ELEMENTARY+INTERMEDIATE ALGEBRA
6th Edition
ISBN: 9781285074672
Author: Larson
Publisher: CENGAGE L
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Question
Chapter 1.1, Problem 40E
a.
To determine
Determine the variable expression for a student’s final score.
a.
Expert Solution
Answer to Problem 40E
The variable expressions for a student’s final score are as follows:
Final score of student A is
Final score of student B is
Final score of student C is
Explanation of Solution
Given information:
Consider, the bar graph provided in the question,
Student A earn 30 points for technique and 40 points for interpretation.
Student B earn 40 points for technique and 20 points for interpretation.
Student A earn 30 points for technique and 50 points for interpretation.
Calculation:
Let,
The variable expressions for a student’s final score are as follows:
Final score of student A is
Final score of student B is
Final score of student C is
b.
To determine
To find the points which is needed to earn for interpretation so that the final score will be greatest of all students.
b.
Expert Solution
Answer to Problem 40E
The final score is obtained as:
Final score of student A is
Final score of student B is
Final score of student C is
Explanation of Solution
Given information:
Consider, the bar graph provided in the question,
Student A earn 30 points for technique and 40 points for interpretation.
Student B earn 40 points for technique and 20 points for interpretation.
Student A earn 30 points for technique and 50 points for interpretation.
Calculation:
Let,
The final score is obtained as:
Final score of student A is:
Final score of student B is:
Final score of student C is:
c.
To determine
To find the final score of each student.
c.
Expert Solution
Answer to Problem 40E
The student must have to earn point greater than 45 for interpretation so as to obtain highest final score.
Explanation of Solution
Given information:
Consider, the bar graph provided in the question,
Student A earn 30 points for technique and 40 points for interpretation.
Student B earn 40 points for technique and 20 points for interpretation.
Student A earn 30 points for technique and 50 points for interpretation.
Calculation:
Let,
The final score is obtained as:
Final score of student A is
Final score of student B is
Final score of student C is
The highest final score is earned by student C which is 80.
Therefore,
The student must have to earn point greater than
Chapter 1 Solutions
ELEMENTARY+INTERMEDIATE ALGEBRA
Ch. 1.1 - Prob. 1CCh. 1.1 - Prob. 2CCh. 1.1 - Prob. 3CCh. 1.1 - Prob. 4CCh. 1.1 - Prob. 1ECh. 1.1 - Prob. 2ECh. 1.1 - Prob. 3ECh. 1.1 - Prob. 4ECh. 1.1 - Prob. 5ECh. 1.1 - Prob. 6E
