EP PRECALC.GRAPHING APPR.-WEBASSIGN-1YR
8th Edition
ISBN: 9780357022078
Author: Larson
Publisher: CENGAGE L
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Question
Chapter 2.8, Problem 13E
a
To determine
To draw
a
Expert Solution
Explanation of Solution
Given information:
Graph:
Interpretation:
Using a graphic utility, the scatter plot for the given data is shown above.
b
To determine
To find if the scatter plot could be modeled by linear model, quadratic model or neither.
b
Expert Solution
Answer to Problem 13E
Linear model
Explanation of Solution
Given information:
To determine if the given graph can be modeled by linear model, quadratic model or neither of them, try to draw a straight line or a parabola through the given scatter plot.
If a straight line can be drawn through the points of the scatter plot, it could be modelled by linear model whereas if a parabola can be drawn through the points of the scatter plot, it could be modelled by a quadratic model.
In case if both are not possible, it could neither be modeled.
Here, in the given graph we could draw a straight line. Therefore, the scatter plot could be modeled by a linear model.
Conclusion:
Therefore, given scatter plot is modeled by linear model.
c
To determine
To find a model for the data using regression feature of a graphing utility.
c
Expert Solution
Answer to Problem 13E
Linear model
Explanation of Solution
Given information:
Calculation:
Using the graphic utility to find the regression,
Conclusion:
Therefore, from the above figure. the regression equation for the linear model is
d
To determine
To draw the model with the scatter plot from subpart (a) using a graphic utility.
d
Expert Solution
Explanation of Solution
Given information:
Graph:
Interpretation:
Using a graphic utility, a straight line is formed when the data is kept on a graph.
e
To determine
To draw a table comparing the original data with the data given by the model.
e
Expert Solution
Explanation of Solution
Given information:
Table:
Draw the table comparing the original data and the data given by the model.
| Original data | Data from the model | ||
| x | y | ||
| 1 | 4.0 | 1 | 3.60364 |
| 2 | 6.5 | 2 | 6.08061 |
| 3 | 8.8 | 3 | 8.55758 |
| 4 | 10.6 | 4 | 11.03455 |
| 5 | 13.9 | 5 | 13.51152 |
| 6 | 15.0 | 6 | 15.98849 |
| 7 | 17.5 | 7 | 18.46546 |
| 8 | 20.1 | 8 | 20.94243 |
| 9 | 24.0 | 9 | 23.4194 |
| 10 | 27.1 | 10 | 25.89637 |
Data from the model is obtained by substituting the values of x as
Interpretation:
When the original data and the data from the model are compared with each other, it is found that the values are nearly equal.
Chapter 2 Solutions
EP PRECALC.GRAPHING APPR.-WEBASSIGN-1YR
