Can any genius explain me about well-ordering principle - Proofs using well-ordering principle. with some examples.

Determine whether the members of the given set of vectors are linearly independent. If they are linearly dependent, find a linear relation among them of the form c1x(1) + c2x(2) + c3x(3) = 0. (Give c1, c2, and c3 as real numbers. If the vectors are linearly independent, enter INDEPENDENT.) x(1) = 9 1 0 , x(2) = 0 1 0 , x(3) = −1 9 0

Topology

Prove or disprove ( with a counterexample)

(a) The continuous image of a Hausdorff space is Hausdorff.

(b)  The continuous image of a connected space is connected.

Topology

(a) Prove that the interval [0,1] with the subspace topology is connected from basic principles.

(b) Prove that the interval [0,1] with the subspace topology is compact from basic principles.

A coffee company uses a​ data-based approach to improving the quality and customer satisfaction of its products. When survey data indicated that the coffee company needed to improve its​ package-sealing process, an experiment was conducted to determine the factors in the​ bag-sealing equipment that might be affecting the ease of opening the bag without tearing the inner liner of the bag. One factor that could affect the rating of the ability of the bag to resist tears was the plate gap on the​ bag-sealing equipment. Data were collected on 19 bags in which the plate gap was varied. Complete parts​ (a) through​ (e) below.

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Click the icon to view the data table.

a. Construct a scatter plot. Choose the correct graph below.

A.

-15-15Tear RatingPlate Gap

Edit Coordinates(0,0)

A scatter plot has a horizontal axis labeled Tear Rating from negative 1 to 5 to in increments of 1 and a vertical axis labeled Plate Gap from negative 2 to 5 in increments of 1. A series of plotted points range from horizontal coordinates 0.0 to 0.9 and vertical coordinates negative 1.2 to 2.2 with no obvious pattern. Also, the following individual points are plotted: (1.9,2.2), (4.0,0.0), (4.3,2.2). All coordinates are approximate.

B.

-15-11Tear RatingPlate Gap

Edit Coordinates(0,0)

A scatter plot has a horizontal axis labeled Tear Rating from negative 1 to 5 to in increments of 1 and a vertical axis labeled Plate Gap from negative 1.25 to 1 in increments of 0.25. A series of plotted points range from horizontal coordinates 0.0 to 0.9 and vertical coordinates negative 1.2 to 0.5 with no obvious pattern. Also, the following individual points are plotted: (1.9,0.1), (4.0,0.6), (4.3,0.2). All coordinates are approximate.

C.

-11-15Plate GapTear Rating

Edit Coordinates(0,0)

A scatter plot has a horizontal axis labeled Plate Gap from negative 1.25 to 1 in increments of 0.25 and a vertical axis labeled Tear Rating from negative 1 to 5 to in increments of 1. A series of plotted points range from horizontal coordinates negative 1.2 to 0.5 and vertical coordinates 0.0 to 0.9 with no obvious pattern. Also, the following individual points are plotted: (0.1,1.9), (0.6,4.0), (0.2,4.3). All coordinates are approximate.

D.

-15-15Plate GapTear Rating

Edit Coordinates(0,0)

A scatter plot has a horizontal axis labeled Plate Gap from negative 2 to 5 in increments of 1 and a vertical axis labeled Tear Rating from negative 1 to 5 to in increments of 1. A series of plotted points range from horizontal coordinates negative 1.2 to 2.2 and vertical coordinates 0.0 to 0.9 with no obvious pattern. Also, the following individual points are plotted: (0.4,1.9), (negative 0.3,4.0), (2.2,4.3). All coordinates are approximate.b. Assuming a linear​ relationship, use the​ least-squares method to find the regression coefficients

b0

and

b1.

c. Interpret the meaning of the​ slope,

b1​,

in this problem. Choose the correct answer below.

A.

For each increase of one in the tear​ rating, the plate gap is expected to increase by

b1

units.

B.

The approximate plate gap when the tear rating is 0 is

b1

units.

C.

The approximate tear rating when the plate gap is 0 units is

b1.

D.

For each​ one-unit increase in the plate​ gap, the tear rating is expected to increase by

b1.

d. Predict the mean tear rating when the plate gap is equal to 0.

nothing

​(Round to four decimal places as​ needed.)

e. What should you tell management of the coffee company about the relationship between the plate gap and the tear​ rating?

A.

The tear rating is not affected by the plate gap.

B.

The tear rating is affected by the plate gap. As the plate gap​ increases, the tear rating

increasesincreases.

C.

As the plate gap​ increases, the tear rating

decreasesdecreases.

The tear rating is affected by the plate gap.

D.

