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Ainat [17]
4 years ago
6

What criteria are typically used to determine which new information systems projects to develop? What arguments might Bob Petros

ki make for developing the proposed customer loyalty system?
Chemistry
1 answer:
GuDViN [60]4 years ago
7 0

Answer: The criteria that are mostly used to determine which new information systems project to develop are as follows:

1. Available Resources

2. Technical Risk

3. Project Size

4. Potential Benefits

5. Customer Retention

The arguments Bob petroski might make for developing the proposed customers loyalty system, is that, it will be the best way for a firm to retain it's customer, and increase the number of times a customer patronize the firm. It is also a way to appreciate the customers that are loyal to the firm. It can also benefit the employees, which patronize the company's goods and services always.

Explanation: a firm that has worked more to have a good customer loyalty system, are likely to not to consider much on customers Retention, because the loyalty bonus also tend to retain customers to the firm.

You might be interested in
How many grams of oxygen are needed to react with 4.6 grams of titanium(IV) chloride
velikii [3]
The answer is: 0.78g
6 0
3 years ago
The surface area of an object to be gold plated is 49.8 cm2 and the density of gold is 19.3 g/cm-3. A current of 3.15. A is appl
garik1379 [7]

Answer: Time required to deposit an even layer of gold with given thickness is 5.3 \times 10^{2} sec.

Explanation:

The given data is as follows.

     Surface area = 49.8 cm^{2},

     Density of gold = 19.3 g/cm^{3},

     Current = 3.15 A,       thickness of gold layer = 1.2 \times 10^{-3} cm

It is known that relation between volume, area and thickness is as follows.

           V = Surface area × Thickness

               = 49.8 \times 1.2 \times 10^{-3} cm

               = 0.05988 cm^{3}

Therefore, we will calculate the time required to deposit an even layer of gold with given thickness is calculated as follows.

  0.05988 \times cm^{3} \times \frac{19.3 g Au}{1 cm^{3}} \times \frac{1 mol Au}{197 g Au} \times \frac{3 mol e^{-}}{1 mol Au} \times \frac{96485}{1 mol e^{-}} \times \frac{1 As}{1 C} \times \frac{1}{3.20 A}

        = 5.3 \times 10^{2} sec

Thus, we can conclude that time required to deposit an even layer of gold with given thickness is 5.3 \times 10^{2} sec.

4 0
3 years ago
Give the electron configuration of vanadium (V), atomic number 23
saveliy_v [14]

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d³ or [Ar] 4s²3d³

<h3>Further explanation</h3>

Given

Vanadium, atomic number 23

Required

The electron configuration

Solution

In an atom, there are levels of energy in the shell and subshell  

This energy level is expressed in the form of electron configurations.  

Charging electrons in the subshell uses the following sequence:  

<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.  </em>

The electron configuration is based on the atomic number which indicates the number of electrons of the atom

Vanadium has the atomic number 23, so the number of electrons = 23

Electron configuration:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d³

or we can write it using the noble gas notation

[Ar] 4s²3d³

4 0
3 years ago
The fundamental unit used to measure mass is??
Soloha48 [4]

The fundamental unit used to measure mass is Kg (kilograms).

6 0
3 years ago
Read 2 more answers
Suppose you need to make the following two precipitates (by two different precipitation reactions): MgCO3 and Ca3(PO4)2. For eac
Setler [38]

Answer:

For the first reaction the reagents are: MgCl2 and Na2CO3

For the second reaction the reagents are: Na2HPO4 and CaCl2

Explanation:

Precipitation reactions lie in the production of a compound that is not soluble, which is called a precipitate, this precipitate is produced when two different solutions are combined, each of which will contribute an ion for the formation of the precipitate. In the first reaction you have:

MgCl2 + Na2CO3 = MgCO3 + 2 NaCl

Type of reaction: double displacement

The second reaction is as follows:

4Na2HPO4 + 3CaCl2 → Ca3 (PO4) 2 + 2NaH2PO4 + 6NaCl

It is the reaction of sodium hydrochlorophosphate and calcium chloride

3 0
3 years ago
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