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vivado [14]
4 years ago
6

To add and remove chart elements, you can use the add chart element button in the charts layout group on the ____ tab.

Computers and Technology
1 answer:
avanturin [10]4 years ago
3 0
This question has multiple choices;
 
<span><span>a. </span>CHART TOOLS LAYOUT</span>
<span><span>b. </span>CHART TOOLS DESIGN</span> <span><span>
c.  </span>INSERT</span>
<span><span>d.  </span>HOME</span>

The answer is A) CHART TOOLS LAYOUT

Once you have created a chart, you can always adjust its look. Instead of changing or adding the chart elements manually, you can apply a predefined layout to your chart. To do so, you will click anywhere in the chart or the chart element and this will display the charts tool. You will then be able to use the chart layout that you want to use on the Design tab, in the Charts layouts group menu.     




You might be interested in
Write a C function that takes an STL vector of int values and determines if all the numbers are different from each other (that
eduard

Answer:

Here is the function:

#include <iostream>  //to use input output functions

#include <vector>  // to use vector (sequence containers)

#include <algorithm>   //use for sequence operations

#include<iterator>  //to move through the elements of sequence or vector

using namespace std;  //to identify objects cin cout

 void DistinctNumbers(std::vector<int> values) { /*function that takes  an STL vector of int values and determines if all the numbers are different from each other */

     sort(values.begin(), values.end());  //sorts the vector elements from start to end

     bool isDistinct = std::adjacent_find(values.begin(), values.end()) == values.end();  //checks for occurrence of two consecutive elements in vector

if(isDistinct==true)  // if all numbers are different from each other

{cout<<"Numbers are distinct";}

else  //if numbers are duplicate or same

{cout<<"Numbers are not distinct";}   }  

int main(){      //start of main function

std::vector<int> v = {1,2,3,4,5};  // vector of int values

DistinctNumbers(v); }  //function call to passing the vector v to check if its elements are distinct

Explanation:

The program that takes an STL vector of int values and the function DistinctNumbers determines if all the numbers are different from each other. It first sorts the contents of the vector in ascending order using sort() method. Then it used the method adjacent_find() to searches the range means from the start to the end of the vector elements, for first occurrence of two consecutive elements that match, and returns an iterator to the first of these two elements, or last if no such pair is found. The result is assigned to a bool type variable isDistinct. It then checks if all the numbers are different and no two adjacent numbers are same. If all the numbers are distinct then this bool variable evaluates to true otherwise false. If the value of isDistinct is true then the message :Numbers are distinct is displayed on screen otherwise message: Numbers are not distinct is displayed in output screen.

7 0
3 years ago
Among the aforementioned parts, the FLASH DRIVE is the most common. Why do you think a flash drive will be essential for you as
zhuklara [117]

Answer:

They can be used to store homework, presentations, research, papers, essays and etc. They can be used to hand out homework assignments, course information or notes

4 0
3 years ago
Assume a 15 cm diameter wafer has a cost of 12, contains 84 dies, and has0.020 defects /cm2Assume a 20 cm diameter wafer has a c
Vitek1552 [10]

Answer:

1. yield_1=0.959 and yield_2=0.909

2. Cost_1=0.148 and Cost_2=0.165

3. New area per die=1.912 cm^2 and yield_1=0.957

   New area per die=2.85 cm^2  and yield_2=0.905

4. defects=0.042 per cm^2 and defects=0.026 per cm^2

Explanation:

1. Find the yield for both wafers.

yield= 1/(1+(defects per unit area*dies per unit area/2))^2

Wafer 1:

Radius=Diameter/2=15/2=7.5 cm

Total Area=pi*r^2=pi(7.5)^2=176.71 cm^2

Area per die= 176.71/84=2.1 cm^2

yield_1= 1/(1+(0.020*2.1/2))^2

yield_1=1/1.04244=0.959

Wafer 2:

Radius=Diameter/2=20/2=10 cm

Total Area=pi*r^2=pi(10)^2=314.159 cm^2

Area per die= 314.159/100=3.14 cm^2

yield_2= 1/(1+(0.031*3.14/2))^2

yield_2=1/1.0997=0.909

2. Find the cost per die for both wafers.

Cost per die= cost per wafer/Dies per wafer*yield

Wafer 1:

Cost_1=12/84*0.959=0.148

Wafer 2:

Cost_2=15/100*0.909=0.165

3. If the number of dies per wafer is increased by 10% and the defects per area unit increases by 15%, find the die area and yield.