As the tear rating​ increases, the plate gap decreases. The plate gap is affected by the tear rating.

use the method of order two to approximate the solution to the following initial value problem y'=e^(t-y),0<=t<=1, y(0)=1, with h=0.5

8. The manager of a diner wants to re-evaluate his staffing needs depending on variations in customer traffic during the day. He collects data on the number of customers served along with four dummy variables representing the morning, afternoon, evening, and night shifts. The dummy variable Morning equals 1 if the information was from the morning shift and 0 otherwise. The dummy variable Afternoon equals 1 if the information was from the afternoon shift and 0 otherwise. The dummy variable Evening equals 1 if the information was from the evening shift and 0 otherwise. The dummy variable Night equals 1 if the information was from the night shift and 0 otherwise. Data is posted on Webcampus in excel file titled Shifts Data.

Estimate a regression model using the number of customers as the response (dependent) variable and the shift dummy variables as the explanatory (independent) variables.

1. What is the regression equation? Attach the Excel output.

2. What is the predicted number of customers during the morning shifts?

3. What is the predicted number of customers during the afternoon shifts?

4. What is the predicted number of customers during the evening shifts?

5. What is the predicted number of customers during the night shifts?

Use the Matlab code discussed in class to build a simulator for the digital modulation scheme PPM Coherent. Once finished, run the simulations required to plot BER curves with a minimum probability of bit-error of 10.

1. Plots:

a. ??? vs ??? (??) plot – in the same graph, plot at least 4 BER curves corresponding to different values of symbol periods and pulse widths.

b. Transmitted signal plot – plot the transmitted signal corresponding to 3 bits.

c. Received signal plots – for the transmitted signal (3 bits), plot the received noisy signal for at least 4 values of ??? (e.g. 0 ??, 10 ??, 20 ??, 30 ??) .

Code in Matlab:

% Description: Simple simulator for digital (binary) communications

%% Main Function
function runComm_code()
clear;clc;
  
SNRdB = 0:0.5:26;
nBits = 1e6;
SNRdBLength = length(SNRdB);
Pe = ones(1,SNRdBLength);
clc; disp('Simulation running: 0.0% completed.');
for i = 1:SNRdBLength
Pe(i) = transmitReceive(nBits,SNRdB(i));
clc;fprintf('%s: %.1f%% completed','Simulation running',i/SNRdBLength*100);
end
hold on; semilogy(SNRdB,Pe); grid on;
clc; disp('Simulation finished.');
return

%% Bit Transmission Function
function [Perror,nErrors,TxBits,RxBits] = transmitReceive(nBits,SNRdB)
% TxBits = [0 1 1 0 1 0 0 1 1 1];
nBitsMax = 100000;
nBitsOrig = nBits;
TxBits = [];
RxBits =[];
if(nBits > nBitsMax)
n = ceil(nBits/nBitsMax);
nErrorsArr = zeros(1,n);
for i = 1:n
if(nBits < nBitsMax)
[~,nErrorsArr(i),TxBitsTmp,RxBitsTmp] = transmitReceive(nBits,SNRdB);
TxBits = [TxBits,TxBitsTmp];
RxBits = [RxBits,RxBitsTmp];
break;
else
[~,nErrorsArr(i),TxBitsTmp,RxBitsTmp] = transmitReceive(nBitsMax,SNRdB);
TxBits = [TxBits,TxBitsTmp];
RxBits = [RxBits,RxBitsTmp];
nBits = nBits - nBitsMax;
end
end
nErrors = sum(nErrorsArr);
Perror = nErrors/nBitsOrig;
return;
end
  
TxBits = round(rand(1,nBits));
TxSignals = modulation(TxBits);
[RxSignals] = addAWGNoise(TxSignals,SNRdB);
RxBits = demodulation(RxSignals);
errorBits = TxBits ~= RxBits;
nErrors = sum(errorBits);
Perror = nErrors/nBits;
  

return

%% Modulation and Demodulation Functions
function [outSignals,time] = modulation(inBits)

% This code is for On-Off Keying (OOK) modulation
% You need to substitute this with your own Matlab code according
% to your assigned modulation scheme.

pulseWidth = 1e-9;
timeWindow = 10.5e-9;
carrierFreq = 4e9;
  
signalBW = 1.5/pulseWidth;
osFactor = 1 + carrierFreq/signalBW;
  
[signal1,time] = pulseSignal(pulseWidth,timeWindow,osFactor);
signal1 = signal1.*squareSignal(time,carrierFreq);
signalLength = length(signal1);
signal0 = zeros(signalLength,1);
  
inBits = inBits(:)';
nBits = length(inBits);
outSignals = zeros(signalLength,nBits);
col0 = inBits == 0;
col1 = inBits == 1;
outSignals(:,col0) = repmat(signal0,1,sum(col0));
outSignals(:,col1) = repmat(signal1,1,sum(col1));

return
function [outBits] = demodulation(inSignals)
  
% This code is for On-Off Keying (OOK) modulation
% You need to substitute this with your own Matlab code according
% to your assigned modulation scheme.

