Data Science Math Skill By Coursera All Quiz Answers

Data Science Math Skill By Coursera All Quiz Answers pdf. You will get All 4 week Answer. Total 120 Mcq Question With Answers 5/25/2020 Practice quiz on Sets | Coursera Practice quiz on Sets TOTAL POINTS 3 1. Let A = {1, 3, 5}. Is the following statement: 3 ∈ A. True or false? 1 / 1 point True False Correct The symbol ∈ stands for “is an element of” and it is true that 3 is an element of A. The other two elements of A are 1 and 5. 2. Let E = {−1, −2, −3}. Compute the cardinality ∣E∣ of E: 1 / 1 point −3 E 0 3 Correct Recall that the cardinality of a set is the number of elements in it. Since E has three elements (which are −1, −2, −3), the cardinality of E is ∣E∣ = 3. 3. Let A = {1, 3, 5} and B = {3, 5, 10, 11, 14}. Which of the following sets is equal to the intersection A ∩ B? {3, 5} {3} {1, 3, 5} {3, 5, 10} Correct The intersection of two sets consists precisely of the elements they share in common. The elements 3 and 5 are in both A and B. Practice quiz on the Number Line, including Inequalities TOTAL POINTS 8 1. Which of the following real numbers is not an integer? 1 / 1 point −3 7 0 4.3 Correct 4.3 is a decimal that is between two consecutive integers (4 and 5). 2. Which of the following is the absolute value ∣ − 7∣ of the number −7? 1 −7 7 0 Correct The absolute value of a number x is the distance along the number line from x to 0. In this case, −7 is 7 units away from 0, and so ∣ − 7∣=7. 3. Suppose I tell you that x and y are two real numbers which make the statement x y true. Which pair of numbers cannot be values for x and y? 1 / 1 point 1 / 1 point x = −1 and y = 0 x = 5 and y = 3.3 x = 1 and y = 7.3 x = −17.3 and y = −17.1 Correct The statement x y means that x is to the left of y on the real number line. Since 5 is to the right of 3.3, these cannot be values for x and y. 4. Suppose I tell you that w is a real number which makes both of the following 1 / 1 point statements true: w 1 and w 1.2. Which of the following numbers could be w? w = 1.05 w = 11 w = 1.2 w = 0 Correct 1.05 1 is true since 1.05 is to the right of 1 on the real number line, and 1.05 1.2 is also true, since 1.05 is to the left of 1.2 on the real number line. 5. Suppose that x and y are two real numbers which satisfy x+ 3 = 4y + 1. Which 1 / 1 point of the following statements are false? 2x + 6 = 8y + 2 x = 4y −2 x + 2 = 4y x = 4y Correct The equation x = 4y cannot be derived from the given equation. 6. Which of the following real numbers is in the open interval (2, 3)? 1 / 1 point 3 2.1 2 1 Correct Recall that the open interval satisfy 2 x 3. Since 2.1 open interval. (2, 3) consists of all real numbers xwhich 2 and 2.1 3, the number 2.1 is in this 7. Which of the following real numbers are in the open ray (3.1, ∞)? 1 / 1 point 0 3.1 4.75 −5 Correct Recall that (3.1, ∞) = {x ∈ R| x 3.1}. Since 4.75 3.1 is true, 4.75 ∈ (3.1, ∞). 8. Which of the following values for x solves the equation −3x + 2 = −4 x = −2 x = 2 3 x = 2 All values of x such that x ≤ 2 Correct First we subtract 2 from both sides of the given equation, to obtain −3x = −6. Finally, to isolate x we divide both sides of the equation by −3 to obtain x = 2. Practice quiz on Simplification Rules and Sigma Notation TOTAL POINTS 6 1. Which of the numbers below is equal to the following 1 / 1 point summation: Σ i 3 =1 i 2 ? 30 14 1 9 Correct We compute Σ 3 i=1i 2 = 12 + 23 + 32 = 14 2. / 1 point Suppose that A = Σk=1k and B = Σj=1 j Which of the following statements is true? B A There is not enough information to do the problem A B A = B Correct A = B. Both summations evaluate to the same number, since k and j are just dummy indices. 