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statuscvo [17]
3 years ago
13

1.

Computers and Technology
1 answer:
ludmilkaskok [199]3 years ago
8 0

Answer:

D. Are concerned with environmental issues is the answer

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This form of analysis is an extension of what-if analysis and is the study of the impact on other variables when one variable is
Elenna [48]

Answer: (C) Sensitivity analysis

Explanation:

 The sensitivity analysis is also known as simulation analysis or "What-if" analysis as, it is basically used for the outcome prediction of the decision making in various range of the given variable.

It is used by making a given arrangement of factors, an investigator can decide that how changing in a single variable influence the final result.

The sensitivity analysis is the process for investigation of how the vulnerability in the yield of a scientific model or framework can be isolated in the system.  

Theretofore, Option (C) is correct as all the other options does not involve in the study of variable and also others are not the extension of what-if analysis.

3 0
3 years ago
Write the method makeNames that creates and returns a String array of new names based on the method’s two parameter arrays, arra
Gwar [14]

Answer:

  1. import java.util.Arrays;
  2. public class Main {
  3.    public static void main(String[] args) {
  4.        String [] first = {"David", "Mike", "Katie", "Lucy"};
  5.        String [] middle = {"A", "B", "C", "D"};
  6.        String [] names = makeNames(first, middle);
  7.        System.out.println(Arrays.toString(names));
  8.    }
  9.    public static String [] makeNames(String [] array1, String [] array2){
  10.           if(array1.length == 0){
  11.               return array1;
  12.           }
  13.           if(array2.length == 0){
  14.               return array2;
  15.           }
  16.           String [] newNames = new String[array1.length];
  17.           for(int i=0; i < array1.length; i++){
  18.               newNames[i] = array1[i] + " " + array2[i];
  19.           }
  20.           return newNames;
  21.    }
  22. }

Explanation:

The solution code is written in Java.

Firstly, create the makeNames method by following the method signature as required by the question (Line 12). Check if any one the input string array is with size 0, return the another string array (Line 14 - 20). Else, create a string array, newNames (Line 22). Use a for loop to repeatedly concatenate the string from array1 with a single space " " and followed with the string from array2 and set it as item of the newNames array (Line 24-26). Lastly, return the newNames array (Line 28).

In the main program create two string array, first and middle, and pass the two arrays to the makeNames methods as arguments (Line 5-6). The returned array is assigned to names array (Line 7). Display the names array to terminal (Line 9) and we shall get the sample output: [David A, Mike B, Katie C, Lucy D]

5 0
3 years ago
Find the root using bisection method with initials 1 and 2 for function 0.005(e^(2x))cos(x) in matlab and error 1e-10?
frutty [35]

Answer:

The root is:

c=1.5708

Explanation:

Use this script in Matlab:

-------------------------------------------------------------------------------------

function  [c, err, yc] = bisect (f, a, b, delta)

% f the function introduce as n anonymous function

%       - a y b are the initial and the final value respectively

%       - delta is the tolerance or error.

%           - c is the root

%       - yc = f(c)

%        - err is the stimated error for  c

ya = feval(f, a);

yb = feval(f, b);

if  ya*yb > 0, return, end

max1 = 1 + round((log(b-a) - log(delta)) / log(2));

for  k = 1:max1

c = (a + b) / 2;

yc = feval(f, c);

if  yc == 0

 a = c;

 b = c;

elseif  yb*yc > 0

 b = c;

 yb = yc;

else

 a = c;

 ya = yc;

end

if  b-a < delta, break, end

end

c = (a + b) / 2;

err = abs(b - a);

yc = feval(f, c);

-------------------------------------------------------------------------------------

Enter the function in matlab like this:

f= @(x) 0.005*(exp(2*x)*cos(x))

You should get this result:

f =

 function_handle with value:

   @(x)0.005*(exp(2*x)*cos(x))

Now run the code like this:

[c, err, yc] = bisect (f, 1, 2, 1e-10)

You should get this result:

c =

   1.5708

err =

  5.8208e-11

yc =

 -3.0708e-12

In addition, you can use the plot function to verify your results:

fplot(f,[1,2])

grid on

5 0
3 years ago
To format an individual sparkline, select the sparkline you want to format, and then click the Ungroup button in the Group group
vichka [17]

Answer:

The answer is "Option a"

Explanation:

A Sparkline is a small graph, which is available inside a sheet, that displays information visually. It displays the information is in the form of patterns on several values. It also uses underline to display the low and high value, use with sparklines, and certain alternatives were wrong, which can be defined as follows:

  • In option b, It is used to define the overall view of data, that's why it is wrong.
  • Option c and Option d both are wrong because, create command is used to create data, or view is used to show that data, they both don't use sparkline option.  
3 0
4 years ago
Which key do programmers use to end running programs?
Aleksandr [31]
<span>Pause/Break   i would say</span>
3 0
3 years ago
Read 2 more answers
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