course mth 163 assignment #010010. `query 10
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18:09:07 query the family of linear functions, Problem 2. Describe the graphs of y = A f(x) for A = -.3 and A = 1.3 and compare; explain the comparison.
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RESPONSE --> The graph for y = Af(x) for A = -.3 will cross the origin and each point will be -.3 times as close to the x axis as in the y=x graph. It will also be decreasing from left to right. The graph for the y = A f(x) for A = 1.3 will cross the origin and each point will be 1.3 times as far away from the x axis. This graph will be increasing from left to right. In comparison of the two, the y = -.3f(x) graph is going the opposite direction as the y = 1.3f(x). confidence assessment: 2
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18:09:58 ** For the basic linear function f(x) = x the A = -.3 graph is obtained by vertically stretching the y = x function by factor -.3, resulting in a straight line thru the origin with slope -.3, basic points (0,0) and (1, -.3), and the A = 1.3 graph is obtained by vertically stretching the y = x function by factor 1.3, resulting in is a straight line thru the origin with slope 1.3, basic points (0,0) and (1, 1.3). **
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RESPONSE --> I did not include the basic points except for the origin. self critique assessment: 2
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19:18:23 describe the graphs of y = f(x) + c for c = .3 and c = -2.7 and compare; explain the comparison.
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RESPONSE --> The graph of y = f(x) + .3 is vertically shifted .3 units higher than the graph for y = x + b. The graph for y = f(x) - 2.7 is vertically shifted -2.7 units from the graph y = x + b confidence assessment: 2
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19:23:22 ** The graphs will have slopes identical to that of the original function, but their y intercepts will vary from -2.7 to .3. **
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RESPONSE --> I did not mention that the two would have identical slopes, but I do understand that they would have this and that the y intercepts would vary from -2.7 to .3 self critique assessment: 2
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19:26:54 query problem 4. linear function y = f(x) = -1.77 x - 3.87 What are your symbolic expressions, using x1 and x2, for the corresponding y coordinates y1 and y2.
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RESPONSE --> x1 & x2 y1 = -1.77 x1 - 3.87 y2 = -1.77 x2 - 3.87 confidence assessment: 2
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19:28:43 ** y1 = f(x1) = -1.77 x1 - 3.87 y2 = f(x2) = -1.77 x2 - 3.87. `dy = y2 - y1 = -1.77 x2 - 3.87 - ( -1.77 x1 - 3.87) = -1.77 x2 + 1.77 x1 - 3.87 + 3.87 = -1.77 ( x2 = x1). Thus slope = `dy / `dx = -1.77 (x2 - x1) / (x2 - x1) = -1.77. This is the slope of the straight line, showing that these symbolic calculations are consistent. **
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RESPONSE --> The question did not ask specifically for slope as did another section of this question. I know how to arrive at the slope, but again it just asked for the expressions to arrive at x1, x2, y1, and y2. self critique assessment: 2
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19:36:56 query problem 5. graphs of families for y = mx + b. Describe your graph of the family: m = 2, b varies from -3 to 3 by step 1.
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RESPONSE --> Each graph will have the same slope of 2 but will be vertically shifted one unit higher with each graph. confidence assessment: 2
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19:37:07 ** The graphs will all have slope 2 and will pass thru the y axis between y = -3 and y = 3. The family will consist of all such graphs. **
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RESPONSE --> self critique assessment: 3
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19:49:15 ** This is of the form y = mx + b with b= 1. So the y intercept is (0, 1). The point 1 unit to the right is (1, 1.5). The x-intercept occurs when y = 0, which implies .5 x + 1 = 0 with solution x = -2, so the x-intercept is (-2, 0). **
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RESPONSE --> Again, this answered before I had a chance..having not looked and the explaination here is my answer: the y intercept is what y is when x = 0: y = (.5 * 0) + 1 y = 1 y intercept is (0, 1) The point one unit to the right of this point is (1, 1.5) The x intercept is what x is when y = 0: -b/m or -1/.5 or -2 ( -2, 0) self critique assessment: 2
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19:52:25 query problem 6. three basic points graph of y = .5 x + 1 What are your three basic points?