Wafer 1:

There is a 10% increase in the number of dies

10% of 84 =8.4

New number of dies=84.4+8=92.4

There is a 15% increase in the defects per cm^2

15% of 0.020=0.003

New defects per area= 0.020 + 0.003=0.023 defects per cm^2

New area per die= 176.71/92.4=1.912 cm^2

yield_1= 1/(1+(0.023*1.912/2))^2=0.957

Wafer 2:

There is a 10% increase in the number of dies

10% of 100=10

New number of dies=100+10=110

There is a 15% increase in the defects per cm^2

15% of 0.031=0.0046

New defects per area= 0.031 + 0.00465=0.0356 defects per cm^2

New area per die= 314.159/110=2.85 cm^2

yield_2= 1/(1+(0.0356*2.85/2))^2=0.905

4. Assume a fabrication process improves the yield from 0.92 to 0.95. Find the defects per area unit for each version of the technology given a die area of

Assuming a die area of 2cm^2

We have to find the defects per unit area for a yield of 0.92 and 0.95

Rearranging the yield equation,

yield= 1/(1+(defects*die area/2))^2

defects=2*(1/sqrt(yield) - 1)/die area

For 0.92 technology

defects=2*(1/sqrt(0.92) - 1)/2

defects=0.042 per cm^2

For 0.95 technology

defects=2*(1/sqrt(0.95) - 1)/2

defects=0.026 per cm^2

6 0
3 years ago
You have created shared folders for all your companies departments and assigned the appropriate permissions. everyone can access
larisa86 [58]
I would suggest the use of a disk quota.

Using a disk quota would limit the available space that users could have used to store data. The New Technology File system (NTFS) has several features like disk quotas to see which users are uploading the most space.  The windows interface Open Computer Management is also another option that limits the number of users of a shared folder.



8 0
4 years ago
Write a C++ program that creates a map containing course numbers and the room numbers of the rooms where the courses meet. The d
swat32

Answer:

Program approach:-

  • Using the necessary header file.
  • Using the standard namespace I/O.
  • Define the integer main function.
  • Mapping course numbers and room numbers.

Explanation:

//header file

#include<iostream>

#include<map>

//using namespace

using namespace std;

//main function

int main(){

               //creating 3 required maps

               map <string,int> rooms;

               map <string,string> instructors;

               map <string,string> times;

               //mapping course numbers and room numbers

               rooms.insert(pair<string,int>("CS101",3004));

               rooms.insert(pair<string,int>("CS102",4501));

               //mapping course numbers and instructor names

               instructors.insert(pair<string,string>("CS101","Haynes"));

               instructors.insert(pair<string,string>("CS102","Alvarado"));

               //mapping course numbers and meeting times

               times.insert(pair<string,string>("CS101","8:00am"));

               times.insert(pair<string,string>("CS102","9:00am"));

               

               char choice='y';

               //looping until user wishes to quit

               while(choice=='y' || choice=='Y'){

                               cout<<"Enter a course number: ";

                               string course;

                               cin>>course;//getting course number

                               //searching in maps for the required course number will return

                               //an iterator

                               map<string, int>::iterator it1=rooms.find(course);

                               map<string, string>::iterator it2=instructors.find(course);

                               map<string, string>::iterator it3=times.find(course);

               

                               if(it1!=rooms.end()){

                                               //found

                                               cout<<"Room: "<<it1->second<<endl;

                               }else{

                                               //not found

                                               cout<<"Not found"<<endl;

                               }

               

                               if(it2!=instructors.end()){

                                               cout<<"Instructor: "<<it2->second<<endl;

                               }

               

                               if(it3!=times.end()){

                                               cout<<"Meeting Time: "<<it3->second<<endl;

                               }

                               //prompting again

                               cout<<"\nDo you want to search again? (y/n): ";

                               cin>>choice;

               }

7 0
3 years ago
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