3. 10 7? 1 / 1 point Which of the numbers below is equal to the summation Σi=1 70 7 55 0 Correct According to one of our Sigma notation simplification rules, this summation is just equal to 10 copies of the number 7 all added together, and so we get 10 × 7 = 70. 4. Suppose that X = Σ 5 i=1i 3 and Y = Σ 5 i=1i 4 . Which of the following expressions is equal to the summation Σ 5 i=1(2i 3 + 5i 4 )? 2X+5Y X + Y 3375 7 Correct To get here, you apply two of our Sigma notation simplification rules Σ 5 i=12i 3 + 5i 4 = 2 (Σ 5 i=1i 3 ) + 5 (Σ 5 i=1i 4 ) = 2X + 5Y . 5. Which of the following numbers is the mean μZ of the set Z = {−2, 4, 7}? 3 4 1 / 1 point 1 / 1 point 13 3 9 Correct To get the mean of a set of numbers, you need to perform two steps: first add them all up (in this case getting −2 + 4 + 7 = 9), and then divide by the number of elements in the set (in this case that number is 3). So you should obtain μ Z = 9 = 3 , which you did! 3 6. Suppose the set X has five numbers in it: X = {x1, x2, x3, x4, x5}. Which of the following expression represents the mean of the set X? 1 N [∑ x i ] N i=1 1 5 [∑5 (x i − μ X ) 2 ] i=1 1 5 [∑ x i ] 5 i=1 ∑ 5 i=1 xi Correct To obtain the mean of a set of numbers, you first add them all up (which is expressed here by the sigma operation inside the square brackets) and then you divide by the number of numbers in the set (which is expressed here by the 1 5 outside the square brackets). Graded quiz on Sets, Number Line, Inequalities, Simplification, and Sigma Notation LATEST SUBMISSION GRADE 100% 1. Let B = {3, 5, 10, 11, 14}. Is the following statement true or false: 3 ∈/ B False True Correct The symbol ∈/ stands for “is not an element of.” Since 3 is in an element of the set B, the given statement is not true. 2. Let A = {1, 3, 5} and B = {3, 5, 10, 11, 14}. Which of the following sets is equal to the union A ∪ B? 1 / 1 point 1 / 1 point {1, 10, 18} {3, 5, 10, 11, 14} {1, 3, 5, 10, 11, 14} {1, 3, 5, 3, 5, 10, 11, 14} Correct The union of two sets consists precisely of the elements that are in at least one of the two sets. That is precisely what is listed here. 3. How many real numbers are there between the integers 1 and 4? 1 / 1 point 2 None Infinitely many 4 Correct There are in fact infinitely many real numbers between any pair of distinct integers, or indeed any pair of distinct real numbers! 4. Suppose I tell you that x and y are two real numbers which make the statement x 1 / 1 point ≥ y true. Which pair of numbers cannot be values for x and y? x = 2 and y = 1 x = 10 and y = 10 x = −1 and y = 0 x = 5 and y = 3.3 Correct Recall that the statement x ≥ y means that x is either equal to yor x is to the right of y on the real number line. Since −1 is actually to the left of 0, these cannot be values for x and y. 5. Suppose that z and w are two positive numbers with z w. Which of the 1 / 1 point following inequalities is false? −z −w −5z −5w w − 7 z − 7 z + 3 w + 3 Correct If we start with z w and multiply both sides by −5, we need to flip the less-than sign, which would give −5z −5w. For an example, try z = 1 and y = 2 and see what happens! 6. Find the set of all x which solve the inequality −2x + 5 ≤ 7 1 / 1 point x ≥ −6 x = −1 x ≤ −1 x ≥ −1 Correct Subtracting 5 from both sides of the given inequality gives −2x ≤ 2. Then we divide both sides by −2, remembering to flip the inequality sign, and we obtain this answer 7. Which of the following real numbers is not in the closed interval [2, 3] 1 / 1 point 1 2.1 2 3 Correct Recall that the closed interval [2, 3] consists of all real numbers x which satisfy 2 ≤ x ≤ 3. Since 2 ≤ 1 is false, 1 ∈/ [2, 3] 8. Which of the following intervals represents the set of all solutions to: 1 / 1 point −5 ≤ x + 2 10? [−7, 8) [−7, 8] (7, 8) [−5, 10) Correct Subtracting 2 from all sides of the inequalities gives −7 ≤ x 8, and the set of all real numbers x which make that true is exactly the halfopen interval [−7, 8). 