Ch. 1.1 - Prob. 7ECh. 1.1 - Prob. 8ECh. 1.1 - Prob. 9ECh. 1.1 - Prob. 10ECh. 1.1 - Prob. 11ECh. 1.1 - Prob. 12ECh. 1.1 - Prob. 13ECh. 1.1 - Prob. 14ECh. 1.1 - Prob. 15ECh. 1.1 - Prob. 16ECh. 1.1 - Prob. 17ECh. 1.1 - Prob. 18ECh. 1.1 - Prob. 19ECh. 1.1 - Prob. 20ECh. 1.1 - Prob. 21ECh. 1.1 - Prob. 22ECh. 1.1 - Prob. 23ECh. 1.1 - Prob. 24ECh. 1.1 - Prob. 25ECh. 1.1 - Prob. 26ECh. 1.1 - Prob. 27ECh. 1.1 - Prob. 28ECh. 1.1 - Prob. 29ECh. 1.1 - Prob. 30ECh. 1.1 - Prob. 31ECh. 1.1 - Prob. 32ECh. 1.1 - Prob. 33ECh. 1.1 - Prob. 34ECh. 1.1 - Prob. 35ECh. 1.1 - Prob. 36ECh. 1.1 - Prob. 37ECh. 1.1 - Prob. 38ECh. 1.1 - Prob. 39ECh. 1.1 - Prob. 40ECh. 1.1 - Prob. 41ECh. 1.1 - Prob. 42ECh. 1.1 - Prob. 43ECh. 1.1 - Prob. 44ECh. 1.1 - Prob. 45ECh. 1.1 - Prob. 46ECh. 1.1 - Prob. 47ECh. 1.1 - Prob. 48ECh. 1.1 - Prob. 49ECh. 1.1 - Prob. 50ECh. 1.1 - Prob. 51ECh. 1.1 - Prob. 52ECh. 1.1 - Prob. 53ECh. 1.1 - Prob. 54ECh. 1.1 - Prob. 55ECh. 1.1 - Prob. 56ECh. 1.1 - Prob. 57ECh. 1.1 - Prob. 58ECh. 1.1 - Prob. 59ECh. 1.1 - Prob. 60ECh. 1.1 - Prob. 61ECh. 1.1 - Prob. 62ECh. 1.1 - Prob. 63ECh. 1.1 - Prob. 64ECh. 1.1 - Prob. 65ECh. 1.1 - Prob. 66ECh. 1.1 - Prob. 67ECh. 1.2 - Prob. 1CCh. 1.2 - Prob. 2CCh. 1.2 - Prob. 3CCh. 1.2 - Prob. 4CCh. 1.2 - Prob. 5CCh. 1.2 - Prob. 6CCh. 1.2 - Prob. 7CCh. 1.2 - Prob. 8CCh. 1.2 - Prob. 9CCh. 1.2 - Prob. 10CCh. 1.2 - Prob. 11CCh. 1.2 - Prob. 1ECh. 1.2 - Prob. 2ECh. 1.2 - Prob. 3ECh. 1.2 - Prob. 4ECh. 1.2 - Prob. 5ECh. 1.2 - Prob. 6ECh. 1.2 - Prob. 7ECh. 1.2 - Prob. 8ECh. 1.2 - Prob. 9ECh. 1.2 - Prob. 10ECh. 1.2 - Prob. 11ECh. 1.2 - Prob. 12ECh. 1.2 - Prob. 13ECh. 1.2 - Prob. 14ECh. 1.2 - Prob. 15ECh. 1.2 - Prob. 16ECh. 1.2 - Prob. 17ECh. 1.2 - Prob. 18ECh. 1.2 - Prob. 19ECh. 1.2 - Prob. 20ECh. 1.2 - Prob. 21ECh. 1.2 - Prob. 22ECh. 1.2 - Prob. 23ECh. 1.2 - Prob. 24ECh. 1.2 - Prob. 25ECh. 1.2 - Prob. 26ECh. 1.2 - Prob. 27ECh. 1.2 - Prob. 28ECh. 1.2 - Prob. 29ECh. 1.2 - Prob. 30ECh. 1.2 - Prob. 31ECh. 1.2 - Prob. 32ECh. 1.2 - Prob. 33ECh. 1.2 - Prob. 34ECh. 1.2 - Prob. 35ECh. 1.2 - Prob. 36ECh. 1.2 - Prob. 37ECh. 1.2 - Prob. 38ECh. 1.2 - Prob. 39ECh. 1.2 - Prob. 40ECh. 1.2 - Prob. 41ECh. 1.2 - Prob. 42ECh. 1.2 - Prob. 43ECh. 1.2 - Prob. 44ECh. 1.2 - Prob. 45ECh. 1.2 - Prob. 46ECh. 1.2 - Prob. 47ECh. 1.2 - Prob. 48ECh. 1.2 - Prob. 49ECh. 1.3 - Prob. 1CCh. 1.3 - Prob. 2CCh. 1.3 - Prob. 3CCh. 1.3 - Prob. 4CCh. 1.3 - Prob. 5CCh. 1.3 - Prob. 6CCh. 1.3 - Prob. 7CCh. 1.3 - Prob. 8CCh. 1.3 - Prob. 9CCh. 1.3 - Prob. 10CCh. 1.3 - Prob. 11CCh. 1.3 - Prob. 12CCh. 1.3 - Prob. 1ECh. 1.3 - Prob. 2ECh. 1.3 - Prob. 3ECh. 1.3 - Prob. 4ECh. 1.3 - Prob. 5ECh. 1.3 - Prob. 6ECh. 1.3 - Prob. 7ECh. 1.3 - Prob. 8ECh. 1.3 - Prob. 9ECh. 1.3 - Prob. 10ECh. 1.3 - Prob. 11ECh. 1.3 - Prob. 12ECh. 1.3 - Prob. 13ECh. 1.3 - Prob. 14ECh. 1.3 - Prob. 15ECh. 1.3 - Prob. 16ECh. 1.3 - Prob. 17ECh. 1.3 - Prob. 18ECh. 1.3 - Prob. 19ECh. 1.3 - Prob. 20ECh. 1.3 - Prob. 21ECh. 1.3 - Prob. 22ECh. 1.3 - Prob. 23ECh. 1.3 - Prob. 24ECh. 1.3 - Prob. 25ECh. 1.3 - Prob. 26ECh. 1.3 - Prob. 27ECh. 1.3 - Prob. 28ECh. 1.3 - Prob. 29ECh. 1.3 - Prob. 30ECh. 1.3 - Prob. 