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6 Pages

### exam2_sp07_stu

Course: ENGR ENGR126, Fall 2009
School: Purdue
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Word Count: 1102

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_________________________________________ Name: Problem #1 (20 points total) A company is evaluating the ultimate strength (Su) of two types of cement as it cures over time. Consider the raw data, the best-fit lines, and the r-squared values determined using the method of least-squares for Cement A (A) and Cement B (B) below. Ultimate Strength, Su (GPa) Su = 0.0386*Time + 2.0571 3 2 1 0 0 1 2 3 4 5 6 Ultimate...

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_________________________________________ Name: Problem #1 (20 points total) A company is evaluating the ultimate strength (Su) of two types of cement as it cures over time. Consider the raw data, the best-fit lines, and the r-squared values determined using the method of least-squares for Cement A (A) and Cement B (B) below. Ultimate Strength, Su (GPa) Su = 0.0386*Time + 2.0571 3 2 1 0 0 1 2 3 4 5 6 Ultimate Strength, Su (GPa) 4 4 Su = 0.6671*Time + 0.5821 3 2 1 0 0 1 2 3 Curing Time (h) 4 5 6 Curing Time (h) Cement A: Su = 2.25 and r 2 = 0.388 Cement B: Su = 2.25 and r 2 = 0.949 1.1 (4 points) The computed r-squared value for A is much less than that of B because: (Clearly circle one response.) a. The Sum of Squares of Errors (SSE) for A is considerably less than for B b. The Sum of Squares of Errors (SSE) for A is considerably greater than for B c. The Sum of Squares of Deviations (SST) for A is considerably less than for B d. The Sum of Squares of Deviations (SST) for A is considerably greater than for B e. Both B and C 1.2 (8 points) Which of the following statements is/are true? (Circle all that are true.) a. The slope of the regression line for B is greater than the slope of the regression line for A. b. The "goodness of fit" (as measured by r2) of the regression line for B is greater than "goodness of fit" of the regression line for A. c. The value of intercept of the regression line for A is greater than the value of the intercept of the regression line for B. d. The mean Su value for A is equal to the mean Su value for B. 1.3 (4 points) The intercept for Cement B is: (Clearly circle one response) a. 0.75 b. 0.5821 c. 0.6671 d. 0 1.4 (2 points) The r2 value of B is 0.949. This means that 94.9% of the variation of the data around the regression line is due to unexplained error. (Clearly circle one response) True False 1.5 (2 points) By comparing A and B, we can tell that the regression line for A fits the data for A better than the regression line for B fits the data for B. (Clearly circle one response) True False 3 Name: _________________________________________ Problem #2: (12 points) The graph below resulted from a set of data stored in two vectors in Matlab: InvTemp and Diffusivity. In your own words (Matlab code or English), describe the steps below that you would take to get an equation modeling this relationship using Matlab. Carbon diffusivity in alpha Fe versus inverse temperature 1.E-06 Diffusivity (cm^2/s) 1.E-07 1.E-08 1.E-09 0.0008 0.0009 0.001 0.0011 0.0012 0.0013 0.0014 0.0015 Inverse Temperature (1/K) Write down an equation that could model this data (the general form). How would you linearize these data (make them look like a line on a rectilinear graph)? How you would obtain the coefficients for your model? 4 Name: _________________________________________ Refer to the following script files for questions 3 & 4 clear all a=1; b=5; c=10; d=15; e=20; n=input('Enter an integer between 1-20: if n>=a & n<=b disp('Range 1') elseif n>b & n<=c disp('Range 2') elseif n>c & n<d disp('Range 3') elseif n>d & n<e disp('Range 4') end Problem #3: (4 points) If the MATLAB script file above were executed, and the user input a value of 8, what would be displayed to the user? a) b) c) d) e) Range 1 Range 2 Range 3 Range 4 None of the Above '); Problem #4: (4 points) If the MATLAB script file above were executed, and the user input a value of 15, what would be displayed to the user? a) b) c) d) e) Range 1 Range 2 Range 3 Range 4 None of the Above 5 Name: _________________________________________ Problem #5: (12 points total) I am forming teams some for an engineering project subject to some rules designed to avoid schedule incompatibilities and staffing shortages on other projects. In each case, circle which Matlab logic phrase matches the verbal description without imposing additional constraints. The five questions are independent--the conditions of the first do not carry over to the others. 