9. Which of the numbers below is equal to the following summation: Σ 5 k=22k? 14 4 10 28 Correct We compute Σ 5 k=22k = 4 + 6 + 8 + 10 = 28. 10. Suppose we already know that Σ 20k=1k = 210. Which of the numbers below is equal to Σ 20k=12k? 210 1 / 1 point 1 / 1 point 2 420 40 Correct By applying one of our Sigma notation simplification rules, we can rewrite the summation in question as 2 (Σ 20 k=1k) = 2 × 210 = 420. 11. Which of the numbers below is equal to the summation Σ 10 i=27? 1 / 1 point 48 70 7 63 Correct According to one of our Sigma notation simplification rules, this summation is just equal to 9 copies of the number 7 all added together, and so we get 9 ⋅ 7 = 63. 12. Which of the following numbers is the variance of the set Z = {−2, 4, 7}? 1 / 1 point 42 69 14 14 Correct To get the variance of a set of numbers, you need to perform four steps: First compute the mean (which is 3) Then calculate all the squared differences between the numbers in the set and this mean (here you get 25, 1, 16) Then add all these up (here you get 42) Then divide by the number of elements in the set (which is 3). Therefore, the variance of Z = 1 3 [(−2 − 3)2 + (4 − 3)2 + (7 − 3)2 ] = 1 3 [25+1+16]= 42 3 =14 13. Which of the following sets does not have zero variance? (hint: don't do any calculation here, just think!) {0, 0, 0, 0, 0, 0, 0} {1, 1, 1, 1} {2, 5, 9, 13} {5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5} Correct Intuitively, the numbers in this set are spread out. Practice quiz on the Cartesian Plane TOTAL POINTS 5 1. Which of the following points in the Cartesian Plane is on the y-axis? 1 / 1 point (0, −5) (5, 0) (−5, 0) (1, 1) Correct The y-axis is defined to be all points in the Cartesian plane with zero as x-coordinate. The point (0, −5) meets that requirement. 2. Find the distance between the points A = (2, 2) and C = (3, 3): 1 / 1 point 1 2 0 2 Correct Recall that the distance between points (a, b) and (c, d) is (c − a) 2 + (d − b) 2 . In this case (a, b) = (2, 2) and (c, d) = (3, 3), so the distance is (3−2)2 +(3−2)2 = 2. 3. Find the point-slope form of the equation of the line that goes between A = (1, 1) and B = 1 / 1 point (5, 3): y − 1 = 1 2 (x − 1) y = 1 x 2 y − 1 = 1 (x − 5) 2 y − 3 = 1 (x − 1) 2 Correct The point-slope form for the equation of a line with slope m that goes through the point (x0, y0) is y − y0 = m(x − x0) In this case, the slope m = 3 − 1 = 1 5 − 1 2 We can choose either A or B for the point on the line, but in neither case do we get this chosen answer. 4. Which of the following points is on the line with equation: 1 / 1 point y − 1 = 2(x − 2)? (3, 2) (0, 0) (2, 3) (2, 1) Correct If we plug in 1 for y and 2 for x in the equation of the line, we make a true statement, 0 = 0, so this point lies on the line. 5. Suppose that a line ℓ has slope 2 and goes through the point (−1, 0). What is the y-intercept of ℓ? 0 1 2 −1 Correct Recall that the y-intercept of ℓ is the y-coordinate of where ℓ hits the y -axis. Since (−1, 0) ∈ ℓ, the point on ℓ with x = 0 is obtained by running one unit from (−1, 0) while rising two units. This gives y = 2 as the y-intercept. 