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RESPONSE --> the y intercept is what y is when x = 0: y = (.5 * 0) + 1 y = 1 y intercept is (0, 1) The point one unit to the right of this point is (1, 1.5) The x intercept is what x is when y = 0: -b/m or -1/.5 or -2 ( -2, 0) confidence assessment: 3
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19:52:37 ** The y intercept occurs when x = 0, at which point we have y = .5 * 0 + 1 = 1. So one basic point is (0, 1). The point 1 unit to the right of the y axis occurs at x = 1, where we get y = .5 * 1 + 1 = 1.5 to give us the second basic point (1, 1.5) }The third point, which is not really necessary, is the x intercept, which occurs when y = 0. This gives us the equation 0 = .5 x + 1, with solution x = -2. So the third basic point is (-2, 0). **
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RESPONSE --> self critique assessment: 3
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20:01:30 query problem 7. simple pendulum force vs. displacement What are your two points and what line do you get from the two points?
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RESPONSE --> I chose (.21, 1.1) and (1.04, 4.1) the equation of the line I get is y = 3.61x + .34. I arrived at that equation by substituting in and making two equations and solving: 1.1 = .21m + b 4.1 = 1.04m + b 4.1 = 1.04m + b -1.1 = -.21m - b 3 = .83m m = 3.61 Substitute this back in : 1.1 = 3.61 * .21 + b 1.1 = .76 + b .34 = b confidence assessment: 2
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20:10:14 STUDENT RESPONSE: The two points are (1.1, .21) and (2.0, .54). These points give us the two simultaneous equations .21- m(1.1) + b .54= m(2.0) +b. If we solve for m and b we will get our y = mx + b form. INSTRUCTOR COMMENT: I believe those are data points. I doubt if the best-fit line goes exactly through two data points. In the future you should use points on your sketched line, not data points. However, we'll see how the rest of your solution goes based on these points. **
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RESPONSE --> I too used data points and seem to have written them in reverse. confidence assessment: 2
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20:16:46 what equation do you get from the slope and y-intercept?
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RESPONSE --> With the pointsput in the right place the line is: y = .28x - .1 **Just for clarification** My slope would be .28 the y intercept would be : y = .28(0) - .1 y = 0 - .1 y = -.1 (0, -.1) confidence assessment: 2
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20:17:11 STUDENT RESPONSE: b= .21 m=.19 INSTRUCTOR COMMENT: ** b would be the y intercept, which is not .21 since y = .21 when x = 1.1 and the slope is nonzero. If you solve the two equations above for m and b you obtain m = .367 and b = -.193. This gives you equation y = mx + b or y = .367 x - .193. **
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RESPONSE --> self critique assessment: 3
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20:26:36 what is your linear regression model?
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RESPONSE --> I did not use the DERIVE program. I do not know exactly what is meant by the linear regression model. It was listed on the DERIVE section. confidence assessment: 2
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20:26:59 ** Your linear regression model would be obtained using a graphing calculator or DERIVE. As a distance student you are not required to use these tools but you should be aware that they exist and you may need to use them in other courses. **
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RESPONSE --> self critique assessment: 3
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20:33:24 What force would be required to hold the pendulum 47 centimeters from its equilibrium position? what equation did you solve to obtain this result?
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RESPONSE --> I used my equation y = .28x - .1 47 = .28x - .1 47.1 = .28x 168 pounds confidence assessment: 1
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20:39:36 ** If your model is y = .367 x - .193, with y = force and x= number of cm from equilibrium, then we have x = 47 and we get force = y = .367 * 47 - .193 = 17 approx. The force would be 17 force units. **
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RESPONSE --> I am just off tonight....I understand what I need to do, I just was somewhere out in left field: y = .28(47) - .1 y = 13.16 - .1 y = 13.06 self critique assessment: 2
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20:43:03 Why would it not make sense to ask what force would be necessary to hold the pendulum 80 meters from its equilibrium position? what equation did you solve to obtain this result?
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RESPONSE --> Honestly, I don't know why it wouldn't make sense to ask. But if it doesn't make sense to ask, why would you use an equation to solve. I don't understand. confidence assessment: 1
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20:45:42 STUDENT RESPONSE: I used the equation f= .10*47+.21 and got the answer 15.41 which would be to much force to push or pull INSTRUCTOR COMMENT: ** The problem is that you can't hold a pendulum further at a distance greater than its length from its equilibrium point--the string isn't long enough. **
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RESPONSE --> I understand now after reading the explaination. I was wondering how you would get an 8 m pendulum to go 80 m, but I wasn't sure if I was just having another ""duh moment"" or not. You can't have a distance greater than the length of the pendulum. self critique assessment: 2
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20:53:10 How far could you hold the pendulum from its equilibrium position using a string with a breaking strength of 25 pounds? what equation did you solve to obtain this result?
