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crimeas [40]
3 years ago
5

A process in which computer software that uses statistical analysis, database technology, and artificial intelligence finds hidd

en patterns, trends, and connections in data so that business owners can make better marketing decisions and predictions about customers' behavior is known as ________.
A. Cusomter Relationship Management
B. Stimulation
C. Data mining
D. Data discovery
Computers and Technology
2 answers:
lyudmila [28]3 years ago
8 0

Answer:

The correct answer is C ) Data Mining.

Explanation:

Data mining is also defined as a process used to extract usable data from a larger set of any raw data. It implies analysing data patterns in large batches of data using one or more software. Data mining is also known as Knowledge Discovery in Data (KDD).

Cheers!

krek1111 [17]3 years ago
3 0
<h2>Answer:</h2>

C. Data mining

<h2>Explanation:</h2>

Data mining has to do with exploration and analysis of large data in order to figure out certain trends and relationships in the data. In data mining, powerful and sophisticated algorithms are used for the sorting, segmentation and analysis of these data. Companies and businesses use the results of these analyses to get knowledge-based information and in turn, make logical decisions and predictions about customer's behaviour.

Common data mining techniques are;

i. anomaly detection

ii. clustering analysis

iii. regression analysis

iv. classification analysis

v. association rule learning.

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Who benefits the most from billing by the second for cloud resources, such as virtual machines?.
scoundrel [369]

Clients that run numerous virtual machines.

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7 0
1 year ago
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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} }

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3 years ago
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