SNRmin = 10^(12/10);
Ps = sum(modulation(1).^2);
Po = Ps./SNRmin;
  
Psignals = sum(inSignals.^2);
outBits = Psignals > Po;
return

%% Noise Adding Function
function [outSignals,noiseSignals] = addAWGNoise(signals,SNRdB)
signalsSize = size(signals);
SNR = 10^(SNRdB/10);
Ps = max(sum(signals.^2));
Po = Ps./SNR;
noiseSignals = sqrt(Po/2).*randn(signalsSize(1),signalsSize(2));
outSignals = signals+noiseSignals;
return

%% Signal Creation Functions
function [outSignal,time] = pulseSignal(Tp,Tw,osFactor)
if(nargin == 2), osFactor = 1; end
if(osFactor < 1), osFactor = 1; end
B = 1.5/Tp;
fs = osFactor*2*B;
Nw = ceil(fs*Tw);
Np = ceil(fs*Tp);
outSignal = [0; ones(Np,1); zeros(Nw-Np-1,1)];
outSignal(end)=0;
time = (0:length(outSignal)-1)'/fs;
return
function [outSignal] = squareSignal(time,freq)
tsample = time(2)-time(1);
tperiod = 1/freq;
n = ceil(tperiod/tsample);
npos = ceil(n/2);
nneg = n - npos;
Nperiods = ceil(time(end)*freq);
signalTmp = [ones(npos,1); -1*ones(nneg,1)];
outSignal = repmat(signalTmp,Nperiods+1,1);
outSignal = [ 0; outSignal(1:length(time)-1) ];
return

Use the Matlab code discussed in class to build a simulator for the digital modulation scheme PPM Coherent. Once finished, run the simulations required to plot BER curves with a minimum probability of bit-error of 10.

1. Plots:

a. ??? vs ??? (??) plot – in the same graph, plot at least 4 BER curves corresponding to different values of symbol periods and pulse widths.

b. Transmitted signal plot – plot the transmitted signal corresponding to 3 bits.

c. Received signal plots – for the transmitted signal (3 bits), plot the received noisy signal for at least 4 values of ??? (e.g. 0 ??, 10 ??, 20 ??, 30 ??) .

Code in Matlab:

% Description: Simple simulator for digital (binary) communications

%% Main Function
function runComm_code()
clear;clc;
  
SNRdB = 0:0.5:26;
nBits = 1e6;
SNRdBLength = length(SNRdB);
Pe = ones(1,SNRdBLength);
clc; disp('Simulation running: 0.0% completed.');
for i = 1:SNRdBLength
Pe(i) = transmitReceive(nBits,SNRdB(i));
clc;fprintf('%s: %.1f%% completed','Simulation running',i/SNRdBLength*100);
end
hold on; semilogy(SNRdB,Pe); grid on;
clc; disp('Simulation finished.');
return

%% Bit Transmission Function
function [Perror,nErrors,TxBits,RxBits] = transmitReceive(nBits,SNRdB)
% TxBits = [0 1 1 0 1 0 0 1 1 1];
nBitsMax = 100000;
nBitsOrig = nBits;
TxBits = [];
RxBits =[];
if(nBits > nBitsMax)
n = ceil(nBits/nBitsMax);
nErrorsArr = zeros(1,n);
for i = 1:n
if(nBits < nBitsMax)
[~,nErrorsArr(i),TxBitsTmp,RxBitsTmp] = transmitReceive(nBits,SNRdB);
TxBits = [TxBits,TxBitsTmp];
RxBits = [RxBits,RxBitsTmp];
break;
else
[~,nErrorsArr(i),TxBitsTmp,RxBitsTmp] = transmitReceive(nBitsMax,SNRdB);
TxBits = [TxBits,TxBitsTmp];
RxBits = [RxBits,RxBitsTmp];
nBits = nBits - nBitsMax;
end
end
nErrors = sum(nErrorsArr);
Perror = nErrors/nBitsOrig;
return;
end
  
TxBits = round(rand(1,nBits));
TxSignals = modulation(TxBits);
[RxSignals] = addAWGNoise(TxSignals,SNRdB);
RxBits = demodulation(RxSignals);
errorBits = TxBits ~= RxBits;
nErrors = sum(errorBits);
Perror = nErrors/nBits;
  

return

%% Modulation and Demodulation Functions
function [outSignals,time] = modulation(inBits)

