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telo118 [61]
2 years ago
9

A company wants to use online methods to target more customers. It decides to conduct a market research by collecting the data o

f a few
customers with their consent. They want to track data of the sites that their customers frequently visit. Which software can help the company?
Computers and Technology
1 answer:
Ronch [10]2 years ago
8 0

Answer:

They could use a CRM

A CRM tool lets you store customer and prospect contact information, identify sales opportunities, record service issues, and manage marketing campaigns, all in one central location — and make information about every customer interaction available to anyone at your company who might need it.

They Could Use PDC

The term “primary data” refers to data you collect yourself, rather than data you gather after another party initially recorded it. Primary data is information obtained directly from the source. You will be the first party to use this exact set of data.

They Could Use Quantitative vs. Qualitative Data

Quantitative data comes in the form of numbers, quantities and values. It describes things in concrete and easily measurable terms. Examples include the number of customers who bought a given product, the rating a customer gave a product out of five stars and the amount of time a visitor spent on your website.

Why Should I Track Users Data:

To ensure it's going smoothly, it's important to keep track of data that directly correlates to the potential success of your business. Key performance and financial metrics can help you address any problems with your business before they occur

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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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Answer: layout, section, number, more options

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