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RESPONSE --> I used the equation I got earlier y = .28x - .1 y = .28(25) - .1 y = 7 - .1 y = 6.9 cm confidence assessment: 1
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20:56:45 ** Using the model y = .367 * x - .193 with y = force = 25 lbs we get the equation 25 = .367 x - .193, which we solve to obtain x = 69 (approx.). Note that this displacement is also unrealistic for this pendulum. **
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RESPONSE --> Where did the model come from, the DERIVE exercise? self critique assessment: 2
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20:58:47 What is the average rate of change associated with this model? Explain this average rate in common-sense terms.
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RESPONSE --> The average rate of change is the change in force divided by the change in distance. confidence assessment: 1
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20:59:02 ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium we can use any two (x, y) points to get the rate of change. In all cases we will get rate of change = change in y / change in x = .367. The change in y is the change in the force, while the change in x is the change in position. The rate of change therefore tells us how much the force changes per unit of change in position (e.g., the force increases by 15 pounds for every inch of displacement). **
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RESPONSE --> self critique assessment: 3
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21:05:36 What is the average slope associated with this model? Explain this average slope in common-sense terms.
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RESPONSE --> The slope is the average rate of change which is the change in the force divided by the change in distance confidence assessment: 1
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21:09:13 ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium the average slope is .367. **
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RESPONSE --> I forgot to type in the slope from the model y = .367x - .193 The slope would be .367 self critique assessment: 2
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21:19:20 As you gradually pull the pendulum from a point 30 centimeters from its equilibrium position to a point 80 centimeters from its equilibrium position, what average force must you exert?
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RESPONSE --> Using the slope from the model (.367) in hte slope formula slope = y2-y1 / x2-x1 .367 = (force)/ 80-30 .367 = force/50 18.35 = force confidence assessment: 2
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21:21:47 ** if it was possible to pull the pendulum back this far and if the model applies you will get Force at 30 cm: y = .367 * 30 - .193 = 10.8 approx. and Force at 80 cm: y = .367 * 80 - .193 = 29 approx. so that ave force between 30 cm and 80 cm is therefore (10.8 + 29) / 2 = 20 approx.. **
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RESPONSE --> I didn't even think to substitute the numbers into the equation....but I understand self critique assessment: 2
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21:49:45 query problem 8. flow range What is the linear function range(time)?
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RESPONSE --> using the info given I found the slope: (97 - 34) / (20 - 80) = (m) slope 63/-60 = m -1.05 = slope then substitute: y = mx + b 97 = -1.05(20) + b 97 = -21 + b 118 = b y = -1.05x + 118 confidence assessment: 2
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21:51:47 ** STUDENT RESPONSE: I obtained model one by drawing a line through the data points and picking two points on the line and finding the slope between them. I then substituted this value for m and used one of my data points on my line for the x and y value and solved for b. the line I got was range(t) = -.95t + 112.38. y = -16/15x + 98 INSTRUCTOR COMMENT: This looks like a good model. According to the instructions it should however be expressed as range(time) = -16/15 * time + 98. **
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RESPONSE --> I did not express in range(time) self critique assessment: 2
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21:53:16 What is the significance of the average rate of change? Explain this average rate in common-sense terms.
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RESPONSE --> The average rate of change is thechange in time divided by the rate of change in the depth confidence assessment: 2
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21:53:27 ** the average rate of change is change in range / change in clock time. The average rate of change indicates the average rate at which range in cm is changing with respect to clock time in sec, i.e., the average number of cm / sec at which the range changes. Thus the average rate tells us how fast, on the average, the range changes. **
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RESPONSE --> self critique assessment: 3
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22:00:48 What is the average slope associated with this model? Explain this average slope in common-sense terms.
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RESPONSE --> By looking at the model y = -16/15x + 98 the slope is -16/15 THe slope is the change in the graph from one point to the next. In this case, the rise is -16 units and the run is 15 units. confidence assessment: 2
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22:01:07 ** it's the average rate at which the range of the flow changes--the average rate at which the position of the end of the stream changes. It's the speed with which the point where the stream reaches the ground moves across the ground. **
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RESPONSE --> self critique assessment: 3
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22:07:17 query problem 9. If your total wealth at clock time t = 0 hours is $3956, and you earn $8/hour for the next 10 hours, then what is your total wealth function totalWealth( t ), where t is time in hours?