31ECh. 1.3 - Prob. 32ECh. 1.3 - Prob. 33ECh. 1.3 - Prob. 34ECh. 1.3 - Prob. 35ECh. 1.3 - Prob. 36ECh. 1.3 - Prob. 37ECh. 1.3 - Prob. 38ECh. 1.3 - Prob. 39ECh. 1.3 - Prob. 40ECh. 1.3 - Prob. 41ECh. 1.3 - Prob. 42ECh. 1.3 - Prob. 43ECh. 1.3 - Prob. 44ECh. 1.4 - Prob. 1CCh. 1.4 - Prob. 2CCh. 1.4 - Prob. 3CCh. 1.4 - Prob. 4CCh. 1.4 - Prob. 5CCh. 1.4 - Prob. 6CCh. 1.4 - Prob. 7CCh. 1.4 - Prob. 8CCh. 1.4 - Prob. 9CCh. 1.4 - Prob. 1ECh. 1.4 - Prob. 2ECh. 1.4 - Prob. 3ECh. 1.4 - Prob. 4ECh. 1.4 - Prob. 5ECh. 1.4 - Prob. 6ECh. 1.4 - Prob. 7ECh. 1.4 - Prob. 8ECh. 1.4 - Prob. 9ECh. 1.4 - Prob. 10ECh. 1.4 - Prob. 11ECh. 1.4 - Prob. 12ECh. 1.4 - Prob. 13ECh. 1.4 - Prob. 14ECh. 1.4 - Prob. 15ECh. 1.4 - Prob. 16ECh. 1.4 - Prob. 17ECh. 1.4 - Prob. 18ECh. 1.4 - Prob. 19ECh. 1.4 - Prob. 20ECh. 1.4 - Prob. 21ECh. 1.4 - Prob. 22ECh. 1.4 - Prob. 23ECh. 1.4 - Prob. 24ECh. 1.4 - Prob. 25ECh. 1.4 - Prob. 26ECh. 1.4 - Prob. 27ECh. 1.4 - Prob. 28ECh. 1.4 - Prob. 29ECh. 1.4 - Prob. 30ECh. 1.4 - Prob. 31ECh. 1.4 - Prob. 32ECh. 1.4 - Prob. 33ECh. 1.4 - Prob. 34ECh. 1.4 - Prob. 35ECh. 1.4 - Prob. 36ECh. 1.4 - Prob. 37ECh. 1.4 - Prob. 38ECh. 1.4 - Prob. 39ECh. 1.4 - Prob. 40ECh. 1.4 - Prob. 41ECh. 1.4 - Prob. 42ECh. 1.4 - Prob. 43ECh. 1.4 - Prob. 44ECh. 1.4 - Prob. 45ECh. 1.4 - Prob. 46ECh. 1.4 - Prob. 47ECh. 1.4 - Prob. 48ECh. 1.4 - Prob. 49ECh. 1.4 - Prob. 50ECh. 1.4 - Prob. 51ECh. 1.4 - Prob. 52ECh. 1.4 - Prob. 53ECh. 1.4 - Prob. 54ECh. 1.4 - Prob. 55ECh. 1.4 - Prob. 56ECh. 1.4 - Prob. 57ECh. 1.4 - Prob. 58ECh. 1.4 - Prob. 59ECh. 1.4 - Prob. 60ECh. 1.4 - Prob. 61ECh. 1.4 - Prob. 62ECh. 1.4 - Prob. 63ECh. 1.4 - Prob. 64ECh. 1.4 - Prob. 65ECh. 1.4 - Prob. 66ECh. 1.4 - Prob. 67ECh. 1.4 - Prob. 68ECh. 1.4 - Prob. 69ECh. 1.4 - Prob. 70ECh. 1.4 - Prob. 71ECh. 1.4 - Prob. 72ECh. 1.4 - Prob. 73ECh. 1.4 - Prob. 74ECh. 1.4 - Prob. 75ECh. 1.4 - Prob. 76ECh. 1.5 - Prob. 1CCh. 1.5 - Prob. 2CCh. 1.5 - Prob. 3CCh. 1.5 - Prob. 4CCh. 1.5 - Prob. 5CCh. 1.5 - Prob. 6CCh. 1.5 - Prob. 7CCh. 1.5 - Prob. 8CCh. 1.5 - Prob. 9CCh. 1.5 - Prob. 1ECh. 1.5 - Prob. 2ECh. 1.5 - Prob. 3ECh. 1.5 - Prob. 4ECh. 1.5 - Prob. 5ECh. 1.5 - Prob. 6ECh. 1.5 - Prob. 7ECh. 1.5 - Prob. 8ECh. 1.5 - Prob. 9ECh. 1.5 - Prob. 10ECh. 1.5 - Prob. 11ECh. 1.5 - Prob. 12ECh. 1.5 - Prob. 13ECh. 1.5 - Prob. 14ECh. 1.5 - Prob. 15ECh. 1.5 - Prob. 16ECh. 1.5 - Prob. 17ECh. 1.5 - Prob. 18ECh. 1.5 - Prob. 19ECh. 1.5 - Prob. 20ECh. 1.5 - Prob. 21ECh. 1.5 - Prob. 22ECh. 1.5 - Prob. 23ECh. 1.5 - Prob. 24ECh. 1.5 - Prob. 25ECh. 1.5 - Prob. 26ECh. 1.5 - Prob. 27ECh. 1.5 - Prob. 28ECh. 1.5 - Prob. 29ECh. 1.5 - Prob. 30ECh. 1.5 - Prob. 31ECh. 1.5 - Prob. 32ECh. 1.5 - Prob. 33ECh. 1.5 - Prob. 34ECh. 1.5 - Prob. 35ECh. 1.5 - Prob. 36ECh. 1.5 - Prob. 37ECh. 1.5 - Prob. 38ECh. 1.5 - Prob. 39ECh. 1.5 - Prob. 40ECh. 1.5 - Prob. 41ECh. 1.5 - Prob. 42ECh. 1.5 - Prob. 43ECh. 1.5 - Prob. 44ECh. 1.5 - Prob. 45ECh. 1.5 - Prob. 46ECh. 1.5 - Prob. 47ECh. 1.5 - Prob. 48ECh. 1.5 - Prob. 49ECh. 1.5 - Prob. 50ECh. 1.5 - Prob. 51ECh. 1.5 - Prob. 52ECh. 1.5 - Prob. 53ECh. 1.5 - Prob. 54ECh. 1.5 - Prob. 55ECh. 1.5 - Prob. 56ECh. 1.5 - Prob. 57ECh. 1.5 - Prob. 58ECh. 1.5 - Prob. 59ECh. 1.5 - Prob. 60ECh. 1.5 - Prob. 61ECh. 1.5 - Prob. 62ECh. 1.5 - Prob. 63ECh. 1.5 - Prob. 64ECh. 1.5 - Prob. 65ECh. 1.5 - Prob. 66ECh. 1.5 - Prob. 67ECh. 1.5 - Prob. 68ECh. 1.5 - Prob. 69ECh. 1.6 - Prob. 1CCh. 1.6 - Prob. 2CCh. 1.6 - Prob. 3CCh. 1.6 - Prob. 4CCh. 1.6 - Prob. 5CCh. 1.6 - Prob. 6CCh. 1.6 - Prob. 7CCh. 1.6 - Prob. 8CCh. 1.6 - Prob. 9CCh. 1.6 - Prob. 10CCh. 1.6 - Prob. 11CCh. 1.6 - Prob. 12CCh. 1.6 - Prob. 1ECh. 1.6 - Prob. 2ECh. 1.6 - Prob. 3ECh. 1.6 - Prob. 4ECh. 1.6 - Prob. 5ECh. 1.6 - Prob. 6ECh. 1.6 - Prob. 7ECh. 1.6 - Prob. 8ECh. 1.6 - Prob. 9ECh. 1.6 - Prob. 10ECh. 1.6 - Prob. 11ECh. 1.6 - Prob. 12ECh. 1.6 - Prob. 13ECh. 1.6 - Prob. 14ECh. 1.6 - Prob. 15ECh. 1.6 - Prob. 16ECh. 1.6 - Prob. 17ECh. 1.6 - Prob. 18ECh. 1.6 - Prob. 19ECh. 1.6 - Prob. 20ECh. 1.6 - Prob. 21ECh. 1.6 - Prob. 22ECh. 1.6 - Prob. 23ECh. 1.6 - Prob. 24ECh. 1.6 - 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Prob. 26CTCh. 1 - Prob. 27CTCh. 1 - Prob. 28CTCh. 1 - Prob. 29CTCh. 1 - Prob. 30CTCh. 1 - Prob. 31CTCh. 1 - Prob. 32CTCh. 1 - Prob. 33CTCh. 1 - Prob. 34CTCh. 1 - Prob. 35CTCh. 1 - Prob. 36CTCh. 1 - Prob. 37CTCh. 1 - Prob. 38CTCh. 1 - Prob. 39CTCh. 1 - Prob. 1CSTCh. 1 - Prob. 2CSTCh. 1 - Prob. 3CSTCh. 1 - Prob. 4CSTCh. 1 - Prob. 5CSTCh. 1 - Prob. 6CSTCh. 1 - Prob. 7CSTCh. 1 - Prob. 8CSTCh. 1 - Prob. 9CSTCh. 1 - Prob. 10CSTCh. 1 - Prob. 11CSTCh. 1 - Prob. 12CSTCh. 1 - Prob. 13CSTCh. 1 - Prob. 1PSPCh. 1 - Prob. 2PSPCh. 1 - Prob. 3PSPCh. 1 - Prob. 4PSPCh. 1 - Prob. 5PSPCh. 1 - Prob. 6PSPCh. 1 - Prob. 7PSP
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- How much is the circumference of a circle whose diameter is 7 feet?C =π darrow_forwardHow to solve 2542/64.132arrow_forwardAssume that you fancy polynomial splines, while you actually need ƒ(t) = e²/3 – 1 for t€ [−1, 1]. See the figure for a plot of f(t). Your goal is to approximate f(t) with an inter- polating polynomial spline of degree d that is given as sa(t) = • Σk=0 Pd,k bd,k(t) so that sd(tk) = = Pd,k for tk = −1 + 2 (given d > 0) with basis functions bd,k(t) = Σi±0 Cd,k,i = • The special case of d 0 is trivial: the only basis function b0,0 (t) is constant 1 and so(t) is thus constant po,0 for all t = [−1, 1]. ...9 The d+1 basis functions bd,k (t) form a ba- sis Bd {ba,o(t), ba,1(t), bd,d(t)} of the function space of all possible sα (t) functions. Clearly, you wish to find out, which of them given a particular maximal degree d is the best-possible approximation of f(t) in the least- squares sense. _ 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -0.1 -0.2 -0.3 -0.4 -0.5 -0.6 -0.7 -0.8 -0.9 -1 function f(t) = exp((2t)/3) - 1 to project -1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5…arrow_forward
