All categories

2 years ago

How does data mining help interactive marketing for a business?

1 answer:

7 0

The way in which data mining helps interactive marketing for a business firm is: D. By predicting what each individual accessing a website is most likely interested in seeing.

<h3>What is data mining?</h3>

Data mining can be defined as a strategic process that typically involves extracting and analyzing data that are contained in large databases, especially through the use of specialized software to generate new information.

In this context, a business firm can adopt the use of data mining to enhance and facilitate interactive marketing by creating a process that can effectively and efficiently predict what each individual (customer) accessing their website is most likely interested in seeing.

Read more on a website here: brainly.com/question/26324021

You might be interested in

A well-designed product will increase?

Colt1911 [192]

Answer:

true

Explanation:

A well designed product will increase in sells and in stock.

8 0

2 years ago

Chad is working on a design that uses the pressure of steam to control a valve in order to increase water pressure in showers. W

Natalija [7]

Answer:

C: Viscosity, the resistance to flow that fluids exhibit

Explanation:

Did it on Edge :)

8 0

3 years ago

Electric heater wires are installed in a solid wall having a thickness of 8 cm and k=2.5 W/m.°C. The right face is exposed to an

Svet_ta [14]

Answer:

2.46 * 10⁵ W/m³

Explanation:

See attached pictures for detailed explanation.

6 0

4 years ago

Read 2 more answers

HELP PLEASE

Sveta_85 [38]

..........23÷357=0.0644257703........

8 0

2 years ago

Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the

alex41 [277]

Answer:

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

Explanation:

A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:

$\dot Q_{ref} - \dot Q_{w} = 0$

$\dot Q_{ref} = \dot Q_{w}$

$\dot m_{ref}\cdot (h_{ref, in} - h_{ref,out}) = \dot m_{w}\cdot (h_{w, out} - h_{w,in})$

The mass flow rate of the cooling water is now cleared:

$\dot m_{w} = \dot m_{ref }\cdot \frac{h_{ref,in}-h_{ref,out}}{h_{w,out}-h_{w,in}}$

Given that $h_{ref,in} = 808.34\,\frac{kJ}{kg}$ , $h_{ref, out} = 88.82\,\frac{kJ}{kg}$ , $h_{w,out} = 104.83\,\frac{kJ}{kg}$ and $h_{w,in} = 62.98\,\frac{kJ}{kg}$ , the mass flow of the cooling water is:

$\dot m_{w} = \left(7.2\,\frac{kg}{min} \right)\cdot \left(\frac{808.34\,\frac{kJ}{kg}-88.82\,\frac{kJ}{kg} }{104.83\,\frac{kJ}{kg}-62.98\,\frac{kJ}{kg} } \right)$

$\dot m_{w} = 123.788\,\frac{kg}{min}$

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

4 0

3 years ago