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RESPONSE --> totalWealth(10) = mt + 3956 If you solve it: totalWealth (10) = (8)(10) + 3956 totalWealth = 80+3956 totalWealth = $4036 confidence assessment: 3
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22:10:48 ** Total wealth has to be expressed in terms of t. A graph of total wealth vs. t would have y intercept 3956, since that is the t = 0 value, and slope 8, since slope represents change in total wealth / change in t, i.e., the number of dollars per hour. A graph with y-intercept b and slope m has equation y = m t + b. Thus we have totalWealth(t) = 8 * t + 3956 . **
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RESPONSE --> I forgot that I needed to add the 8 in the equation. I thought it would be a ""general"" equation, but I understand. self critique assessment: 2
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22:17:14 At what clock time will your total wealth reach $4000? what equation did you solve to obtain this result?
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RESPONSE --> I used the equation totalWealth = 8t + 3956 4000 = 8t +3956 44 = 8t t = 5.5 hours confidence assessment: 3
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22:17:45 STUDENT RESPONSE: To find the clock time when my total wealth will reach 4000 I solved the equation totalWealth(t) = 4000. The value I got when I solved for t was t = 5.5 hours. 4.4 hours needed to reach 4000 4000 = 10x + 3956 INSTRUCTOR COMMENT: Almost right. You should solve 4000 = 8 x + 3956, obtaining 5.5 hours. This is equivalent to solving totalWealth(t) = 4000 = 8 t + 3956, which is the more meaningful form of the relationship. **
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RESPONSE --> self critique assessment: 3
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22:19:22 What is the meaning of the slope of your graph?
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RESPONSE --> the slope in the graph shows a constant wealth confidence assessment: 2
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22:19:30 GOOD STUDENT RESPONSE: The slope of the graph shows the steady rate at which money is earned on an hourly basis. It shows a steady increase in wealth.
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RESPONSE --> self critique assessment: 3
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22:37:27 query problem 10. Experience shows that when a certain widget is sold for $30, a certain store can expect to sell 200 widgets per week, while a selling price of $28 increases the number sold to 300. What linear function numberSold(price) describes this situation?
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RESPONSE --> This one was a little harder to understand. I wasn't totally sure how to go about setting up the equation. numberSold(price) = 50 price confidence assessment: 1
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22:38:52 query problem 10. Experience shows that when a certain widget is sold for $30, a certain store can expect to sell 200 widgets per week, while a selling price of $28 increases the number sold to 300. What linear function numberSold(price) describes this situation?
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RESPONSE --> This one was a little harder to understand. I wasn't totally sure how to go about setting up the equation. numberSold(price) = 50 price ***Didn't this one just come up?????** confidence assessment: 1
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22:45:31 If you make a graph of y = numberSold vs. x = price you have graph points (30, 200) and (28, 300). You need the equation of the straight line through these points. You plug these coordinates into the form y = m x + b and solve for m and b. Or you can use another method. Whatever method you use you get y = -50 x + 1700. Then to put this into the notation of the problem you write numberSold(price) instead of y and price instead of x. You end up with the equation numberSold(price) = -50 * price + 1700. **
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RESPONSE --> Like I said before I wasn't exatly sure how to go about working on this problem. I do understand now that I should have substituted into a y = mx + b formula to find m & b confidence assessment: 2
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22:47:58 If the store must meet a quota by selling 220 units per week, what price should they set? what equation did you solve to obtain this result?
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RESPONSE --> numberSold(price) = -50 * price + 1700 220 = -50 (price) + 1700 -1480 = -50p Price = 29.60 confidence assessment: 3
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22:48:12 ** If the variables are y and x, you know y so you can solve for x. For the function numberSold(price) = -50 * price + 1700 you substitute 220 for numbersold(price) and solve for price. You get the equation 220 = -50 * price + 1700 which you can solve to get price = 30, approx. **
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RESPONSE --> self critique assessment: 3
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22:50:58 If each widget costs the store $25, then how much total profit will be expected from selling prices of $28, $29 and $30? what equation did you solve to obtain this result?
