Thinking Like an Engineer: An Active Learning Approach (3rd Edition)
Thinking Like an Engineer: An Active Learning Approach (3rd Edition)
3rd Edition
ISBN: 9780133593211
Author: Elizabeth A. Stephan, David R. Bowman, William J. Park, Benjamin L. Sill, Matthew W. Ohland
Publisher: PEARSON
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Chapter 20, Problem 1ICA

Answer the question with each of the following code segments. If an error will occur, write ERROR and explain what caused the error.

a.    What is stored in Count by the following code?

Chapter 20, Problem 1ICA, Answer the question with each of the following code segments. If an error will occur, write ERROR , example  1

b.    What is displayed on the screen by the following code?

Chapter 20, Problem 1ICA, Answer the question with each of the following code segments. If an error will occur, write ERROR , example  2

c.    What is displayed on the screen by the following code?

Chapter 20, Problem 1ICA, Answer the question with each of the following code segments. If an error will occur, write ERROR , example  3

d.    What is stored in P and C after the following code executes?

Chapter 20, Problem 1ICA, Answer the question with each of the following code segments. If an error will occur, write ERROR , example  4

a.

Expert Solution
Check Mark
To determine

What variable is stored in Count.

Answer to Problem 1ICA

The variable stored in Count is 8.

Explanation of Solution

Given data:

Count =0;

for N=10:-0.2:8.5;

Count=Count+1;

end

Program execution:

Steps to execute the MATLAB program:

  1. 1. Open the command window and paste the given program.
  2. 2. Press enter.
  3. 3. Output is displayed in command window.

Enter Count in the command window and execute the program.

>> Count

Count =

     8

Conclusion:

Thus, the variable stored in Count is 8.

b.

Expert Solution
Check Mark
To determine

What is displayed on the screen after executing the given program.

Answer to Problem 1ICA

The output displayed on the screen:

6

12

20

30

42

56

Explanation of Solution

Given data:

for m=3:8

fprintf('%0.0f\n',m^2-m)

end

Program execution:

Steps to execute the MATLAB program:

  1. 1. Open the command window and paste the given program.
  2. 2. Press enter.
  3. 3. Output is displayed in command window.

MATLAB Output after execution of program is displayed as follows.

6

12

20

30

42

56

Conclusion:

Thus, the Output displayed on the screen:

6

12

20

30

42

56

c.

Expert Solution
Check Mark
To determine

What is displayed on the screen after executing the given program.

Answer to Problem 1ICA

The output displayed in the screen is 56.

Explanation of Solution

Given data:

for m=3:8;

M2=m^2-m;

end

fprintf('%0.0f\n',M2)

Program execution:

Steps to execute the MATLAB program:

  1. 1. Open the command window and paste the given program.
  2. 2. Press enter.
  3. 3. Output is displayed in command window.

MATLAB Output after execution of program is displayed as follows.

56

Conclusion:

Thus, the output displayed in the screen is 56.

d.

Expert Solution
Check Mark
To determine

What variables are stored in p and C.

Answer to Problem 1ICA

The variables stored in in p and C are p=945 and C=4.

Explanation of Solution

Given data:

S=3;

I=2;

P=1;

C=0;

for K=S:I:S^I

P=P*K;

C=C+1;

end

Program execution:

Steps to execute the MATLAB program:

  1. 1. Open the command window and paste the given program.
  2. 2. Press enter.
  3. 3. Output is displayed in command window.

Enter P and C in the command window and execute the program.

>> P

P =

   945

>> C

C =

     4

Conclusion:

Thus, the variables stored in in p and C are p=945 and C=4.

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Chapter 20 Solutions

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