1. Suppose that A = {1, 2, 10} and B = {4, 8, 40}. Which of the following formulae do not define a function f : A → B? 1 / 1 point f(a) = 4a, for each a ∈ A f(1) = 4, f(2) = 40, and f(10) = 8. f(1) = 4, f(2) = 4, and f(10) = 4. f(1) = 5, f(2) = 8, and f(10) = 40. Correct A function f : A → B is a rule which assigns an element f(a) ∈ B to each a ∈ A. In this case, unfortunately, f(1) = 5 ∈/ B. 2. Suppose that A contains every person in the VBS study (see the second video in 1 / 1 point the course if you're confused here!). Suppose that Y = {+, −} and Z = {H, S} Suppose that T : A → Y is the function which gives T (a) = + if person atests positive and T (a) = − if they test negative. Suppose that D : A → Z is the function which gives D(a) = H does not actually have VBS and D(a) = S if the person actually has VBS. Which of the following must be true of person a if we have a false positive? T (a) = − and D(a) = H T (a) = − and D(a) = S T (a) = + and D(a) = S Practice quiz on Types of Functions TOTAL POINTS 65/25/2020 Practice quiz on Types of Functions | Coursera T (a) = + and D(a) = H Correct Recall that a false positive is a positive test result (so T (a) = +) which is misleading because the person actually does not have the disease ( D(a) = H) 3. Consider the function g : R → R defined by g(x) = x 2 − 1. Which of the 1 / 1 point following points are not on the graph of g? (0, −1) (−1, 0) (1, 0) (2, −1) Correct Recall that the graph of g consists of all points (x, y) such that y = g(x). Here g(2) = 3 = −1, so the point (2, −1) is emph{not} on the graph of g. 4. Let the point A = (2, 4). Which of the following graphs does not contain the 1 / 1 point point A? The graph of f(x) = 2x The graph of g(x) = x + 2 The graph of h(x) = x − 1 The graph of s(x) = x 2 Correct The graph of h consists of all points (x, y) such that y = h(x). Here h(2) = 1 = 4, so the point (2, 4) is not on the graph of h. 5. Suppose that h(x) = −3x + 4. Which of the following statements is true? 1 / 1 point All statements are correct h is neither a strictly increasing function nor a strictly decreasing function. h is a strictly decreasing function h is a strictly increasing function Correct A function h is called strictly decreasing if whenever a b, then h(a) h(b) Since the graph of h is a line with negative slope, this is in fact true! 6. Suppose that f : R → R is a strictly increasing function, with f(3) = 15 Which of the following is a possible value for f(3.7)? 17 3 14.7 −3 Correct A function f is called strictly increasing if whenever a b, then f(a) f(b). Since f(3) = 15 is given and 3 3.7, it must be that 15 f(3.7), and this answer satisfies that. Graded quiz on Cartesian Plane and Types of Function LATEST SUBMISSION GRADE 100% 1. Which of the following points in the Cartesian Plane have positive x-coordinate 1 / 1 point and negative y-coordinate? (5, 7) (−4, 5) (0, 0) (7, −1) Correct The x-coordinate, 7, is positive, and the y-coordinate, −1, is negative. 2. Which of the following points is in the first quadrant of the Cartesian Plane? 1 / 1 point (−5, 1) (5, −1) (7, 11) (−4, −7) Correct The first quadrant is defined to be all points in the Cartesian plane whose coordinates are both positive. 3. Let A, B, C, D be points in the Cartesian Plane, and let the set S = {B, C, D} 1 / 1 point Suppose that the distances from A to B, C, D are 5.3, 2.1, and 11.75, respectively. Which of the following points is the nearest neighbor to the point A in the set S? A B D C Correct The distance from A to C is 2.1 and that is smaller than the distance from A to any other element of S. 4. Find the distance between the points A = (2, 2) and B = (−1, −2). 1 / 1 point −25 5 25 1 Correct Recall that the distance between points (a, b) and(c, d) is (c − a) 2 + (d − b) 2 In this case we have: (−1 − 2)2 + (−2 − 2)2 = (−3)2 + (−4)2 = 25 = 5 5. Find the slope of the line segment between the points A = (0, 1) and B = (1, 0). 