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RESPONSE --> numberSold(price) = -50 * price + 1700. numberSold(price) = -50 * 28 + 1700 numberSold(price) = -1400 + 1700 numberSold(price) = 300 numberSold(price) = -50 * 29 + 1700 numberSold(price) = -1450 + 1700 numberSold(price) = 250 numberSold(price) = -50 * 30 + 1700 numberSold(price) = -1500 + 1700 numberSold(price) = 200 confidence assessment: 3
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22:51:38 STUDENT RESPONSE: If each widget costs the store $25, then they should expect to earn a profit of 300 dollars from a selling price of $28, 250 dollars from a price of $29 and 200 dollars from a price of $30. To find this I solved the equations numberSold(28); numberSold(29), and numberSold(30). Solving for y after putting the price values in for p. They will sell 300, 250 and 200 widgets, respectively (found by solving the given equation). To get the total profit you have to multiply the number of widgets by the profit per widget. At $28 the profit per widgit is $3 and the total profit is $3 * 300 = $900; at $30 the profit per widgit is $5 and 200 are sold for profit $1000; at $29 what happens? **
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RESPONSE --> self critique assessment: 3
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23:20:12 query problem 11. quadratic function depth(t) = .01 t^2 - 2t + 100 representing water depth vs. What is the equation of the straight line connecting the t = 20 point of the graph to the t = 60 point?
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RESPONSE --> Substitute for t with 20 and 60 to find the y coordinates: y = .01 t^2 - 2t + 100 y = .01 (20)^2 - 2(20) + 100 y = .01(400) - 40 + 100 y = 4 + 60 y = 64 y = .01 (60)^2 - 2(60) + 100 y = .01(3600) - 120 + 100 y = 36 - 20 y = 16 Use theese to make 2 y = mx + b equations to solve for m & b: 64 = m20 + b 16 = m60 + b 64 = 20m + b -16 = -60m - b 48 = -40m m = 1.2 Substitute: 64 = (1.2)(20) + b 64 = 24 + b 40 = b This would leave y = 1.2x + 40 confidence assessment: 3
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23:22:23 ** The t = 20 point is (20,64) and the t = 60 point is (60, 16), so the slope is (-48 / 20) = -1.2. This can be plugged into the form y = m t + b to get y = -1.2 t + b. Then plugging in the x and y coordinates of either point you get b = 88. y = -1.2 t + 88 **
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RESPONSE --> I missed putting the negative sign in front of 1.2. That does make b = 88 self critique assessment: 2
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23:32:06 query problem 13. quadratic depth function y = depth(t) = .01 t^2 - 2t + 100. What is `dy / `dt based on the two time values t = 30 sec and t = 40 sec.
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RESPONSE --> Substitute in to get y: y = .01(30)^2 - 2(30) + 100 y = .01(900) - 60 + 100 y = 9 + 40 y = 49 y = .01 (40)^2 - 2(40) + 100 y = .01(1600) - 80 + 100 y = 16 + 20 y = 36 Use this to get the slope: (49-36) / (30-40) 13/-10 -1.3 confidence assessment: 3
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23:32:14 ** For t = 30 we have y = 49 and for t = 40 we have y = 36. The slope between (30, 49) and (40,36) is (36 - 49) / (40 - 30) = -1.3. This tells you that the depth is changing at an average rate of -1.3 cm / sec. **
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RESPONSE --> self critique assessment: 3
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23:35:54 what is `dy / `dt based on t = 30 sec and t = 31 sec.
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RESPONSE --> We already have the (30, 49) y = .01 (31)^2 - 2(31) + 100 y = 9.61-62+100 y = 47.61 (49-47.61) / (30-31) 1.39/-1 -1.39 confidence assessment: 3
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23:35:59 ** Based on t = 30 and t = 31 the value for `dy / `dt is -1.39, following the same steps as before **
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RESPONSE --> self critique assessment: 3
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23:39:17 what is `dy / `dt based on t = 30 sec and t = 30.1 sec.
