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garri49 [273]
3 years ago
9

In the value chain concept, upgrading IT is considered what kind of activity?

Computers and Technology
1 answer:
ANTONII [103]3 years ago
6 0

Answer:

Upgrading IT is considered as B. Support activity

Explanation:

It is considered as support activities because they help the primary activities move smoothly and consistently.

For example, efficient communication between a particular firm and its subsidiaries will be extremely difficult without the necessary IT infrastructure, which will slow down the primary activities that are to be carried out.

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MA_775_DIABLO [31]

Answer:

problems or struggles

Explanation:

i believe this because with an effective team most problems aren't there and you can understand and work out problems or struggles

4 0
3 years ago
Why is a memory hierarchy of different memory types used instead of only one kind of memory?
Kipish [7]

Answer: Memory hierarchy is the hierarchy that is created on the basis of the response time of different memories. The performance obtained by the memory helps in creating a computer storage space in distinguished form. The factors considered for the creating of the hierarchy structure are usually response time, storage capacity, complexity etc.

Usage of different kind of memories take place due to different kind of requirements from the system which cannot be fulfilled using one memory device.The requirement is based on saving time, decreasing complexity , improving performance etc.Example of requirements can be like some functions and files do not require much space , some might require quick accessing,etc.

Thus hierarchy of any particular system is in the form of fast to slow order from registers,cache memory, Random access memory(RAM) and secondary memory.

5 0
3 years ago
​What file system below does not support encryption, file based compression, and disk quotas, but does support extremely large v
Elena L [17]

Answer:

D.  ReFS

Explanation:

File system is simply a management system for files that controls how and where data are stored, where they can be located and how data can be accessed. It deals with data storage and retrieval.

Examples of file system are NTFS, FAT(e.g FAT 16 and FAT 32), ReFS.

ReFS, which stands for Resilient File System, is designed primarily to enhance scalability by allowing for the storage of extremely large amounts of data and efficiently manage the availability of the data. It is called "resilient" because it ensures the integrity of data by offering resilience to data corruption. It does not support transaction, encryption, file based compression, page file and disk quotas, to mention a few.

6 0
3 years ago
Rubbing two sticks together to make a fire is an example what
Jet001 [13]
Rubbing two sticks together will cause friction
7 0
3 years ago
Read 2 more answers
A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co
tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k  ...(1)

probability of 2 hops = k(1-k)  ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

                                                       = (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

                                                       = k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops                = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

     ∞                             n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

    n-1

   = 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet =  k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet =  k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

6 0
3 years ago
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