1 / 1 point −1 1 2 0 Correct The slope of this line segment is 0 − 1 1 − 0 = −1 6. Find the point-slope form of the equation of the line with slope −2 that goes through the point (5, 4). (5, 4) y − 4 = −2(x − 5) y − 5 = −2(x − 4) y − 4 = 2(x − 5) Correct The point-slope form for the equation of a line with slope m that goes through the point (x0, y0) is y − y0 = m(x − x0). In this case, the slope m = −2 is given and the point (5, 4) on the line is given. 7. Which of the following equations is for a line with the same slope as y = −3x + 2 ? 1 / 1 point 1 / 1 point y = 8x − 3 y = 5x + 2 y = −3x − 8 y = 5x Correct The slope-intercept formula for a line is y = mx + b, where m is the slope and b is the y-coordinate of the point where the line hits the yaxis. This line has slope m = −3 which is the same slope as the given line. 8. Which of the following equations is for a line with the same y-intercept as y = 1 / 1 point −3x + 2? y = 5x + 2 y = 8x − 3 y = 5x y = −3x − 8 Correct The the slope-intercept formula for a line is y = mx + b, where mis the slope and b is the y-coordinate of the point where the line hits the yaxis. This line has a y-intercept of 2 which is the same as the given line. 9. How many lines contain both the point A = (1, 1) and the point B = (2, 2)? 1 / 1 point None infinitely many 1 2 Correct The line with equation y = x is the one and only line that meets the stated requirements. 10. Suppose that we have two sets, A = {a, b} and Z = {x, y}. How many different 1 / 1 point functions F : A → Z are possible? 4 There are none 1 There are infinitely many Correct A function F : A → Z is a rule which assigns an element F (a) ∈ Z to each element a ∈ A. There are two elements in A; namely, a and b. For each of these elements, there are two assignment choices we could make: x and y. Here are the four possible functions: F (a) = x, F (b) = y, OR F (a) = y, F (b) = x, OR F (a) = x, F (b) = x, OR F (a) = y, F (b) = y. 11. How many graphs contain both the point A = (0, 0) and the point B = (1, 1) 1 / 1 point None 1 2 Infinitely many Correct The graphs of f(x) = x, g(x) = x 2 , h(x) = x 3 , s(x) = x 4 , …all contain both A and B 12. Suppose that g : R → R is a continuous function whose graph intersects the x- 1 / 1 point axis more than once. Which of the following statements is true? g is strictly increasing. All of the above. g is strictly decreasing. g is neither strictly increasing nor strictly decreasing. Correct The function g fails the horizontal line test, so it can neither be strictly increasing nor strictly decreasing. 13. Find the slope of the line segment between the points A = (1, 1) and B = (5, 3). 20 4 1 2 2 Correct The slope of this line segment is 3 − 1 = 1 5 − 1 2 , where 3 − 1 is the rise and 5 − 1 is the run. Practice quiz onTangent Lines to Functions TOTAL POINTS 2 1. Suppose that f : R → R is a function. Which of the following expressions 1 / 1 point corresponds to f ′ (2), the slope of the tangent line to the graph of f (x) at x = 2? f ′ (2) = 2 f ′ (2) = mx + b f ′ (2) = limh→0 f ′ (2) = limh→0 f (2+h)−f (2) h f (a+h)−f (a) h Correct This expression can be obtained from the first screen of our video by plugging in 2 for a. 2. Suppose that h : R → R is a function whose graph is shown as the blue curve in the figure. For how many values of a is h ′ (a) = 0? 3 Never Always 2 Correct h ′ (a) gives the slope of the tangent line to the graph of h at the point x = a. When h ′ (a) = 0, this means that the tangent line is horizontal. There are two places (one on each side of the y-axis) where this tangent line is horizontal, so this answer is correct. Practice quiz on Exponents and Logarithms TOTAL POINTS 12 1. Re write the number 784 = 2 × 2 × 2 × 2 × 7 × 7 using exponents. (26 )(76 ) (2 × 7)6 (24 )(72 ) (164 )(492 ) Correct For this type of problem, count the number of times each relevant factor appears in the product. That number is the exponent for that factor. 