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RESPONSE --> We know (30, 49) y= .01 (30.1)^2 - 2(30.1) + 100 y = 9.06 - 60.2 + 10 y = 48.86 (49-48.86)/(30-30.1) .14/-.1 -1.4 confidence assessment: 3
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23:39:34 ** STUDENT RESPONSE: The value for 'dy / `dt based on t = 30 sec and t = 30.1 sec is -1.4 INSTRUCTOR COMMENT: ** Right if you round off the answer. However the answer shouldn't be rounded off. Since you are looking at a progression of numbers (-1.3, -1.39, and this one) and the differences in these numbers get smaller and smaller, you have to use a precision that will always show you the difference. Exact values are feasible here and shoud be used. I believe that this one comes out to -1.399. **
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RESPONSE --> self critique assessment: 3
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23:41:53 What do you think you would get for `dy / `dt if you continued this process?
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RESPONSE --> All responses would be in the -1.39 range confidence assessment: 2
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23:42:07 STUDENT RESPONSE: An even more and more accurate slope value. I don't think it would have to continue to decrease. INSTRUCTOR COMMENT **If you look at the sequence -1.3, -1.39, -1.399, ..., what do you think happens? It should be apparent that the limiting value is -1.4 **
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RESPONSE --> self critique assessment: 3
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23:45:24 What does the linear function tell you?
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RESPONSE --> linear function tells youi the slope of the line. confidence assessment: 2
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23:45:34 ** The function tells you that at any clock time t the rate of depth change is given by the function .02 t - 2. For t = 30, for example, this gives us .02 * 30 - 2 = -1.4, which is the rate we obtained from the sequence of calculations above. **
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RESPONSE --> self critique assessment: 3
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23:46:35 query problem 14. linear function y = f(x) = .37 x + 8.09
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RESPONSE --> confidence assessment:
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23:51:53 what are the first five terms of the basic sequence {f(n), n = 1, 2, 3, ...} for this function.
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RESPONSE --> After substituting: When n=1: 8.46 When n=2: 8.83 When n=3: 9.20 When n=4: 9.57 When n=5: 9.94 confidence assessment: 3
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23:52:00 ** The first five terms are 8.46, 8.83, 9.2, 9.57, and 9.94 **
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RESPONSE --> self critique assessment: 3
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23:53:29 What is the pattern of these numbers?
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RESPONSE --> They increase by .37 confidence assessment: 3
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23:53:41 ** These numbers increase by .37 at each interval. **
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RESPONSE --> self critique assessment: 3
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23:57:45 If you didn't know the equation for the function, how would you go about finding the 100th member of the sequence? How can you tell your method is valid?
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RESPONSE --> We would just insert 100 in the function. Work the problem. confidence assessment: 2
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23:58:52 ** You could find the 100th member by noting that you have 99 ?umps?between the first number and the 100 th, each ?ump?being .37. Multiplying 99 times .37 and then adding the result to the 'starting value' (8.46). STUDENT RESPONSE: simply put 100 as the x in the formula .37x +8.09 INSTRUCTOR COMMENT: That's what you do if you have the equation. Given just the numbers you could find the 100th member by multiplying 99 times .37 and then adding the result to the first value 8.46. **
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RESPONSE --> I now understand how to go about finding the 100th member of the set. self critique assessment: 2
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00:04:07 for quadratic function y = g(x) = .01 x^2 - 2x + 100 what are the first five terms of the basic sequence {g(n), n = 1, 2, 3, ...}?
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RESPONSE --> The first five terms are 98.01, 96.04, 94.09, 92.16, 90.25 confidence assessment: 3
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00:04:23 ** We have g(1) = .01 * 1^2 - 2 * 1 + 100 = 98.01 g(2) = .01 * 2^2 - 2 * 2 + 100 = 96.04, etc. The first 5 terms are therefore {98.01, 96.04, 94.09, 92.16, 90.25}
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RESPONSE --> self critique assessment: 3
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00:05:47 What is the pattern of these numbers?
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RESPONSE --> They each decrease 1.97, 1.95, 1.93, 1.91 confidence assessment: 2
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00:05:59 ** The changes in these numbers are -1.97, -1.95, -1.93, -1.91. With each interval of x, the change in y is .02 greater than for the previous interval. **
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RESPONSE --> self critique assessment: 3
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00:07:39 ** According to the pattern established above, the next three changes are -1.89, -1.87, -1.85. This gives us g(6) = g(5) - 1.89, g(7) = g(6) - 1.87, g(7) = g(6) - 1.85. **
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RESPONSE --> Based on the explaination, I understand how to find the next three members. self critique assessment: 2
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00:08:38 How can you verify that your method is valid?
