All categories

2 years ago

Which type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope?

Engineering

1 answer:

seraphim [82]2 years ago

7 0

Narrow AI is the type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope.

Natural Language Understanding (NLU) is used to describe extracting information from unstructured text using algorithms.

<h3>What is Narrow AI ?</h3>

This is known to be a form or a type of AI that is said to be able to carry out a dedicated task as a result of its intelligence.

Note that Narrow AI is the type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope.

Natural Language Understanding (NLU) is used to describe extracting information from unstructured text using algorithms.

Learn more about Artificial Intelligence from

brainly.com/question/25523571

#SPJ1

You might be interested in

I am trying to test out the software Classroom relay and I am just ask if there is any way kids can stop Classroom relay form se

Inessa05 [86]

Answer:

What is classroom relay?

Plz answer in ch-at

Explanation:

3 0

3 years ago

Read 2 more answers

An electric winch operates on 140 volts [V] and draws 6 amperes [A] of current. The winch has an efficiency of 64%. The winch

Elis [28]

Answer:

Explanation:

efficiency = (energy output / energy input)

energy input = ivt where i is current = 6 A, v is volts = 140 and t time = 11 seconds

energy input = 6 × 140 × 11 = 9240 J

energy output = 0.64 × 9240 = 5913.6 J

510 pound-force = 2268.593 N

at height h, potential energy = mgh = 5913.6 J

2268.593 × h = 5913.6

h = 5913.6 / ( 2268.593) = 2.61 m

3 0

3 years ago

Determine whether or not each of the following signals is periodic.

Sloan [31]

Answer:

a) periodic (N = 1)

b) not periodic

c) not periodic

d) periodic (N = 8)

e) periodic (N = 16)

Explanation:

For function to be a periodic: f(n) = f(n+N)

$a) x[n]=sin(\frac{8\pi}{2}n+1)\\\\sin(\frac{8\pi}{2}n+1)=sin(4\pi n+1)$

It is periodic with fundamental period N = 1

$b) x[n]=cos(\frac{n}{8} -\pi)\\\\\frac{1}{8} N=2\pi k$

N must be integer. So it is nor periodic

$c) x[n]=cos(\frac{\pi}{8} n^2)\\\\cos(\frac{\pi}{8} (n+N)^2)=cos(\frac{\pi}{8} (n^2+N^2+2nN)\\\\N^2 = 16 \:\:or\:\:2nN=16$

Since N is dependent to n. So it is not periodic.

$d) x[n]=cos(\frac{\pi }{2} n) cos(\frac{\pi }{4} n)\\\\x[n] = \frac{1}{2} cos(\frac{3\pi }{4} n) + \frac{1}{2} cos(\frac{\pi }{4} n)\\\\N_1=8\:\:and\:\:N_2=8\\$

So it is periodic with fundamental period N = 8.

$e) x[n]=2cos(\frac{\pi }{4} n)+sin(\frac{\pi }{8} n)-2cos(\frac{\pi }{2} n+\frac{\pi }{6} )\\\\N_1=8\:\:and\:\:N_2=16\:\:and\:\:N_3=4$

So it is periodic with N = 16.

3 0

3 years ago

What steel type and ASTM designation is preferred for W-shapes?

Ulleksa [173]

Answer:

The preferred steel type for W-shapes is structural steel and the its preferred ASTM designation is ASTM A992.

Explanation:

The ASTM A992 is a structural steel and it's the most available for w-shapes; besides its availabilty, its ductility improvements makes it the preferred choice; other common designations for this shapes are ASTM A572 Grade 50,0r ASTM A36, but this designations aren't as available as ASTM A992, and it has to be confirmed prior to their specification.

3 0

4 years ago

1. If a bolt is size 1/2" or larger, then its corresponding wrench size should be____ larger than the bolt size

Mrac [35]

Answer:

C. 1/4"
B. 3/16"

Explanation:

1. For hex bolts, lag bolts, and square bolts, the wrench size is 1/4" larger than the bolt size for 1/2" and 9/16" bolts. For 5/8" bolts and larger, the wrench size is <em>50% larger than the bolt size</em>.

2. For 7/16" bolts, the wrench size is 5/8", so is 3/16" larger than the bolt. This holds down to 1/4" bolts, where the wrench size may be 3/8" or 7/16".

3 0

4 years ago