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garik1379 [7]
3 years ago
9

Which of the following represent emerging classes of application software? Check all of the boxes that apply.

Computers and Technology
3 answers:
soldi70 [24.7K]3 years ago
7 0
Word processing and databases are the most common type of application software. In this case, application software, often called productivity programs, are programs designed to specifically attain specific tasks such as creating documents, databases, sending email, designing graphics and even playing games. A mobile app is also an emerging type of application software designed to run on portable devices like mobiles and PDAs. To some extent, email software is an application software since it can also be distinguished on the basis of usage.






Snezhnost [94]3 years ago
5 0

Answer:

medical robotics, machine translation, and mobile phone apps

Explanation:

just took it

yea i dont get it2 years ago
0 0

medical robotics
machine translation
mobile phone apps

You might be interested in
A pointing device uses light to determine hand motion.
kirill [66]
The answer to this is Optical mouse. 

The reason the answer is optical mouse is because the optical mouse is a computer mouse which uses a light source, typically a light-emitting diode, and a light detector, such as an array of photodiodes, to detect movement relative to a surface. 

Hope this helped :)
have a great day 
3 0
3 years ago
The data below represents the yearly earnings (in $1000s of dollars) that high school and college (BS) graduates earn at a small
Taya2010 [7]

Answer:

library(moments)

library(readxl)

Exam1Q1 <- read_excel("Exam1Q1.xlsx")

View(Exam1Q1)

## Warning in system2("/usr/bin/otool", c("-L", shQuote(DSO)), stdout = TRUE):

## running command ''/usr/bin/otool' -L '/Library/Frameworks/R.framework/

## Resources/modules/R_de.so'' had status 1

plot(density(Exam1Q1$Highschool))

gostino.test(Exam1Q1$Highschool)

##  

##  D'Agostino skewness test

##  

## data:  Exam1Q1$Highschool

## skew = -0.24041, z = -0.69269, p-value = 0.4885

## alternative hypothesis: data have a skewness

agostino.test(Exam1Q1$BS)

##  

##  D'Agostino skewness test

##  

## data:  Exam1Q1$BS

## skew = 0.53467, z = 1.49330, p-value = 0.1354

## alternative hypothesis: data have a skewness

anscombe.test(Exam1Q1$Highschool)

##  

##  Anscombe-Glynn kurtosis test

##  

## data:  Exam1Q1$Highschool

## kurt = 2.56950, z = -0.29677, p-value = 0.7666

## alternative hypothesis: kurtosis is not equal to 3

anscombe.test(Exam1Q1$BS)

##  

##  Anscombe-Glynn kurtosis test

##  

## data:  Exam1Q1$BS

## kurt = 3.6449, z = 1.2599, p-value = 0.2077

## alternative hypothesis: kurtosis is not equal to 3

t.test(Exam1Q1$Highschool,Exam1Q1$BS, var.equal = T)

##  

##  Two Sample t-test

##  

## data:  Exam1Q1$Highschool and Exam1Q1$BS

## t = 0.16929, df = 76, p-value = 0.866

## alternative hypothesis: true difference in means is not equal to 0

## 95 percent confidence interval:

##  -1.131669  1.341926

## sample estimates:

## mean of x mean of y  

##  39.51282  39.40769

Explanation:

Summary:

The data set include two dependent variables which states High School annual earnings and BS annual earnings in the same small firm. Firstly, I used density plot function to view the distribution of two sets of variables. Secondly, I tested the skewness and kurtosis.Skewness of High School is -0.24041 which is left tail. Skewness of BS is 0.53476 which is right tail.Kurtosis of High School is 2.5695 which lower than 3 is platykurtic. Kurtosis of BS is 3.6449 which larger than 3 is leptokurtic. Thirdly, the sample population is 39 for each and has unknow standard deviation, so that I’m going to use t-test to perform the comparing analysis. The result shows p-value which equal to 0.866 is larger than significant value(alpha = 0.05). To the concludsion, there is no difference between yearly earnings of High School and of BS at a small firm.

3 0
3 years ago
1. In the Entity-Relationship Model, relationships can have attributes. T/F
Anestetic [448]

Answer:

 1. True  2. True 3. Conceptual 4. 3 tables.

Explanation:

1. In the ER Model relationships can have attributes associated to them. For example in an organisation there is a one-to-one relationship between Employee and Department entities i.e. an Employee manages a Department and each Department is managed by some Employee. If you want to store the Start_Date from which the employee began to manage the department then you can give the Start_Date attribute to the relationship manages.

2. Attributes are the smallest division of data in the ER diagram. Attributes are used to describe an entity.For example an Employee entity can have the attributes Name, DateofBirth, Address, Salary. Attribute is represented by an oval with a name inside in ER diagram.

3. Conceptual  

Conceptual level is the highest level ER model as it contains less but defines the overall scope of the model.  It defines the data entities that are commonly used by the organization. A model on conceptual level of abstraction captures and represents business scope of the problem. It depicts the most important entities and the relationships between these entities.  A conceptual ER model can be used as the base for one or more logical data models. Conceptual level describes or defines the structure of the whole database. It hides details about the physical storage and concentrates on describing entities, relationships, etc.  It is part of the conceptual stage of database design and it provides a high-level data description.

4. 3 tables can be used to represent this relationship. One is a Student table. Second is the Class table. Now if we see the relationship. A student can take many classes. A class can be taken by many students. This is a many-to-many relationship between these two entities. It is not directly possible to implement this type of relationship in a database. To implement this relationship it has to be broken into 2 one to many relationships.  In order to do this, a new entity has to be created. So create one more table to establish a relationship between Student and Class entities. Primary keys of Student and Class tables are added into new table to create a composite primary key. This table can be named as Enrollment. Hence 3 tables are used.

5 0
3 years ago
What was the biggest problem with the earliest version of the internet in the late 1960’s?
posledela

Answer:

Explanation:

1.security issue

2. Computers were too big

3. Not very reliable

4. Networks couldn't talk to each other

5. Only be used in universities, governments, and businesses

8 0
3 years ago
Using a microphone to record a sound on your computer is an example of:
maks197457 [2]

Answer:

digital to analog conversion output device

5 0
1 year ago
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