Ch. 2.1 - Prob. 1ECh. 2.1 - Prob. 2ECh. 2.1 - Prob. 3ECh. 2.1 - Prob. 4ECh. 2.1 - Prob. 5ECh. 2.1 - Prob. 6ECh. 2.1 - Prob. 7ECh. 2.1 - Prob. 8ECh. 2.1 - Prob. 9ECh. 2.1 - Prob. 10E
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Prob. 18ECh. 2.7 - Prob. 19ECh. 2.7 - Prob. 20ECh. 2.7 - Prob. 21ECh. 2.7 - Prob. 22ECh. 2.7 - Prob. 23ECh. 2.7 - Prob. 24ECh. 2.7 - Prob. 25ECh. 2.7 - Prob. 26ECh. 2.7 - Prob. 27ECh. 2.7 - Prob. 28ECh. 2.7 - Prob. 29ECh. 2.7 - Prob. 30ECh. 2.7 - Prob. 31ECh. 2.7 - Prob. 32ECh. 2.7 - Prob. 33ECh. 2.7 - Prob. 34ECh. 2.7 - Prob. 35ECh. 2.7 - Prob. 36ECh. 2.7 - Prob. 37ECh. 2.7 - Prob. 38ECh. 2.7 - Prob. 39ECh. 2.7 - Prob. 40ECh. 2.7 - Prob. 41ECh. 2.7 - Prob. 42ECh. 2.7 - Prob. 43ECh. 2.7 - Prob. 44ECh. 2.7 - Prob. 45ECh. 2.7 - Prob. 46ECh. 2.7 - Prob. 47ECh. 2.7 - Prob. 48ECh. 2.7 - Prob. 49ECh. 2.7 - Prob. 50ECh. 2.7 - Prob. 51ECh. 2.7 - Prob. 52ECh. 2.7 - Prob. 53ECh. 2.7 - Prob. 54ECh. 2.7 - Prob. 55ECh. 2.7 - Prob. 56ECh. 2.7 - Prob. 57ECh. 2.7 - Prob. 58ECh. 2.7 - Prob. 59ECh. 2.7 - Prob. 60ECh. 2.7 - Prob. 61ECh. 2.7 - Prob. 62ECh. 2.7 - Prob. 63ECh. 2.7 - Prob. 64ECh. 2.7 - Prob. 65ECh. 2.7 - Prob. 66ECh. 2.7 - Prob. 67ECh. 2.7 - Prob. 68ECh. 2.7 - Prob. 69ECh. 2.7 - Prob. 70ECh. 2.7 - Prob. 71ECh. 2.7 - Prob. 72ECh. 2.7 - Prob. 73ECh. 2.7 - Prob. 74ECh. 2.7 - Prob. 75ECh. 2.7 - Prob. 76ECh. 2.7 - Prob. 77ECh. 2.7 - Prob. 78ECh. 2.7 - Prob. 79ECh. 2.7 - Prob. 80ECh. 2.7 - Prob. 81ECh. 2.7 - Prob. 82ECh. 2.7 - Prob. 83ECh. 2.7 - Prob. 84ECh. 2.7 - Prob. 85ECh. 2.7 - Prob. 86ECh. 2.7 - Prob. 87ECh. 2.7 - Prob. 88ECh. 2.7 - Prob. 89ECh. 2.7 - Prob. 90ECh. 2.7 - Prob. 91ECh. 2.7 - Prob. 92ECh. 2.7 - Prob. 93ECh. 2.7 - Prob. 94ECh. 2.7 - Prob. 95ECh. 2.7 - Prob. 96ECh. 2.7 - Prob. 97ECh. 2.7 - Prob. 98ECh. 2.7 - Prob. 99ECh. 2.7 - Prob. 100ECh. 2.7 - Prob. 101ECh. 2.8 - Prob. 1ECh. 2.8 - Prob. 2ECh. 2.8 - Prob. 3ECh. 2.8 - Prob. 4ECh. 2.8 - Prob. 5ECh. 2.8 - Prob. 6ECh. 2.8 - Prob. 7ECh. 2.8 - Prob. 8ECh. 2.8 - Prob. 9ECh. 2.8 - Prob. 10ECh. 2.8 - Prob. 11ECh. 2.8 - Prob. 12ECh. 2.8 - Prob. 13ECh. 2.8 - Prob. 14ECh. 2.8 - Prob. 15ECh. 2.8 - Prob. 16ECh. 2.8 - Prob. 17ECh. 2.8 - Prob. 18ECh. 2.8 - Prob. 19ECh. 2.8 - Prob. 20ECh. 2.8 - Prob. 21ECh. 2.8 - Prob. 22ECh. 2.8 - Prob. 23ECh. 2.8 - Prob. 24ECh. 2.8 - Prob. 25ECh. 2.8 - Prob. 26ECh. 2.8 - Prob. 27ECh. 2.8 - Prob. 28ECh. 2.8 - Prob. 29ECh. 2.8 - Prob. 30ECh. 2.8 - Prob. 31ECh. 2.8 - Prob. 32ECh. 2.8 - Prob. 33ECh. 2.8 - Prob. 34ECh. 2.8 - Prob. 35ECh. 2.8 - Prob. 36ECh. 2.8 - Prob. 37ECh. 2.8 - Prob. 38ECh. 2.8 - Prob. 39ECh. 2.8 - Prob. 40ECh. 2 - Prob. 1CRCh. 2 - Prob. 2CRCh. 2 - Prob. 3CRCh. 2 - Prob. 4CRCh. 2 - Prob. 5CRCh. 2 - Prob. 6CRCh. 2 - Prob. 7CRCh. 2 - Prob. 8CRCh. 2 - Prob. 9CRCh. 2 - Prob. 10CRCh. 2 - Prob. 11CRCh. 2 - Prob. 12CRCh. 2 - Prob. 13CRCh. 2 - Prob. 14CRCh. 2 - Prob. 15CRCh. 2 - Prob. 16CRCh. 2 - Prob. 17CRCh. 2 - Prob. 18CRCh. 2 - Prob. 19CRCh. 2 - Prob. 20CRCh. 2 - Prob. 21CRCh. 2 - Prob. 22CRCh. 2 - Prob. 23CRCh. 2 - Prob. 24CRCh. 2 - Prob. 25CRCh. 2 - Prob. 26CRCh. 2 - Prob. 27CRCh. 2 - Prob. 28CRCh. 2 - Prob. 29CRCh. 2 - Prob. 30CRCh. 2 - Prob. 31CRCh. 2 - Prob. 32CRCh. 2 - Prob. 33CRCh. 2 - Prob. 34CRCh. 2 - Prob. 35CRCh. 2 - Prob. 36CRCh. 2 - Prob. 37CRCh. 2 - Prob. 38CRCh. 2 - Prob. 39CRCh. 2 - Prob. 40CRCh. 2 - Prob. 41CRCh. 2 - Prob. 42CRCh. 2 - Prob. 43CRCh. 2 - Prob. 44CRCh. 2 - Prob. 45CRCh. 2 - Prob. 46CRCh. 2 - Prob. 47CRCh. 2 - Prob. 48CRCh. 2 - Prob. 49CRCh. 2 - Prob. 50CRCh. 2 - Prob. 51CRCh. 2 - Prob. 52CRCh. 2 - Prob. 53CRCh. 2 - Prob. 54CRCh. 2 - Prob. 55CRCh. 2 - Prob. 56CRCh. 2 - Prob. 57CRCh. 2 - Prob. 58CRCh. 2 - Prob. 59CRCh. 2 - Prob. 60CRCh. 2 - Prob. 61CRCh. 2 - Prob. 62CRCh. 2 - Prob. 63CRCh. 2 - Prob. 64CRCh. 2 - Prob. 65CRCh. 2 - Prob. 66CRCh. 2 - Prob. 67CRCh. 2 - Prob. 68CRCh. 2 - Prob. 69CRCh. 2 - Prob. 70CRCh. 2 - Prob. 71CRCh. 2 - Prob. 72CRCh. 2 - Prob. 73CRCh. 2 - Prob. 74CRCh. 2 - Prob. 75CRCh. 2 - Prob. 76CRCh. 2 - Prob. 77CRCh. 2 - Prob. 78CRCh. 2 - Prob. 79CRCh. 2 - Prob. 80CRCh. 2 - Prob. 81CRCh. 2 - Prob. 82CRCh. 2 - Prob. 83CRCh. 2 - Prob. 84CRCh. 2 - Prob. 85CRCh. 2 - Prob. 86CRCh. 2 - Prob. 87CRCh. 2 - Prob. 88CRCh. 2 - Prob. 89CRCh. 2 - Prob. 90CRCh. 2 - Prob. 91CRCh. 2 - Prob. 92CRCh. 2 - Prob. 93CRCh. 2 - Prob. 94CRCh. 2 - Prob. 95CRCh. 2 - Prob. 96CRCh. 2 - Prob. 97CRCh. 2 - Prob. 98CRCh. 2 - Prob. 99CRCh. 2 - Prob. 100CRCh. 2 - Prob. 101CRCh. 2 - Prob. 102CRCh. 2 - Prob. 103CRCh. 2 - Prob. 104CRCh. 2 - Prob. 105CRCh. 2 - Prob. 106CRCh. 2 - Prob. 107CRCh. 2 - Prob. 108CRCh. 2 - Prob. 109CRCh. 2 - Prob. 110CRCh. 2 - Prob. 111CRCh. 2 - Prob. 112CRCh. 2 - Prob. 113CRCh. 2 - Prob. 114CRCh. 2 - Prob. 115CRCh. 2 - Prob. 116CRCh. 2 - Prob. 117CRCh. 2 - Prob. 118CRCh. 2 - Prob. 119CRCh. 2 - Prob. 120CRCh. 2 - Prob. 121CRCh. 2 - Prob. 122CRCh. 2 - Prob. 123CRCh. 2 - Prob. 124CRCh. 2 - Prob. 125CRCh. 2 - Prob. 126CRCh. 2 - Prob. 127CRCh. 2 - Prob. 128CRCh. 2 - Prob. 129CRCh. 2 - Prob. 130CRCh. 2 - Prob. 131CRCh. 2 - Prob. 132CRCh. 2 - Prob. 133CRCh. 2 - Prob. 134CRCh. 2 - Prob. 135CRCh. 2 - Prob. 136CRCh. 2 - Prob. 137CRCh. 2 - Prob. 138CRCh. 2 - Prob. 139CRCh. 2 - Prob. 140CRCh. 2 - Prob. 141CRCh. 2 - Prob. 142CRCh. 2 - Prob. 143CRCh. 2 - Prob. 144CRCh. 2 - Prob. 145CRCh. 2 - Prob. 146CRCh. 2 - Prob. 147CRCh. 2 - Prob. 148CRCh. 2 - Prob. 149CRCh. 2 - Prob. 150CRCh. 2 - Prob. 151CRCh. 2 - Prob. 152CRCh. 2 - Prob. 153CRCh. 2 - Prob. 154CRCh. 2 - Prob. 155CRCh. 2 - Prob. 156CRCh. 2 - Prob. 157CRCh. 2 - Prob. 158CRCh. 2 - Prob. 159CRCh. 2 - Prob. 160CRCh. 2 - Prob. 