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RESPONSE --> You can actually test it out and plug in the next three numbers. confidence assessment: 2
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00:08:47 ** You can verify the result using the original formula; if you evaluate it at 5, 6 and 7 it should confirm your results. That's the best answer that can be given at this point. You should understand, though that even if you verified it for the first million terms, that wouldn't really prove it (who knows what might happen at the ten millionth term, or whatever). It turns out that to prove it would require calculus or the equivalent. **
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RESPONSE --> self critique assessment: 3
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00:14:01 query problem 15. The difference equation a(n+1) = a(n) + .4, a(1) = 5 If you substitute n = 1 into a(n+1) = a(n) + .4, how do you determine a(2)?
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RESPONSE --> Substitute 2 into the equation and get 6 confidence assessment: 1
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00:16:48 ** You get a(1+1) = a(1) + .4, or a(2) = a(1) + .4. Knowing a(1) = 5 you get a(2) = 5.4. **
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RESPONSE --> I missed the fact that it was .4 and not 4 That is what caused problem self critique assessment: 2
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00:22:43 If you substitute n = 2 into a(n+1) = a(n) + .4 how do you determine a(3)?
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RESPONSE --> You get a(2+1) = a(2) + .4 or a(3) = a(2) +.4 Knowing that a(2) = 5.4 then a(3) = 5.8 confidence assessment: 2
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00:22:48 ** You have to do the substitution. You get a(2+1) = a(2) + .4, or since 2 + 1 = 3, a(3) = a(2) + .4 Then knowing a(2) = 5.4 you get a(3) = 5.4 + .4 = 5.8. **
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RESPONSE --> self critique assessment: 3
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00:24:25 If you substitute n = 3 into a(n+1) = a(n) + .4, how do you determine a(4)?
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RESPONSE --> Just keep adding .4 at this rate a(4) will be 6.2 confidence assessment: 2
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00:24:30 ** We get a(4) = a(3) +.4 = 5.8 + .4 = 6.2 **
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RESPONSE --> self critique assessment: 3
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00:25:14 What is a(100)?
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RESPONSE --> (.4 * 100) + 5 40 + 5 45 confidence assessment: 2
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00:25:32 ** a(100) would be equal to a(1) plus 99 jumps of .4, or 5 + 99*.4 = 44.6. **
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RESPONSE --> I went one too far. self critique assessment: 3
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00:27:10 query problem 17. difference equation a(n+1) = a(n) + 2 n, a(1) = 4. What is the pattern of the sequence?
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RESPONSE --> Not sure since this shows an extra multiplication step confidence assessment: 1
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00:28:05 ** You get a(2) = a(1) + 2 * 1 = 4 + 2 = 6, then a(3) = a(2) + 2 * 3 = 6 + 6 = 12 then a(4) = a(3) + 2 * 4 = 12 + 8 = 20; etc. The sequence is 6, 12, 20, 30, 42, ... . **
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RESPONSE --> After reviewing the explaination I understand that I would just multiply by whatever n is. self critique assessment: 2
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00:28:25 What kind of function do you think a(n) is (e.g., linear, quadratic, exponential, etc.)?
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RESPONSE --> exponential confidence assessment: 2
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00:28:56 ** The differences of the sequence are 6, 8, 10, 12, . . .. The difference change by the same amount each time, which is a property of quadratic functions. **
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RESPONSE --> I was wrong. I thought that since they seemed to grow that it would be exponential. self critique assessment: 1
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00:30:53 query the slope = slope equation Explain the logic of the slope = slope equation (your may take a little time on this one)
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RESPONSE --> This is used if you only have one data point. confidence assessment: 1
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00:32:45 ** The slope = slope equation sets the slope between two given points equal to the slope between one of those points and the variable point (x, y). Since all three points lie on the same straight line, the slope between any two of the three points must be equal to the slope between any other pair. **
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RESPONSE --> This is what we have been saying all along. self critique assessment: 1
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00:39:21 query problem 7. streamRange(t), 50 centimeters at t = 20 seconds, range changes by -10 centimeters over 5 seconds. what is your function?