- An image processor considered a 750×750 pixels large subset of an image and converted it into gray-scale, resulting in matrix gIn - a false-color visualization of gIn is shown in the top-left below. He prepared a two-dim. box filter f1 as a 25×25 matrix with only the 5×5 values in the middle being non-zero – this filter is shown in the top-middle position below. He then convolved £1 with itself to get £2, before convolving £2 with itself to get f3. In both of the steps, he maintained the 25×25 size. Next, he convolved gIn with £3 to get gl. Which of the six panels below shows g1? Argue by explaining all the steps, so far: What did the image processor do when preparing ₤3? What image processing operation (from gin to g1) did he prepare and what's the effect that can be seen? Next, he convolved the rows of f3 with filter 1/2 (-1, 8, 0, -8, 1) to get f4 - you find a visualization of filter f 4 below. He then convolved gIn with f4 to get g2 and you can find the result shown below. What…arrow_forward3ur Colors are enchanting and elusive. A multitude of color systems has been proposed over a three-digits number of years - maybe more than the number of purposes that they serve... - Everyone knows the additive RGB color system – we usually serve light-emitting IT components like monitors with colors in that system. Here, we use c = (r, g, b) RGB with r, g, bЄ [0,1] to describe a color c. = T For printing, however, we usually use the subtractive CMY color system. The same color c becomes c = (c, m, y) CMY (1-c, 1-m, 1-y) RGB Note how we use subscripts to indicate with coordinate system the coordinates correspond to. Explain, why it is not possible to find a linear transformation between RGB and CMY coordinates. Farbenlehr c von Goethe Erster Band. Roſt einen Defte mit fergen up Tübingen, is et 3. Cotta'fden Babarblung. ISIO Homogeneous coordinates give us a work-around: If we specify colors in 4D, instead, with the 4th coordinate being the homogeneous coordinate h so that every actual…arrow_forwardCan someone provide an answer & detailed explanation please? Thank you kindly!arrow_forward
- Given the cubic function f(x) = x^3-6x^2 + 11x- 6, do the following: Plot the graph of the function. Find the critical points and determine whether each is a local minimum, local maximum, or a saddle point. Find the inflection point(s) (if any).Identify the intervals where the function is increasing and decreasing. Determine the end behavior of the graph.arrow_forwardGiven the quadratic function f(x) = x^2-4x+3, plot the graph of the function and find the following: The vertex of the parabola .The x-intercepts (if any). The y-intercept. Create graph also before solve.arrow_forwardwhat model best fits this dataarrow_forward
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