5.1 (4 points) Bob and Ajay cannot be on the same team. The team must have at least one of Bob or Guili, and the team must have at least one of Francesca or Dewayne. (A|B) & xor(B,G) & xor(F,D) ~(A&B) & (B|G) & (F|D) (~A|~B) & xor(B,G) & xor(F,D) xor(A,B) & (B|G) & (F|D) 5.2 (4 points) At least 2 of the people on the team must be from a particular workgroup consisting of Ajay, Bob, Clara, Dewayne, Elie, Francesca, and Guili. (A&B&C&D&E&F&G)>=2 (A|B|C|D|E|F|G)>1 (A+B+C+D+E+F+G)>1 ~((A&B&C&D&E&F&G)<2) 5.3 (4 points) Neither Dewayne nor Francesca can be on the team. ~(D&F) ~D&~F ~xor(D,F) D*F==0 6 Name: _________________________________________ Problem #6: (24 points total 8 points per phase region) A phase diagram for Carbon and Platinum is shown below. To the right of the figure, write a conditional statement that displays the phases present in this system for any temperature and composition on this diagram. Call the phases "Pt + G", "L" and "L + G" for simplicity. Hints: Do the lower phase first and the phases extend (extrapolate) beyond the boundaries of graph shown. If it is assumed the lines shown are linear, the mixture has the following characteristics: Below 1700 C, it is a mixture of solid platinum and graphite. Above 1700 C, there are two possible phases: a Liquid (L) phase and a Liquid (L) + Graphite phase. The end points of the division line between these two phases are labeled on the diagram. All the phases present in regions bordering a boundary can be present on the boundary. L L + Graphite Pt + Graphite 7 Name: _________________________________________ Problem #7: (24 points total) The following graphs showing different functions will be used to answer the questions. The origin of each graph is in the lower left of the graph window except for Graph A. In each case, x is independent, y is dependent, and constants K, L, V, W, g may be positive or negative. 3 60 120 2 50 100 1 Graph 0 origin 0 -1 A 1 2 3 4 5 6 7 8 9 10 40 B 80 30 60 C 0 1 2 3 4 5 6 7 8 9 10 20 40 -2 10 20 -3 -4 Graph 0 origin 3.5 0 1 2 3 4 5 6 7 8 9 10 Graph origin 0 5 16 4.5 3 4 14 12 3.5 2.5 3 10 2.5 2 D 1 2 3 4 5 6 7 8 9 10 2 F 1.5 E 1 2 3 4 5 6 7 8 9 10 8 6 1.5 1 4 1 0.5 0.5 2 Graph 0 origin 0 Graph 0 origin 0 Graph 0 origin 0 1 2 3 4 5 6 7 8 9 10 7.1 (2 points each) Write the letter that best represents the shape of that graph. Your letter choice must be clear and match the letter of the graph you choose (even if you usually draw that letter differently. y = e^(gx) y = W*x^(0.5) y = ln(x) y = x^2 y = (V/x) y = K*x + C ___________________________ ___________________________ ___________________________ ___________________________ ___________________________ ___________________________ 7.2 (2 points each) Write the Matlab command (plot, semilogy, semilogx, or loglog) that would generate a plot of the same function to make it appear linear. y = e^(gx) y = W*x^(0.5) y = ln(x) y = x^2 y = (V/x) y = K*x + C ___________________________ ___________________________ ___________________________ ___________________________ ___________________________ ___________________________ 8
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Purdue - ENGR - ENGR126
Name: _Solution_ Problem #1 (20 points total) A company is evaluating the ultimate strength (Su) of two types of cement as it cures over time. Consider the raw data, the best-fit lines, and the r-squared values determined using the method of least-squares
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ENGINEERING 126 Spring 2008 Exam 3 Name: _ Student ID: _ Division/Section:_INSTRUCTIONS: Duration: 60 minutes Keep your eyes on your own work! Keep your work covered at all times! 1. Each student is responsible for following directions. Read carefully. 2
