Compute the correlation of X and Y. Compute the covariance of X and .pdf
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Compute the correlation of X and Y. Compute the covariance of X and Y. (Refer to the Homework 14 solutions for the marginal PDFs.) Compute E(5X(1 + Y)+3Y( l - X)) from your work in part (b) (i.e., don\'t do any new integration). Compute the correlation coefficient rho of X and Y. Are X and Y uncorrelated? Justify your answer. Solution correlation of X willbe 3x/2+1/4 correlation of Y will be 3/4(y^2+1).
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Complete the PEM Assembly Flow using the processes listed below Wafer rightarrow
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Solution
Order of processes is as follows:
(a) Wafer
(g) Dice
(h) Die Attach
(c) Wire Bond
(b) Encapsulation
(d) Deflash
(f) Trim/Form
(e) Ship.
classify the give differential equation as to type and order. classif.pdf
classify the give differential equation as to type and order. classify the ordinary differential
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Solution
Its first order nonlinear ordinary diff.equation. And this in non-homogeneous
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Classify the data as either qualitative or quantitative. A large reco.pdf
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Solution
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College students often make up a substantial portion of the popul.pdf
College students often make up a substantial portion of the population of college cities and
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students. What is the probability that in a random sample of 150 people from State College, more
than 50 are not college students?
Solution
Given p=1-0.711=0.289
By using Normal approximation,
X~Normal(mean=n*p=150*0.289=43.35, s=n*p*(1-p)=sqrt(150*0.289*0.711)=5.55)
The probability is
P(X>50)=P((X-mean)/s >(50-43.35)/5.55)
=P(Z> 1.2)
=0.1151 (check standard normal table).
Cisco Systems, Inc. (CSCO) has a current dividend of $0.84 per share..pdf
Cisco Systems, Inc. (CSCO) has a current dividend of $0.84 per share. The current stock price
price of Home Depot is $24.45. Cisco has a dividend payout of 43.97% and an expected return
on equity of 16.42%. What is the cost of Cisco common equity? (1)
Solution
Dividend payout ratio = Dividend per share/Earnings available for equity share holders
43.97% = 0.84/Earning available for equity share holders
Earnings available for equity share holders = 0.84/43.97%
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Class C and Class F fly ashes are used extensively as pozzolanic mate.pdf
Class C and Class F fly ashes are used extensively as pozzolanic materials in concrete. Discuss
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Solution
Pozzolanic reaction is the chemical reaction which occurs between the pozzolans ( Fly ash ) and
the calcium hydroxide to give CSH gel.
Ca(OH)2 + H4SiO4 CaH2SiO4·2 H2O
In short, CH + SH C-S-H.
City A+ has a population of 314,159 people on January 1, 2010. The po.pdf
City A+ has a population of 314,159 people on January 1, 2010. The population is growing at a
rate of 1,500 people per year. Find a linear population model P(t) = P0 + at for the population of
City A+. Reset the clock so that t = 0 on January 1, 2010. Be sure to include in your answer
appropriate labels for the units for t and P.
Solution
p(0)=314,159.
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classify the give differential equation as to type and order. classif.pdf
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by the information in the second column?
Solution
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College students often make up a substantial portion of the population of college cities and
towns. State College, Pennsylvania, ranks first with 71.1% of its population made up of college
students. What is the probability that in a random sample of 150 people from State College, more
than 50 are not college students?
Solution
Given p=1-0.711=0.289
By using Normal approximation,
X~Normal(mean=n*p=150*0.289=43.35, s=n*p*(1-p)=sqrt(150*0.289*0.711)=5.55)
The probability is
P(X>50)=P((X-mean)/s >(50-43.35)/5.55)
=P(Z> 1.2)
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Cisco Systems, Inc. (CSCO) has a current dividend of $0.84 per share..pdf
Cisco Systems, Inc. (CSCO) has a current dividend of $0.84 per share. The current stock price
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on equity of 16.42%. What is the cost of Cisco common equity? (1)
Solution
Dividend payout ratio = Dividend per share/Earnings available for equity share holders
43.97% = 0.84/Earning available for equity share holders
Earnings available for equity share holders = 0.84/43.97%
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Consider the following function Reduce all fractions to lowest terms..pdf
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Solution
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City A+ has a population of 314,159 people on January 1, 2010. The po.pdf
City A+ has a population of 314,159 people on January 1, 2010. The population is growing at a
rate of 1,500 people per year. Find a linear population model P(t) = P0 + at for the population of
City A+. Reset the clock so that t = 0 on January 1, 2010. Be sure to include in your answer
appropriate labels for the units for t and P.