2. What is (x 2 − 5)0 ? (x 2 ) −4 1 (x 2 ) − 5 Correct Any real number (except zero) raised to the"zeroith'' power = 1. 3. Simplify ((x − 5)2 ) −3 (x − 5)−6 1 / 1 point 1 / 1 point 1 / 1 point (x − 5) (x − 5)−1 (x − 5)−5 Correct By Rule 2, "Power to a Power,'' multiply the exponents and get: (x − 5)(2×−3) = (x − 5)−6 1 By the definition of negative exponents, this is equal to (x − 5)6 4. ( 2 2 1 / 1 point 87 ) Simplify 8 8−10 8−4 8−1 8−5 Correct We can first simplify what is inside the parenthesis to 8 −5 using the Division and Negative Powers Rule. Then apply division and negative powers-- the result is the same. 8 4 = 8−10 814 5. log 35 = log 7 + log x Solve for x 4 7 28 5 Correct log(x) =log 35 − log 7 log(x) = log ( 35 7 ) By the Quotient Rule log x = log 5 6. log2(x 2 + 5x + 7) = 0 Solve for x x = 2 x = −2 or x = −3 x = 3 x = 2 or x = 3 Correct We use the property that b log b a = a Use both sides as exponent for 2. 1 / 1 point 1 / 1 point 2log2 x 2 +5x+7 = 20 x 2 + 5x + 7 = 1 x 2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 x = −3 OR x = −2 7. Simplify log2 72 − log2 9 log2 63 3 log2 4 4 Correct 72 By the quotient rule, this is log2 = log2 2 3 = 3 9 8. Simplify log3 9 − log3 3 + log3 5 8 15 log3 15 1 / 1 point 1 / 1 point log3 8 Correct 9 × 5 By the Quotient and Product Rules, this is log3 = log3 15 3 9. Simplify log2(38 × 57 ) 56 × log2 15 (8 × log2 3) + (7 × log2 5) (5 × log2 3) + (8 × log2 5) 15 × log2 56 Correct We first apply the Product Rule to convert to the sum: log2(38 ) + log2(57 ). Then apply the power and root rule. 10. If log10 y = 100, what is log2 y =? 332.19 301.03 500 20 Correct Use the change of base formula, loga b = logx b logx a Where the "old'' base is x and the "new'' base is a. 1 / 1 point 1 / 1 point 100 = 100 So log10 (2) 0.30103 = 332.19 11. A tree is growing taller at a continuous rate. In the past 12 years it has grown 1 / 1 point from 3 meters to 15 meters. What is its rate of growth per year? 12.41% 11.41% 10.41% 13.41% Correct ln 15 123 = 0.1341 12. Bacteria can reproduce exponentially if not constrained. Assume a colony grows at a continually compounded rate of 400% per day. How many days before a colony with initial mass of 6.25 X 10−10 grams weights 1000Kilograms? 8.75 days 87.5 days 875 days 0.875 days Correct 6.25 × 10−10 × e 4t = 106 4t = ln ( 106 (6.25 × 10 ) = 35.00878 −10 ) t = ln 106 6.25 × 10 = 8. −10 Graded quiz on Tangent Lines to Functions, Exponents and Logarithms LATEST SUBMISSION GRADE 100% 1. Convert 49 1 to exponential form, using 7 as the factor. 49−1 7−2 7 7 3 (72 ) Correct The rule for a factor to a Negative exponent is to divide by the same factor to a positive exponent with the same absolute value. 2. A light-year (the distance light travels in a vacuum in one year) is 9, 460trillion meters. Express in scientific notation. 9460 × 1012 meters 9.46 × 1015 kilometers 0.946 × 1016 9.46 × 1015 meters. Correct 9, 460 is (9.4 × 103 ) meters and one trillion meters is 1012 meters. 1 / 1 point 1 / 1 point (9.4 × 103 )(1012 ) = 9.4 × 1015 . A kilometer is 1000meters. 