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Math 4653: Elementary Probability: Spring 2007Homework #2. Problems and Solutions (corrected)1. Sec. 1.5: #2: Polyas urn scheme. An urn contains 4 white balls and 6 black balls. A ball is chosen at random, and its color noted. The ball is then replaced,
Berkeley - STAT - 134
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Math 4653: Elementary Probability: Spring 2007Homework #1. Problems and Solutions1. Appendix 1 (vi): Prove that 2n nn=k=0n kn n-kn=k=0n k2.Solution. The left side is the number of all subsets of the set cfw_1, 2, . . . , n-1, n, n+1, . . . ,
Berkeley - STAT - 134
Math 4653: Elementary Probability: Spring 2007Homework #3. Problems and Solutions1. Sec. 2.4: #2: Find Poisson approximations to the probabilities of the following events in 500 independent trials with probability 0.02 of success on each trial: a) 1 suc
Berkeley - STAT - 134
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Berkeley - STAT - 134
Math 4653: Elementary Probability: Spring 2007Homework #7. Problems and Solutions1. Ch. 4, Review: #21: Suppose R1 and R2 are two independent random variables with the 1 same density function f (x) = x exp(- 2 x2 ) for x 0. Find a) the density of Y = mi
Berkeley - STAT - 134
STAT516 Solution to Homework 3Section 2.1 4 1 6 1 = 0.375 2. P (2 boys &amp; 2 girls) = = 2 2 2 16 Hence, P (different number of boys &amp; girls in a family of 4 children) = 1 - P (2 boys &amp; 2 girls) = 1 - 0.375 = 0.625 So, in a family of 4 children, different n
Berkeley - STAT - 134
STAT516Section 3.2FALL 2005Solution to Homework 52. a) average(3rd list) = average(1st list) + average(2nd list) = 5.8 b) average(3rd list) = average(1st list) average(2nd list) = -2.2 c) &amp; d) Can't do it: need to know the order of the numbers in the
Berkeley - STAT - 134
Solution to Homework 8Section 4.2 1. Let X denote the lifetime of an atom. Then X has an exponential distribution with = log 2 (as half life is 1). a) P ( X &gt; 5) = e -5 = 1/ 32 . b) P ( X &gt; t ) = .1 e - t = .1 t = (log10) / t = 3.32 . c) Assume that life
Berkeley - STAT - 134
Section 5.1 1. a) 7/124b) 5/36.411 1 200122. a) 0.1 b) 1 - 2(1/ 2)(0.19) 2 = 0.0975 (0.2) 20.20.01 0.014. a) 1 - ( 3 / 4 ) = 7 /16 = 0.437520.210.250 0.25 1X b) P - 1 Y 4 0.25 P X = 5 Y4 1 4 3 X 1 - + 9 / 40 = 0.225 = = 3 2 5 43 4
Berkeley - STAT - 134
Section 5.4 1. a)r , 12X2 1r1, 11 dx if 0 &lt; z &lt; 1 2 0z0b) f X1 + X 2 ( z ) =f X1 ( x) f X 2 ( z - x)dx=1 1 x&lt;2 2 dx = 0&lt; x&lt;1;zx&lt; z 2 dx = 0 &lt; x &lt;1;0 &lt; z - - 2&lt;dx 20 111if 1 &lt; z &lt; 21 dx if 2 &lt; z &lt; 3 2 z-2 Xz if 0 &lt; z &lt; 1 2 1 = if 1 &lt; z &lt; 2
Berkeley - STAT - 134
STAT516Section 2.4 1.FALL 2005SOLUTION TO HOMEWORK 4b) See text for Poisson(2) histogram. c) See text and &quot;approximate&quot; Poisson(.3284) histogram.2. a) The number of successes in 500 independent trials with success probability 0.02 has Binomial(500,0.
Berkeley - STAT - 134
Solution to Homework 7Section 4.1 1. Treat the density function as constant over those small intervals. 1 a) (0.001)( )=0.000399 2 1 b) (0.001)( )( e -0.5 )=0.00025 2-4 2. a) cx dx = 1 1 1 cc = 3 b) mean = 3/2 c) variance = 3. a) cx(1 - x)dx = 10 1/
Berkeley - STAT - 134
Section 4.5 6. a) 1 - 1/8 = 7/8. f 3x 2 f ( x) = b) 0 c) E ( X ) = 3 / 4. d) P( X Section 4.6 1. Let X i denote the gap (in minutes) between the ith person's arrival time and noon. Then X i 's are i.i.d. N( 0, 52). a) P (1st person arrives before 11:50) =
Berkeley - STAT - 134
STAT516 Solution to Homework 1 1.1.4: We have 18 black, 18 red, 2 green, total 38 a) P(both lose)=2/38 b) P(at least one wins)=1-P(both lose)=36/38 c) P(at least one loses)=1-P(both win)=1-0=1 1.3.2: a)(AB c ) (Ac B) or A B - AB b)(A B C)c or Ac B c C c1
Berkeley - STAT - 134
Section 3.3 19. Let X j be the weight (in lbs) of the j -th person, j = 1, 2, ., 30. Hence, total weight is: S = X 1 + X 2 + . + X 30 As X j 's are independent with E ( X j ) = 150 and SD( X j ) = 55 , we can use CLT to calculate: 5000 - 30 150 P ( S &gt;\$ 5
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