Solution
p(0)=314,159.
Consider the annual maximum series for the US Geological survey gatin.pdf
Consider the annual maximum series for the US Geological survey gating station on Clear
Creek near Pearland, TX The statistics of the transformed flows are for given average of log of
values (y-bar), standard deviation of log values (Sy), skewness (Gs), and sample site (n), Using
the Pearson Type III distribution compare the discharges for given recurrence interval with and
without the US Water Resource Council recommendation on s coefficients. The map skew (Gm)
is given.
Solution
Frequency Factors K for Gamma and log-Pearson Type III Distributions
Recurrence Interval In Years
1.0101
2
5
10
25
50
100
200
SKEW COEFFICIENT
Cs
Percent Chance (>=) = 1-F
99
50
20
10
4
2
1
0.5
3
-0.667
-0.396
0.420
1.180
2.278
3.152
4.051
4.970
2.9
-0.690
-0.390
0.440
1.195
2.277
3.134
4.013
4.904
2.8
-0.714
-0.384
0.460
1.210
2.275
3.114
3.973
4.847
2.7
-0.740
-0.376
0.479
1.224
2.272
3.093
3.932
4.783
2.6
-0.769
-0.368
0.499
1.238
2.267
3.071
3.889
4.718
2.5
-0.799
-0.360
0.518
1.250
2.262
3.048
3.845
4.652
2.4
-0.832
-0.351
0.537
1.262
2.256
3.023
3.800
4.584
2.3
-0.867
-0.341
0.555
1.274
2.248
2.997
3.753
4.515
2.2
-0.905
-0.330
0.574
1.284
2.240
2.970
3.705
4.444
2.1
-0.946
-0.319
0.592
1.294
2.230
2.942
3.656
4.372
2
-0.990
-0.307
0.609
1.302
2.219
2.912
3.605
4.298
1.9
-1.037
-0.294
0.627
1.310
2.207
2.881
3.553
4.223
1.8
-1.087
-0.282
0.643
1.318
2.193
2.848
3.499
4.147
1.7
-1.140
-0.268
0.660
1.324
2.179
2.815
3.444
4.069
1.6
-1.197
-0.254
0.675
1.329
2.163
2.780
3.388
3.990
1.5
-1.256
-0.240
0.690
1.333
2.146
2.743
3.330
3.910
1.4
-1.318
-0.225
0.705
1.337
2.128
2.706
3.271
3.828
1.3
-1.383
-0.210
0.719
1.339
2.108
2.666
3.211
3.745
1.2
-1.449
-0.195
0.732
1.340
2.087
2.626
3.149
3.661
1.1
-1.518
-0.180
0.745
1.341
2.066
2.585
3.087
3.575
1
-1.588
-0.164
0.758
1.340
2.043
2.542
3.022
3.489
0.9
-1.660
-0.148
0.769
1.339
2.018
2.498
2.957
3.401
0.8
-1.733
-0.132
0.780
1.336
1.993
2.453
2.891
3.312
0.7
-1.806
-0.116
0.790
1.333
1.967
2.407
2.824
3.223
0.6
-1.880
-0.099
0.800
1.328
1.939
2.359
2.755
3.132
0.5
-1.955
-0.083
0.808
1.323
1.910
2.311
2.686
3.041
0.4
-2.029
-0.066
0.816
1.317
1.880
2.261
2.615
2.949
0.3
-2.104
-0.050
0.824
1.309
1.849
2.211
2.544
2.856
0.2
-2.178
-0.033
0.830
1.301
1.818
2.159
2.472
2.763
0.1
-2.252
-0.017
0.836
1.292
1.785
2.107
2.400
2.67
0
-2.326
0.000
0.842
1.282
1.751
2.054
2.326
2.576
-0.1
-2.4
0.017
0.846
1.27
1.716
2.000
2.252
2.482
-0.2
-2.472
0.033
0.850
1.258
1.680
1.945
2.178
2.388
-0.3
-2.544
0.050
0.853
1.245
1.643
1.890
2.104
2.294
-0.4
-2.615
0.066
0.855
1.231
1.606
1.834
2.029
2.201
-0.5
-2.686
0.083
0.856
1.216
1.567
1.777
1.955
2.108
-0.6
-2.755
0.099
0.857
1.200
1.528
1.720
1.880
2.016
-0.7
-2.824
0.116
0.857
1.183
1.488
1.663
1.806
1.926
-0.8
-2.891
0.132
0.856
1.166
1.448
1.606
1.733
1.837
-0.9
-2.957
0.148
0.854
1.147
1.407
1.549
1.660
1.749
-1
-3.022
0.164
0.852
1.128
1.366
1.492
1.588
1.664
-1.1
-3.087
0.180
0.848
1.107
1.324
1.435
1.518
1.581
-1.2
-3.149
0.195
0.844
1.086
1.282
1.379
1.449
1.501
-1.3
-3.2.
