(a.)
If it is plausible that a fairly chosen group could be all republicans, or if there is a good reason to be suspicious.
It has been determined that while it is not fully impossible, it is nonetheless implausible that a fairly chosen group could be all republicans, and so there is a good reason to be suspicious.
Given:
The Republican Chairperson of a county legislature had to select
Concept used:
The drawing of names from a hat are not independent events. The subsequent drawing depends on the names already drawn.
Calculation:
It is given that the legislature had
So, the hat had the names of
At this point of time, the hat has the names of
Now, the probability that the first name drawn is of a republican is
At this point of time, the hat has the names of
Now, the probability that the second name drawn is also a republican given that the first name drawn is of a republican is
At this point of time, the hat has the names of
Now, the probability that the third name drawn is also a republican given that the first two names drawn are of republicans is
At this point of time, the hat has the names of
Now, the probability that the fourth name drawn is also a republican given that the first three names drawn are of republicans is
Solving, the probability that the fourth name drawn is also a republican given that the first three names drawn are of republicans is approximately
This implies that while it is not fully impossible, it is nonetheless implausible that a fairly chosen group could be all republicans, and so there is a good reason to be suspicious.
Conclusion:
It has been determined that while it is not fully impossible, it is nonetheless implausible that a fairly chosen group could be all republicans, and so there is a good reason to be suspicious.
(b.)
The assumption, which the previous calculations have been based on and why that assumption may or may not be warranted.
It has been determined that the assumption, which the previous calculations have been based on is that the drawing of each name is equally likely. That is, the probability of drawing each name is equal.
It has been determined that, this assumption may be invalid because according to the calculations based on this assumption, the occurred event is not plausible.
Given:
The Republican Chairperson of a county legislature had to select
Concept used:
The drawing of names from a hat are not independent events. The subsequent drawing depends on the names already drawn.
Calculation:
As determined previously, the probability that the fourth name drawn is also a republican given that the first three names drawn are of republicans is approximately
So, while it is not fully impossible, it is nonetheless implausible that a fairly chosen group could be all republicans, and so there is a good reason to be suspicious.
The assumption, which the previous calculations have been based on is that the drawing of each name is equally likely. That is, the probability of drawing each name is equal.
Now, this assumption may be invalid because according to the calculations based on this assumption, the occurred event is not plausible.
Conclusion:
It has been determined that the assumption, which the previous calculations have been based on is that the drawing of each name is equally likely. That is, the probability of drawing each name is equal.
It has been determined that, this assumption may be invalid because according to the calculations based on this assumption, the occurred event is not plausible.
Chapter 10 Solutions
PRECALCULUS:GRAPHICAL,...-NASTA ED.
- Is the function f(x) continuous at x = 1? (x) 7 6 5 4 3 2 1 0 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -1 -2 -3 -4 -5 -6 -71 Select the correct answer below: The function f(x) is continuous at x = 1. The right limit does not equal the left limit. Therefore, the function is not continuous. The function f(x) is discontinuous at x = 1. We cannot tell if the function is continuous or discontinuous.arrow_forwardQuestion Is the function f(x) shown in the graph below continuous at x = -5? f(z) 7 6 5 4 2 1 0 -10 -6 -5 -4 1 0 2 3 5 7 10 -1 -2 -3 -4 -5 Select the correct answer below: The function f(x) is continuous. The right limit exists. Therefore, the function is continuous. The left limit exists. Therefore, the function is continuous. The function f(x) is discontinuous. We cannot tell if the function is continuous or discontinuous.arrow_forwardThe graph of f(x) is given below. Select all of the true statements about the continuity of f(x) at x = -1. 654 -2- -7-6-5-4- 2-1 1 2 5 6 7 02. Select all that apply: ☐ f(x) is not continuous at x = -1 because f(-1) is not defined. ☐ f(x) is not continuous at x = −1 because lim f(x) does not exist. x-1 ☐ f(x) is not continuous at x = −1 because lim ƒ(x) ‡ ƒ(−1). ☐ f(x) is continuous at x = -1 J-←台arrow_forward
- Let h(x, y, z) = — In (x) — z y7-4z - y4 + 3x²z — e²xy ln(z) + 10y²z. (a) Holding all other variables constant, take the partial derivative of h(x, y, z) with respect to x, 2 h(x, y, z). მ (b) Holding all other variables constant, take the partial derivative of h(x, y, z) with respect to y, 2 h(x, y, z).arrow_forwardints) A common representation of data uses matrices and vectors, so it is helpful to familiarize ourselves with linear algebra notation, as well as some simple operations. Define a vector ♬ to be a column vector. Then, the following properties hold: • cu with c some constant, is equal to a new vector where every element in cv is equal to the corresponding element in & multiplied by c. For example, 2 2 = ● √₁ + √2 is equal to a new vector with elements equal to the elementwise addition of ₁ and 2. For example, 問 2+4-6 = The above properties form our definition for a linear combination of vectors. √3 is a linear combination of √₁ and √2 if √3 = a√₁ + b√2, where a and b are some constants. Oftentimes, we stack column vectors to form a matrix. Define the column rank of a matrix A to be equal to the maximal number of linearly independent columns in A. A set of columns is linearly independent if no column can be written as a linear combination of any other column(s) within the set. If all…arrow_forwardThe graph of f(x) is given below. Select each true statement about the continuity of f(x) at x = 3. Select all that apply: 7 -6- 5 4 3 2 1- -7-6-5-4-3-2-1 1 2 3 4 5 6 7 +1 -2· 3. -4 -6- f(x) is not continuous at a = 3 because it is not defined at x = 3. ☐ f(x) is not continuous at a = - 3 because lim f(x) does not exist. 2-3 f(x) is not continuous at x = 3 because lim f(x) ‡ ƒ(3). →3 O f(x) is continuous at a = 3.arrow_forward
- Is the function f(x) continuous at x = 1? (z) 6 5 4 3. 2 1 0 -10 -9 -7 -5 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -1 -2 -3 -4 -5 -6 -7 Select the correct answer below: ○ The function f(x) is continuous at x = 1. ○ The right limit does not equal the left limit. Therefore, the function is not continuous. ○ The function f(x) is discontinuous at x = 1. ○ We cannot tell if the function is continuous or discontinuous.arrow_forwardIs the function f(x) shown in the graph below continuous at x = −5? f(x) 7 6 5 4 2 1 0 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -1 -2 -3 -4 -5 -6 -7 Select the correct answer below: The function f(x) is continuous. ○ The right limit exists. Therefore, the function is continuous. The left limit exists. Therefore, the function is continuous. The function f(x) is discontinuous. ○ We cannot tell if the function is continuous or discontinuous.arrow_forward4. Evaluate the following integrals. Show your work. a) -x b) f₁²x²/2 + x² dx c) fe³xdx d) [2 cos(5x) dx e) √ 35x6 3+5x7 dx 3 g) reve √ dt h) fx (x-5) 10 dx dt 1+12arrow_forward
- Math 2 question. thxarrow_forwardPlease help on this Math 1arrow_forward2. (5 points) Let f(x) = = - - - x² − 3x+7. Find the local minimum and maximum point(s) of f(x), and write them in the form (a, b), specifying whether each point is a minimum or maximum. Coordinates should be kept in fractions. Additionally, provide in your answer if f(x) has an absolute minimum or maximum over its entire domain with their corresponding values. Otherwise, state that there is no absolute maximum or minimum. As a reminder, ∞ and -∞ are not considered absolute maxima and minima respectively.arrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning