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dmitriy555 [2]
3 years ago
6

Activities provided for the satisfaction of others and consumed at the time of purchase are…

Computers and Technology
1 answer:
Natasha2012 [34]3 years ago
6 0
The answer is services
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A) What is the maximum value that can be represented as an unsigned n-bit binary integer?
My name is Ann [436]

Answer:

The maximum value that are represented as unsigned n -bit binary integer is 2^n-1. The unsigned binary integer refers to the fixed point of the system that does not contain any fractional digits.

The unsigned binary integer contain module system with the power 2. The number of student table in the class is the best example of the unsigned integer. The numbers can be represented by using the binary notation and bits in the computer system.

5 0
3 years ago
____________ refers to the computer-to-computer transmission of business data in a structured format.
motikmotik

Answer:

Electronic Data Interchange (EDI)

Explanation:

Electronic Data Interchange

It is the automated interchange of commercial information using a structured format. A process that allows a company to send information to another company electronically rather than on paper. Commercial entities that conduct business electronically are called business partners.

Numerous commercial documents can be exchanged using electronic data interchange, but the two most common are purchase orders and invoices. At a bottom, EDI interchanges the preparation and handling of mail associated with traditional commercial communication. However, the true power of EDI is that it standardizes the information communicated in commercial documents, which makes a "paperless" exchange possible.

6 0
3 years ago
Describe the user interface in other high-technology devices commonly found in the home or office, such as a smartphone, HD tele
Vadim26 [7]

Answer:

Explanation:

Different technologies use different user interface designs in order to make the user experience as easy and intuitive as possible. This varies drastically from one device to another because of the capabilities and size of each device. If we take a fitness/smart watch into consideration, this device does not use pop up menus or side scrolling menus but instead uses large full screen menus where each option nearly fills the entire screen. That is done because the smart watch screens are very small and making everything full screen makes reading and swiping through options that much easier for the user. If the user interface were the same as in a television or smartphone it would be impossible to navigate through the different options on such a tiny screen.

7 0
3 years ago
Which of the following is the SECOND step of the intuitive​ lowest-cost method​?
coldgirl [10]

The very first step of the lowest cost method is to find the cell with the lowest cost in the entire matrix representing the cost of transportation along with supply and demand.

C. Find the cell with the lowest cost from the remaining​ (not crossed​ out) cells.

<u>Explanation:</u>

The second step in the lowest cost method is to allocate as many units as possible to that cell (having the lowest cost) without exceeding the supply or demand.

Then cross out the row or column (or both) that is exhausted by the assignment made. These two steps are further repeated until all the assignments are made and the total cost of transportation is calculated at the end.  

3 0
3 years ago
Show the array that results from the following sequence of key insertions using a hashing system under the given conditions: 5,
sergejj [24]

Answer:

a) Linear probing is one of the hashing technique in which we insert values into the hash table indexes based on hash value.

Hash value of key can be calculated as :

H(key) = key % size ;

Here H(key) is the index where the value of key is stored in hash table.

----------

Given,

Keys to be inserted are : 5 , 205, 406,5205, 8205 ,307

and size of the array : 100.

First key to be inserted is : 5

So, H(5) = 5%100 = 5,

So, key 5 is inserted at 5th index of hash table.

-----

Next key to inserted is : 205

So, H(205) = 205%100 = 5 (here collision happens)

Recompute hash value as follows:

H(key) =(key+i) % size here i= 1,2,3...

So, H(205) =( 205+1)%100 = 206%100 = 6

So, key 205 is inserted in the 6th index of the hash table.

----------

Next Key to be inserted : 406

H(406) = 406%100 = 6(collision occurs)

H(406) =(406+1) %100 = 407%100 = 7

So, the value 406 is inserted in 7the index of the hash table.

-----------------

Next key : 5205

H(5205) = 5205%100 = 5(collision)

So, H(5205) = (5205+1)%100 = 6( again collision)

So, H(5205) = 5205+2)%100 = 7(again collision)

So, H(5205) = (5205+3)%100 = 8 ( no collision)

So, value 5205 is inserted at 8th index of the hash table.

-------------

Similarly 8205 is inserted at 9th index of the hash table because , we have collisions from 5th to 8th indexes.

-------------

Next key value is : 307

H(307) = 307%100 = 7(collision)

So, (307+3)%100 = 310%100 = 10(no collision)  

So, 307 is inserted at 10th index of the hash table.

So, hash table will look like this:

Key       index

5         5

205         6

406         7

5205 8

8205 9

307         10

b) Quadratic probing:

Quadratic probing is also similar to linear probing but the difference is in collision resolution. In linear probing in case of collision we use : H(key) = (key+i)%size but here we use H(key) =( key+i^2)%size.

Applying Quadratic probing on above keys:.

First key to be inserted : 5.

5 will go to index 5 of the hash table.

-------

Next key = 205 .

H(205) = 205%100 = 5(collision)

So. H(key)= (205+1^2)%100 = 6(no collision)

So, 205 is inserted into 6th index of the hash table.

--------

Next key to be inserted 406:

So, 406 %100 = 6 (collision)

(406+1^2)%100 = 7(no collision)

So, 406 is moved to 7th index of the hash table.

----------

Next key is : 5205

So, 5205%100 = 5 (collision)

So, (5205+1^2)%100 = 6 ( again collision)

So, (5205+2^2)%100 = 9 ( no collision)

So, 5205 inserted into 9th index of hash table.

-----------

Next key is 8205:

Here collision happens at 5the , 6the , 9th indexes,

So H(8205) = (8205+4^2)%100 = 8221%100 = 21

So, value 8205 is inserted in 21st index of hash table.

--------

Next value is 307.

Here there is collision at 7the index.

So, H(307) = (307+1^2)%100 = 308%100= 8.

So, 307 is inserted at 8the index of the hash table.

Key           Index

5                  5

205                  6

406                  7

5205                9

8205               21

307                   8

3 0
3 years ago
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