161CRCh. 2 - Prob. 162CRCh. 2 - Prob. 163CRCh. 2 - Prob. 164CRCh. 2 - Prob. 165CRCh. 2 - Prob. 166CRCh. 2 - Prob. 1CTCh. 2 - Prob. 2CTCh. 2 - Prob. 3CTCh. 2 - Prob. 4CTCh. 2 - Prob. 5CTCh. 2 - Prob. 6CTCh. 2 - Prob. 7CTCh. 2 - Prob. 8CTCh. 2 - Prob. 9CTCh. 2 - Prob. 10CTCh. 2 - Prob. 11CTCh. 2 - Prob. 12CTCh. 2 - Prob. 13CTCh. 2 - Prob. 14CTCh. 2 - Prob. 15CTCh. 2 - Prob. 16CTCh. 2 - Prob. 17CTCh. 2 - Prob. 18CTCh. 2 - Prob. 19CTCh. 2 - Prob. 20CTCh. 2 - Prob. 21CT
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- This question builds on an earlier problem. The randomized numbers may have changed, but have your work for the previous problem available to help with this one. A 4-centimeter rod is attached at one end to a point A rotating counterclockwise on a wheel of radius 2 cm. The other end B is free to move back and forth along a horizontal bar that goes through the center of the wheel. At time t=0 the rod is situated as in the diagram at the left below. The wheel rotates counterclockwise at 1.5 rev/sec. At some point, the rod will be tangent to the circle as shown in the third picture. A B A B at some instant, the piston will be tangent to the circle (a) Express the x and y coordinates of point A as functions of t: x= 2 cos(3πt) and y= 2 sin(3t) (b) Write a formula for the slope of the tangent line to the circle at the point A at time t seconds: -cot(3πt) sin(3лt) (c) Express the x-coordinate of the right end of the rod at point B as a function of t: 2 cos(3πt) +411- 4 -2 sin (3лt) (d)…arrow_forward5. [-/1 Points] DETAILS MY NOTES SESSCALCET2 6.5.AE.003. y y= ex² 0 Video Example x EXAMPLE 3 (a) Use the Midpoint Rule with n = 10 to approximate the integral कर L'ex² dx. (b) Give an upper bound for the error involved in this approximation. SOLUTION 8+2 1 L'ex² d (a) Since a = 0, b = 1, and n = 10, the Midpoint Rule gives the following. (Round your answer to six decimal places.) dx Ax[f(0.05) + f(0.15) + ... + f(0.85) + f(0.95)] 0.1 [0.0025 +0.0225 + + e0.0625 + 0.1225 e0.3025 + e0.4225 + e0.2025 + + e0.5625 €0.7225 +0.9025] The figure illustrates this approximation. (b) Since f(x) = ex², we have f'(x) = 0 ≤ f'(x) = < 6e. ASK YOUR TEACHER and f'(x) = Also, since 0 ≤ x ≤ 1 we have x² ≤ and so Taking K = 6e, a = 0, b = 1, and n = 10 in the error estimate, we see that an upper bound for the error is as follows. (Round your final answer to five decimal places.) 6e(1)3 e 24( = ≈arrow_forward2. [-/1 Points] DETAILS MY NOTES SESSCALCET2 6.5.015. Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.) ASK YOUR TEACHER 3 1 3 + dy, n = 6 (a) the Trapezoidal Rule (b) the Midpoint Rule (c) Simpson's Rule Need Help? Read It Watch Itarrow_forward
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