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RESPONSE --> Even after looking at the notes i am not sure how to seet up the function. confidence assessment: 1
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00:40:24 ** The rate at which streamRange changes is change in streamRange / change in t = -10 cm / (5 sec) = -2 cm/s. This will be the slope m of the graph. Since streamRange is 50 cm when t = 20 sec the point (20, 50) lies on the graph. So the graph passes through (20, 50) and has slope -2. The function is therefore of the form y = m t + b with m = -2, and b such that 50 = -2 * 20 + b. Thus b = 90. The function is therefore y = -2 t + 90, or using the meaningful name of the function steamRange(t) = -2t + 90 You need to use function notation. y = f(x) = -2x + 90 would be OK, or just f(x) = -2x + 90. The point is that you need to give the funcion a name. Another idea here is that we can use the 'word' streamRange to stand for the function. If you had 50 different functions and, for example, called them f1, f2, f3, ..., f50 you wouldn't remember which one was which so none of the function names would mean anything. If you call the function streamRange it has a meaning. Of course shorter words are sometimes preferable; just understand that function don't have to be confined to single letters and sometimes it's not a bad idea to make the names easily recognizable. STUDENT RESPONSE: y = -2x + 50 INSTRUCTOR COMMENT: ** At t = 20 sec this would give you y = -2 * 20 + 50 = 10. But y = 50 cm when t = 20 sec. Slope is -10 cm / (5 sec) = -2 cm/s, so you have y = -2 t + b. Plug in y = 50 cm and t = 20 sec and solve for b. You get b = 90 cm. The equation is y = -2 t + 90, or streamRange(t) = -2t + 90. **
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RESPONSE --> The explaination makes sense after looking at it. self critique assessment: 1
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00:42:34 what is the clock time at which the stream range first falls to 12 centimeters?
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RESPONSE --> 12 = -2t + 90 -78 = -2t 39 sec confidence assessment: 3
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00:42:38 ** Using the correct equation streamRange(t) = -2t + 90, you would set streamRange(t) = 12 and solve 12 = -2t + 90, obtaining t = 39 sec. **
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RESPONSE --> self critique assessment: 3
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00:54:59 query problem 9. equation of the straight line through t = 5 sec and the t = 7 sec points of the quadratic function depth(t) = .01 t^2 - 2t + 100 What is the slope and what does it tell you about the depth function?
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RESPONSE --> y = .01 (5)^2 - 2(5) + 100 y = .25 -10 + 100 y = -9.75 + 100 y = 90.25 y = .01 (7)^2 - 2(7) + 100 y = .49 - 14 +100 y = -13.51 + 100 y = 86.49 sub. 90.25 = 5m + b -86.49 = -7m - b 3.76 = -2m -1.88 m 90.25 = -1.88(5) + b 90.25 = -9.40 + b 99.65 = b y = -1.88x+99.65 The slope is -1.88 and it tells the change in depth over change in time. confidence assessment: 3
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00:56:40 ** You have to get the whole equation. y = m t + b is now y = -1.88 t + b. You have to solve for b. Plug in the coordinates of the t = 7 point and find b. You get 90.9 = -1.88 * 7 + b so b = 104, approximately. Find the correct value. The equation will end up something like y = depth(t) = -1.88 t + 104. **
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RESPONSE --> I don't know where you got 90.9 for the y value of x = 7. I keep getting 86.49 self critique assessment: 2
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00:57:53 ** The deviations are for t = 3, 4, 5, 6, 7, & 8 as follows: .08, .03. 0. -.01, 0, .03. **
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RESPONSE --> To be honest, I didn't do the deviations to know for sure how muh they are ""off"" from the originals. self critique assessment: 1
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00:58:59 at what t value do we obtain the closest values?
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RESPONSE --> I'm not exactly sure where I should plug what in. Maybe if I take another look at it tomorrow, I'll see what I need to do. confidence assessment: 0
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00:59:19 ** Not counting t= 5 and t = 7, which are 0, the next closest t value is t = 6, the deviation for this is -.01. **
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RESPONSE --> self critique assessment: 3
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07:07:04 On which side of the t = 5 and t = 7 points is the linear approximation closer to the quadratic function? On which side does the quadratic function 'curve away' from the linear most rapidly?
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RESPONSE --> It would be closer to the quadratic function at 5 and curve away at 7 confidence assessment: 2
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07:07:12 ** On the t = 4 side the approximation is closer. The quadratic function curves away on the positive x side. **
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RESPONSE --> self critique assessment: 3
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07:10:27 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I did not realize one assignment could be so long. It took me 6 hours to finish just this one assingment! I still have the Assignment 11 query to do. I hope it is not quite as long as this one. This was my surprise during the assignment. self critique assessment: 2
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