromised in the course of improving the machine learning accuracy.

Learn more about Adversarial artificial intelligence here:

brainly.com/question/22190244

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Carbon nanotubes can be a considered toxic agent due to: (a)- Its ability to produce toxic gases (b)Its ability to penetrate ski
SpyIntel [72]

Answer: b)Its ability to penetrate skin quickly due to its very small diameter

Explanation: Carbon nano tubes(CNT) are the material widely used in the medical field due to the atomic structure of it ans also have small size. Toxicity in the carbon nano tubes is because their small sized atomic particles which can enter the skin by penetration or inhalation. But are still preferred in the medicine because having unique properties like mechanical property, chemical property,surface property etc.

4 0
3 years ago
Calculate the theoretical density of FCC iron (eg. austenitic stainless steel). The lattice parameter for FCC iron is 0.357 nm a
Ann [662]

Answer: 12.4 feet

Explanation:

If there is a smooth transition and there is no change in slopes, energy considerations can be used

The cube has a kinetic energy of

ke = mv^2/2 = 10 lbm * 20^2ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2

At the highest point when there is a gain in potential energy

pe = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2

If there is no loss in energies,

pe = ke

322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2

h = 2000 /322 = 6.211 (ft)

= h / sin(30) = 12.4 ft

8 0
3 years ago
What are the factors that influence the power input to the compressor?
Lena [83]

Answer:

option e is correct answer

5 0
3 years ago
In an air compressor the compression takes place at a constant internal energy and 50KJ of heat are rejected to the cooling wate
lozanna [386]

Answer:

work is 50 kj

Explanation:

Given  data

heat (Q) = 50 kj

To find out

work input for the compression stroke per kilogram of air

Solution

we will apply here "first law of thermodynamics" i.e.

The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another  location. i.e.

ΔU = Q – W                        ................1

here ΔU is change in internal energy, Q is heat and W is work done

here U = 0 because air compressor the compression takes place at a constant internal energy in question

so that by equation 1

Q = W

and Q = 50

so work will be 50 kj

8 0
3 years ago
An excited electron in an Na atom emits radiation at a wavelength 589 nm and returns to the ground state. If the mean time for t
11Alexandr11 [23.1K]

Answer:   Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm

                length of the photon emitted: 6.0 m

Explanation:

The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.

Recall the Heisenberg's uncertainty principle:-

                                 ∆t∆E ≈ h ( Planck's Constant)

The transition time ∆t corresponds to the energy that is ∆E

E=h/t = \frac{(1/2\pi)*6.626*10x^{-34} J.s}{20*10x^{-9} } = 5.273*10x^{-27} J =  3.29* 10^{-8} eV.

The corresponding uncertainty in the emitted frequency ∆v is:

∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)=  7.958 × 10^6 s^-1

To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate

                                                    λ = c/v

                                                    dλ/dv = -c/v² = -λ²/c

Therefore,

      ∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)

                                 =  9.20 × 10⁻¹⁵ m or 9.20 fm

     The length of the photon (<em>l)</em> is

l = (light velocity) × (emission duration)

  = (3.0 × 10⁸  m/s)(20 × 10⁻⁹ s) = 6.0 m          

                                                   

6 0
3 years ago
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