% This code is for On-Off Keying (OOK) modulation
% You need to substitute this with your own Matlab code according
% to your assigned modulation scheme.

pulseWidth = 1e-9;
timeWindow = 10.5e-9;
carrierFreq = 4e9;
  
signalBW = 1.5/pulseWidth;
osFactor = 1 + carrierFreq/signalBW;
  
[signal1,time] = pulseSignal(pulseWidth,timeWindow,osFactor);
signal1 = signal1.*squareSignal(time,carrierFreq);
signalLength = length(signal1);
signal0 = zeros(signalLength,1);
  
inBits = inBits(:)';
nBits = length(inBits);
outSignals = zeros(signalLength,nBits);
col0 = inBits == 0;
col1 = inBits == 1;
outSignals(:,col0) = repmat(signal0,1,sum(col0));
outSignals(:,col1) = repmat(signal1,1,sum(col1));

return
function [outBits] = demodulation(inSignals)
  
% This code is for On-Off Keying (OOK) modulation
% You need to substitute this with your own Matlab code according
% to your assigned modulation scheme.

SNRmin = 10^(12/10);
Ps = sum(modulation(1).^2);
Po = Ps./SNRmin;
  
Psignals = sum(inSignals.^2);
outBits = Psignals > Po;
return

%% Noise Adding Function
function [outSignals,noiseSignals] = addAWGNoise(signals,SNRdB)
signalsSize = size(signals);
SNR = 10^(SNRdB/10);
Ps = max(sum(signals.^2));
Po = Ps./SNR;
noiseSignals = sqrt(Po/2).*randn(signalsSize(1),signalsSize(2));
outSignals = signals+noiseSignals;
return

%% Signal Creation Functions
function [outSignal,time] = pulseSignal(Tp,Tw,osFactor)
if(nargin == 2), osFactor = 1; end
if(osFactor < 1), osFactor = 1; end
B = 1.5/Tp;
fs = osFactor*2*B;
Nw = ceil(fs*Tw);
Np = ceil(fs*Tp);
outSignal = [0; ones(Np,1); zeros(Nw-Np-1,1)];
outSignal(end)=0;
time = (0:length(outSignal)-1)'/fs;
return
function [outSignal] = squareSignal(time,freq)
tsample = time(2)-time(1);
tperiod = 1/freq;
n = ceil(tperiod/tsample);
npos = ceil(n/2);
nneg = n - npos;
Nperiods = ceil(time(end)*freq);
signalTmp = [ones(npos,1); -1*ones(nneg,1)];
outSignal = repmat(signalTmp,Nperiods+1,1);
outSignal = [ 0; outSignal(1:length(time)-1) ];
return

pls explain the answer a little bit, thank you.

Let A ∈ Mm,n(F) and let TA : F n → F m be the corresponding linear map (see (1.2)). (a) Prove that A has a left inverse if and only if TA is injective. (b) Prove that A has a right inverse if and only if TA is surjective. (c) Prove that A has a two-sided inverse if and only if TA is an isomorphism.

Dependent T (T-test for paired) Table should look like this. Answer the three parts.

A. Write your research question and hypotheses

B. Test-Statistic (the t-ratio). Determine the degrees of freedom. df = (N – 1) .

C. Refer to the t-values in appendix B with df at the .05 significant level and at the .01 significant level in order to make the final conclusion.

Problem 2 For the same list as above, sort the list in such a way where the first set of numbers are ascending even and the second set of numbers are ascending odd. Your output should look like the following: [ 2, 4, 6, … 1, 3, 5 .. ]. The list is A = [ 6, 9, 0, 4, 2, 8, 0, 0, 8, 5, 2, 1, 3, 20, -1 ]

Numerical Analysis:

Use Muller's method to find a solution in [0.1, 1] accurate to within 600x^4−550x^3+200x^2−20x−1=0. Explain how the algorithm works.

Please show table with all values, not just the final root value.

Proposition 8.59. Suppose that X, Y, W, Z, A, B are sets. Let f : X → Y , W ⊆ X, Z ⊆ X, A ⊆ Y , and B ⊆ Y . Then the following are true:

prove the following ?

(1) f(W ∩ Z) ⊆ f(W) ∩ f(Z).

(2) f(W ∪ Z) = f(W) ∪ f(Z).

(3) f−1(A ∩ B) ⊆ f−1(A) ∪ f−1(B)

4) f−1(A ∪ B) = f−1(A) ∪ f−1(B).

(5) X−f−1(A)⊆f−1(Y −A).

(6) W ⊆ f−1(f(W)

Isabella is filing as the head of a household. Her taxable income is $68,014. Use the Tax Table, Exhibit 18-3 from your text, to find the tax liability.