Essentials of MIS (13th Edition)
13th Edition
ISBN: 9780134802756
Author: Kenneth C. Laudon, Jane Laudon
Publisher: PEARSON
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Expert Solution & Answer
Chapter 9, Problem 2IQ
Explanation of Solution
Enterprise Resource Planning:
- It is nothing but a planning that joins various departments and its functions all over an organization into a single
information system . - This planning helps the employees to make their decisions on viewing the wide enterprise information about various business operations.
- The main motive is to provide common
database that supports various functions used by different business departments. - Example: Consider the departments named sales and accounting that depend on the same data for their own needs.
JDA manufacturing systems:
JDA stands for “James Donald Armstrong who is the founder of the company. JDA software provides supply chain management, planning for manufacturing and retailing, operation and collaboration management, etc...
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Students have asked these similar questions
I need help fixing the minor issue where the text isn't in the proper place, and to ensure that the frequency cutoff is at the right place.
My code:
% Define frequency range for the plot
f = logspace(1, 5, 500); % Frequency range from 10 Hz to 100 kHz
w = 2 * pi * f; % Angular frequency
% Parameters for the filters - let's adjust these to get more reasonable cutoffs
R = 1e3; % Resistance in ohms (1 kΩ)
C = 1e-6; % Capacitance in farads (1 μF)
% For bandpass, we need appropriate L value for desired cutoffs
L = 0.1; % Inductance in henries - adjusted for better bandpass response
% Calculate cutoff frequencies first to verify they're in desired range
f_cutoff_RC = 1 / (2 * pi * R * C);
f_resonance = 1 / (2 * pi * sqrt(L * C));
Q_factor = (1/R) * sqrt(L/C);
f_lower_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) + 1/(2*Q_factor));
f_upper_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) - 1/(2*Q_factor));
% Transfer functions
% Low-pass filter (RC)
H_low = 1 ./ (1 + 1i * w *…
My code is experincing minor issue where the text isn't in the proper place, and to ensure that the frequency cutoff is at the right place.
My code:
% Define frequency range for the plot
f = logspace(1, 5, 500); % Frequency range from 10 Hz to 100 kHz
w = 2 * pi * f; % Angular frequency
% Parameters for the filters - let's adjust these to get more reasonable cutoffs
R = 1e3; % Resistance in ohms (1 kΩ)
C = 1e-6; % Capacitance in farads (1 μF)
% For bandpass, we need appropriate L value for desired cutoffs
L = 0.1; % Inductance in henries - adjusted for better bandpass response
% Calculate cutoff frequencies first to verify they're in desired range
f_cutoff_RC = 1 / (2 * pi * R * C);
f_resonance = 1 / (2 * pi * sqrt(L * C));
Q_factor = (1/R) * sqrt(L/C);
f_lower_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) + 1/(2*Q_factor));
f_upper_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) - 1/(2*Q_factor));
% Transfer functions
% Low-pass filter (RC)
H_low = 1 ./ (1 + 1i * w *…
I would like to know the main features about the following three concepts:
1. Default forwarded
2. WINS Server
3. IP Security (IPSec).
Chapter 9 Solutions
Essentials of MIS (13th Edition)
Ch. 9.2 - Prob. 1CQ1Ch. 9.2 - Prob. 2CQ1Ch. 9.2 - Prob. 3CQ1Ch. 9.2 - Prob. 4CQ1Ch. 9.3 - Prob. 1CQ2Ch. 9.3 - Prob. 2CQ2Ch. 9.3 - Prob. 3CQ2Ch. 9.3 - Prob. 4CQ2Ch. 9 - Prob. 1IQCh. 9 - Prob. 2IQ
Ch. 9 - Prob. 3IQCh. 9 - Prob. 1RQCh. 9 - Prob. 2RQCh. 9 - Prob. 3RQCh. 9 - Prob. 4RQCh. 9 - Prob. 5DQCh. 9 - Prob. 6DQCh. 9 - Prob. 7DQCh. 9 - Prob. 8HMPCh. 9 - Prob. 9HMPCh. 9 - Prob. 11HMPCh. 9 - Prob. 12CTPCh. 9 - Prob. 13CSQCh. 9 - Prob. 14CSQCh. 9 - Prob. 15CSQCh. 9 - Prob. 16CSQCh. 9 - Prob. 17CSQCh. 9 - Prob. 18MLMCh. 9 - Prob. 19MLM
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