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Npr02_16

% file npr02_16.m % Data for problem P2-16minvec3 DV = [A|Ac; A;    B;    C; (A&B)|(A&C)|(B&C); A&B&C; A&C; (A&B)-2*(B&C)];DP = [ 1  0.221 0.209 0.112   0.197           0.045  0.062      0];TV = [A|B|C; (A&Bc&Cc)|(Ac&B&Cc)|(Ac&Bc&C)];disp('Call for mincalc')

Npr02_17

% file npr02_17.m % Data for problem P2-17% A = alignment;  B = brake work;  C = headlight minvec3DV = [A|Ac; A&B&C; (A&B)|(A&C)|(B&C); B&C;    A  ];DP = [ 1    0.100      0.325          0.125 0.550];TV = [A&Bc&Cc; Ac&(~(B&C))];disp('Call for mincalc')

Npr02_18

% file npr02_18.m % Date for problem P2-18minvec3 DV = [A|Ac; A&(B|C); Ac; Ac&Bc&Cc];DP = [ 1     0.3     0.6    0.1];TV = [B|C; (((A&B)|(Ac&Bc))&Cc)|(A&C); Ac&(B|Cc)];disp('Call for mincalc')  % Modification % DV = [DV; Ac&B&C; Ac&B];% DP = [DP   0.2     0.3];

Npr02_19

% file npr02_19.m % Data for problem P2-19% A = computer;  B = monitor;  C = printer minvec3DV = [A|Ac; A&B; A&B&Cc; A&C; B&C; (A&Cc)|(Ac&C); ...           (A&Bc)|(Ac&B); (B&Cc)|(Bc&C)];DP = [1 0.49 0.17 0.45 0.39 0.50 0.43 0.43];TV = [A; B; C; (A&B&Cc)|(A&Bc&C)|(Ac&B&C); (A&B)|(A&C)|(B&C); A&B&C];disp('Call for mincalc')

Npr02_20

% file npr02_20.m % Data for problem P2-20minvec3 DV = [A|Ac; A;     B;  A&B&C; A&C; (A&B)|(A&C)|(B&C); B&C - 2*(A&C)];DP = [  1  0.232 0.228 0.045 0.062      0.197            0];TV = [A|B|C; Ac&Bc&C];disp('Call for mincalc') % Modification% DV = [DV; C];% DP = [DP  0.230 ];

Npr02_21

% file npr02_21.m % Data for problem P2-21minvec3 DV = [A|Ac; A;  A&B; A&B&C;  C;  Ac&Cc];DP = [ 1   0.4 0.3  0.25   0.65  0.3 ];TV = [(A&Cc)|(Ac&C); Ac&Bc; A|B; A&Bc];disp('Call for mincalc') % Modification% DV = [DV; Ac&B&Cc; Ac&Bc];% DP = [DP   0.1      0.3 ];

Npr02_22

% file npr02_22.m % Data for problem P2-22minvec3 DV = [A|Ac; A;  A&B; A&B&C;  C;  Ac&Cc];DP = [ 1   0.4 0.5  0.25   0.65  0.3 ];TV = [(A&Cc)|(Ac&C); Ac&Bc; A|B; A&Bc];disp('Call for mincalc') % Modification% DV = [DV; Ac&B&Cc; Ac&Bc];% DP = [DP   0.1      0.3 ];

Npr02_23

% file npr02_23.m % Data for problem P2-23minvec3 DV = [A|Ac; A;  A&C; A&B&C;  C;  Ac&Cc];DP = [ 1   0.4 0.3  0.25   0.65  0.3 ];TV = [(A&Cc)|(Ac&C); Ac&Bc; A|B; A&Bc];disp('Call for mincalc') % Modification% DV = [DV; Ac&B&Cc; Ac&Bc];% DP = [DP   0.1      0.3 ];

Npr03_01

% file npr03_01.m % Data for problem P3-1minvec3 DV = [A|Ac; A;  A&B; B&C; Ac|(B&C); Ac&B&Cc];DP = [ 1   0.55 0.30 0.20   0.55     0.15  ];TV = [Ac&B; B];disp('Call for mincalc')

Npr04_04

% file npr04_04.m % Data for problem P4-4pm = [0.032 0.016 0.376 0.011 0.364 0.073 0.077 0.051];disp('Minterm probabilities for P4-4 are in pm')

Npr04_05

% file npr04_05.m % Data for problem P4-5pm = [0.084 0.196 0.036 0.084 0.085 0.196 0.035 0.084 ...           0.021 0.049 0.009 0.021 0.020 0.049 0.010 0.021]; disp('Minterm probabilities for P4-5 are in pm')

Npr04_06

% file npr04_06.m % Data for problem P4-6pm =  [0.085 0.195 0.035 0.085 0.080 0.200 0.035 0.085 ...         0.020 0.050 0.010 0.020 0.020 0.050 0.015 0.015]; disp('Minterm probabilities for P4-6 are in pm')

Mpr05_16

% file mpr05_16.m % Data for Problem P5-16A = [51 26  7; 42 32 10; 19 54 11; 24 53  7; 27 52  5;      49 19 16; 16 59  9; 47 32  5; 55 17 12; 24 53  7]; B = [27 34  5; 19 43  4; 39 22  5; 38 19  9; 28 33  5;     19 41  6; 37 21  8; 19 42  5; 27 33  6; 39 21  6];disp('Call for oddsdf')

Questions & Answers

a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
how did I we'll learn this
Noor Reply
f(x)= 2|x+5| find f(-6)
Prince Reply
f(n)= 2n + 1
Samantha Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive
Samson Reply

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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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