Consider the ANOVA table for a simple linear regression given below. .pdf
Consider the ANOVA table for a simple linear regression given below. What is the value of the
mean square error for this regression analysis?
Solution
The error degrees of freedom are 14 since they must sum to 15. MSE = SSE/d.f. =
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Christy Reed has been depositing $2,000 in her savings account every .pdf
Christy Reed has been depositing $2,000 in her savings account every December since 2001.
Her account earns 7 percent compounded annually.
Solution
1) Future Value = PMT(P/A,7%,12)
FV = $2000 x 7.943 = $15,886
2) Nper calculated in excel
Double Amount in year = NPER (Rate, pmt, pv,fv) = NPER(10%,0,1,-2) = 7.2 years
Triple Amount in years = NPER(10%,0,1,-3) = 11.5 years
Using F/P table @10% we can find NPER
year 1 F/P = 1.1
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Consider a catchment that has undergone extensive urbanisation. What .pdf
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would expect to this catchment in the following hydrographic characteristics? Peak discharge
for a given AEP Time to peak discharge Runoff volume for a similar storm event Include in
your discussion a justification for the projects change in the characteristic.
Solution
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Developed lands are often requires Removal of vegetation and soil, grading the land surface, and
constructing drainage networks. This can increase runoff to streams from rainfall. As a result, the
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to the ground surface which results in reduction in seepage. Reduction in seepage leads to
reduction in time for peak discharge.
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1/2 - 1/2 e^-x/b - -1/2 = 1 - 1/2e^-x/b
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1 - Y = 1/2 e^-x/b
2 - 2Y = e^-x/b
ln(2-2Y) = -x/b
x = -ln(2-2Y) b Y >= 1/2
Our solution is
x = ln(2Y) b Y <= 1/2
x = -ln(2-2Y) b Y >= 1/2
where Y ~ U[0,1]
As f(x) = k(x-a)^(k-1)/(b-a)^k a <= x <= b
F(x) = I[a,x] k(x-a)^(k-1)/(b-a)^k =(x-a)^k/(b-a)^k E[a,x] =
(x-a)^k/(b-a)^k - 0 =
(x-a)^k/(b-a)^k
Then, if(x-a)^k/(b-a)^k = y
(x-a)^k =(b-a)^k y
Taking the k\'th root on both sides
x-a = (b-a) y^(1/k)
x = a +(b-a) y^(1/k)
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Consider the annual maximum series for the US Geological survey gatin.pdf
Consider the annual maximum series for the US Geological survey gating station on Clear
Creek near Pearland, TX The statistics of the transformed flows are for given average of log of
values (y-bar), standard deviation of log values (Sy), skewness (Gs), and sample site (n), Using
the Pearson Type III distribution compare the discharges for given recurrence interval with and
without the US Water Resource Council recommendation on s coefficients. The map skew (Gm)
is given.
Solution
Frequency Factors K for Gamma and log-Pearson Type III Distributions
Recurrence Interval In Years
1.0101
2
5
10
25
50
100
200
SKEW COEFFICIENT
Cs
Percent Chance (>=) = 1-F
99
50
20
10
4
2
1
0.5
3
-0.667
-0.396
0.420
1.180
2.278
3.152
4.051
4.970
2.9
-0.690
-0.390
0.440
1.195
2.277
3.134
4.013
4.904
2.8
-0.714
-0.384
0.460
1.210
2.275
3.114
3.973
4.847
2.7
-0.740
-0.376
0.479
1.224
2.272
3.093
3.932
4.783
2.6
-0.769
-0.368
0.499
1.238
2.267
3.071
3.889
4.718
2.5
-0.799
-0.360
0.518
1.250
2.262
3.048
3.845
4.652
2.4
-0.832
-0.351
0.537
1.262
2.256
3.023
3.800
4.584
2.3
-0.867
-0.341
0.555
1.274
2.248
2.997
3.753
4.515
2.2
-0.905
-0.330
0.574
1.284
2.240
2.970
3.705
4.444
2.1
-0.946
-0.319
0.592
1.294
2.230
2.942
3.656
4.372
2
-0.990
-0.307
0.609
1.302
2.219
2.912
3.605
4.298
1.9
-1.037
-0.294
0.627
1.310
2.207
2.881
3.553
4.223
1.8
-1.087
-0.282
0.643
1.318
2.193
2.848
3.499
4.147
1.7
-1.140
-0.268
0.660
1.324
2.179
2.815
3.444
4.069
1.6
-1.197
-0.254
0.675
1.329
2.163
2.780
3.388
3.990
1.5
-1.256
-0.240
0.690
1.333
2.146
2.743
3.330
3.910
1.4
-1.318
-0.225
0.705
1.337
2.128
2.706
3.271
3.828
1.3
-1.383
-0.210
0.719
1.339
2.108
2.666
3.211
3.745
1.2
-1.449
-0.195
0.732
1.340
2.087
2.626
3.149
3.661
1.1
-1.518
-0.180
0.745
1.341
2.066
2.585
3.087
3.575
1
-1.588
-0.164
0.758
1.340
2.043
2.542
3.022
3.489
0.9
-1.660
-0.148
0.769
1.339
2.018
2.498
2.957
3.401
0.8
-1.733
-0.132
0.780
1.336
1.993
2.453
2.891
3.312
0.7
-1.806
-0.116
0.790
1.333
1.967
2.407
2.824
3.223
0.6
-1.880
-0.099
0.800
1.328
1.939
2.359
2.755
3.132
0.5
-1.955
-0.083
0.808
1.323
1.910
2.311
2.686
3.041
0.4
-2.029
-0.066
0.816
1.317
1.880
2.261
2.615
2.949
0.3
-2.104
-0.050
0.824
1.309
1.849
2.211
2.544
2.856
0.2
-2.178
-0.033
0.830
1.301
1.818
2.159
2.472
2.763
0.1
-2.252
-0.017
0.836
1.292
1.785
2.107
2.400
2.67
0
-2.326
0.000
0.842
1.282
1.751
2.054
2.326
2.576
-0.1
-2.4
0.017
0.846
1.27
1.716
2.000
2.252
2.482
-0.2
-2.472
0.033
0.850
1.258
1.680
1.945
2.178
2.388
-0.3
-2.544
0.050
0.853
1.245
1.643
1.890
2.104
2.294
-0.4
-2.615
0.066
0.855
1.231
1.606
1.834
2.029
2.201
-0.5
-2.686
0.083
0.856
1.216
1.567
1.777
1.955
2.108
-0.6
-2.755
0.099
0.857
1.200
1.528
1.720
1.880
2.016
-0.7
-2.824
0.116
0.857
1.183
1.488
1.663
1.806
1.926
-0.8
-2.891
0.132
0.856
1.166
1.448
1.606
1.733
1.837
-0.9
-2.957
0.148
0.854
1.147
1.407
1.549
1.660
1.749
-1
-3.022
0.164
0.852
1.128
1.366
1.492
1.588
1.664
-1.1
-3.087
0.180
0.848
1.107
1.324
1.435
1.518
1.581
-1.2
-3.149
0.195
0.844
1.086
1.282
1.379
1.449
1.501
-1.3
-3.2.
Consider the ANOVA table for a simple linear regression given below. .pdf
Consider the ANOVA table for a simple linear regression given below. What is the value of the
mean square error for this regression analysis?
Solution
The error degrees of freedom are 14 since they must sum to 15. MSE = SSE/d.f. =
38.45/14 = 2.7464.
Consider the AC circuit shown in Figure P75 (a) Find the branch curre.pdf
Consider the AC circuit shown in Figure P75 (a) Find the branch currents I, I2, I3, I4, I5 and I6.
(b) Determine the complex power of each circuit element. (c) Are the complex powers in the
circuit balanced? Explain.
Solution
Please wait. I am uploading the image.
Christy Reed has been depositing $2,000 in her savings account every .pdf
Christy Reed has been depositing $2,000 in her savings account every December since 2001.
Her account earns 7 percent compounded annually.
Solution
1) Future Value = PMT(P/A,7%,12)
FV = $2000 x 7.943 = $15,886
2) Nper calculated in excel
Double Amount in year = NPER (Rate, pmt, pv,fv) = NPER(10%,0,1,-2) = 7.2 years
Triple Amount in years = NPER(10%,0,1,-3) = 11.5 years
Using F/P table @10% we can find NPER
year 1 F/P = 1.1
Year 7 F/P = 1.949 (its approx double the number) = NPER = 7
Year = 12 F/P = 3.138 (its approx triple the number = NPER = 12.
Consider a continuous-time LTI system whose transfer function is give.pdf
Consider a continuous-time LTI system whose transfer function is given by: H(j omega) = sin
(4 omega)/omega If the input to this system is a periodic signal with period T = 8, determine the
corresponding system output y(t)
Solution
the correcsponding output for this functions is sinc(4t).
Consider a catchment that has undergone extensive urbanisation. What .pdf
Consider a catchment that has undergone extensive urbanisation. What are the changes you
would expect to this catchment in the following hydrographic characteristics? Peak discharge
for a given AEP Time to peak discharge Runoff volume for a similar storm event Include in
your discussion a justification for the projects change in the characteristic.
Solution
(A) Peak Discharge for a given AEP: Due to urbaniztion the type of use of land changes.
Developed lands are often requires Removal of vegetation and soil, grading the land surface, and
constructing drainage networks. This can increase runoff to streams from rainfall. As a result, the
peak discharge increases in nearby streams.
(B) Time to Peak Discharge: Time to peak discharge reduces due to urbanisation as more paving
to the ground surface which results in reduction in seepage. Reduction in seepage leads to
reduction in time for peak discharge.
(C) Runoff volume for a similar storm event: Due to urbaniztion, there is chance to increse in to
runoff volume due to following reasons
(i) More paving means lower percolation and seepage in ground surfce - this leads to increase in
runoff volume.
(ii) Urbaniztion creates obstructions to the flow of water, which agin helps in increasing the
runoff volume..
Compute the inverse cdfs of the following density functions and descr.pdf
Compute the inverse cdfs of the following density functions and describe how to generate
random variables from these distributions based on the inverse-cdf method. f(x) = 1 / 2b e -|x|/b,
b > 0, x R. (1) f(x) = k(x - a) k-1 / (b - a)k, k > 0, a x b. (2).
Solution
I will use I[a,b] for the integral from a to b and E[a,b] for evaluated from a to b
For - infinity < x < 0, F(x) = |[-infinity,x] 1/2b e ^ x/b = 1/2 e ^x/b E[-infinity, x] =
1/2 e ^x/b - 0 = 1/2 e ^x/b
For 0 < x < infinity, as the value of integral above = 1/2 at 0, F(x) =
1/2 + |[0,x] 1/2b e^-x/b = 1/2 - 1/2e^-x/bE[0, x] =
1/2 - 1/2 e^-x/b - -1/2 = 1 - 1/2e^-x/b
Then, if 1/2 e^x/b = Y, e^x/b = 2Y, and x/b = ln(2Y) and x = ln(2Y) b Y <= 1/2
if1 - 1/2e^-x/b = Y
1 - Y = 1/2 e^-x/b
2 - 2Y = e^-x/b
ln(2-2Y) = -x/b
x = -ln(2-2Y) b Y >= 1/2
Our solution is
x = ln(2Y) b Y <= 1/2
x = -ln(2-2Y) b Y >= 1/2
where Y ~ U[0,1]
As f(x) = k(x-a)^(k-1)/(b-a)^k a <= x <= b
F(x) = I[a,x] k(x-a)^(k-1)/(b-a)^k =(x-a)^k/(b-a)^k E[a,x] =
(x-a)^k/(b-a)^k - 0 =
(x-a)^k/(b-a)^k
Then, if(x-a)^k/(b-a)^k = y
(x-a)^k =(b-a)^k y
Taking the k\'th root on both sides
x-a = (b-a) y^(1/k)
x = a +(b-a) y^(1/k)
.
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