3. Simplify (x 8 )(y 3 )(x −10 )(y −2 ) (x −80 )(y −6 ) (x −2 )(y) (x 2 )(y) (x)(y −2 ) Correct By the Division and Negative Powers Rule, this is (x (8−10))(y (3−2) ) 4. Simplify [(x 4 )(y −6 )]−1 (x − 4) (y 6 ) (x 3 )(y −7 ) (x −4 )(y 6 ) (x 4 ) (y −6 ) Correct By the Power to a Power Rule, each of the exponents is multiplied by (−1) 5. Solve for x: log2 (39x) − log2 (x − 5) = 4 80 1 / 1 point 1 / 1 point 1 / 1 point 38 23 80 −80 23 39 23 Correct 39x = 4 by the Quotient Rule. log 2 (x − 5) Since both sides are equal, we can use them as exponents in an equation. log 39x 2 2 (x − 5) = 24 39x (x − 5) = 16 39x = 16 × (x − 5) 39x = 16x − 80 23x = −80 x = −80 23 6. Simplify this expression: 1 / 1 point −3 1 ) 2 (x 2 −3 x 4 x−1 4 x 3 1 x 3 Correct We use the Power to a Power Rule -- multiply exponents: 1 × −3 = x −3 x 2 2 4 7. Simplify log2 8 − log2 4 − (log3 4.5 + log3 2) 1 / 1 point 0 1 2 −1 Correct This is equivalent to: log2( 8 4 ) − log3(4.5 × 2) = 1 − 2 = −1 8. If log3 19 = 2.680, what is log9 19? 1 / 1 point 0.8934 0.4347 5.216 1.304 Correct To convert from log3 to log9, divide by log3 9. Which is equal to 2, so the answer is 1.34 9. If log 10 b = 1.8 and log b = 2.5752, what is a? 1 / 1 point a 6 3 4 5 Correct To solve for a in the formula; loga b= log x b logx a loga b = 2.5752 and log10 b = 1.8 Therefore, log10 a must equal to 1.8 2.5752 = 0.69897 Treating both sides of equation log10 a = 0.69897 as exponents of 10 gives a = 100.69897 = 5 10. An investment of 1, 600 is worth 7, 400 after 8.5 years. 1 / 1 point What is the continuously compounded rate of return of this investment? 17.01% 20.01 18.02% 19.01% Correct ln 7400 1600 = 0.18017 8.5 11. A pearl grows in an oyster at a continuously compounded 1 / 1 point rate of .24 per year. If a 25-year old pearl weighs 1 gram, what did it weigh when it began to form? 0.2478 0.02478 0. 0. Correct e(0.24×25) =1x x = 1 (e0.24×25 ) x = 1 403.4288 x = 0. 12. log2 z = 6.754. What is log10(z)? 1 / 1 point 0.49185 2.03316 0.82956 1.3508 Correct log2 z = log2 10 (log10 z) × (log2 10) = 3. Therefore, log10 z = 6.754 = 2.03316 3. 13. Suppose that g : R → R is a function, and that g(1) = 10. Suppose that g ′ (a) is negative for every single value of a.Which of the following could possibly be g(1.5)? g(1.5) = 9.7 g(1.5) = 11 g(1.5) = 103.4 g(1.5) = 10.1 Correct Since the slope of the tangent line to the graph of g is negative everywhere on the graph, we know that g is decreasing function! And therefore we must have g(1.5) g(1). That is the case here, so this value is at least possible. Practice quiz on Probability Concepts TOTAL POINTS 9 1. If x = "It is raining,'' what is ∼ (∼ x)? 1 / 1 point "It is never raining" "It is always raining" "It is not raining" "It is raining" Correct The second negation cancels out the first one. Similarly ∼ (∼ (∼ x)) =∼ x 2. If the statement “I am 25 years old” is assigned probability 0, what probability is 1 / 1 point assigned to the statement “I am not 25 years old”? −1 1 Unknown 0 Correct It is always the case that p(x) + p(∼ x) = 1. 3. If I assign to the statement x = "it will rain today'' a probability of p(x) = 0.35, 1 / 1 point what probability must I assign to the statement "it will not rain today?" .65 .5 0 .35 Correct p(x) + p(∼ x) = 1 4. Is the following collection of statements a probability distribution? 1. I own a Toyota pickup truck 2. I do not own a Toyota pickup truck 3. I own a non-Toyota pickup truck 4. I do not own a non-Toyota pickup truck No Yes Correct The statements are not exclusive:1 and 4 could both be true, 2 and 3 could both be true, 2 and 4 could both be true, and even (1) and (3) could both be true (if I owned more than one pickup truck). 5. I don’t know what it means to be “ingenuous.” What probability would I assign to the statement, “I am ingenuous OR I am not ingenuous"? 1 / 1 point 1 / 1 point 1 0 .5 -1 Correct It is always the case, regardless of the content of the statement x, that p(x or ∼ x) = 1 6. A friend of mine circumscribes a circle inside a square, so that the diameter of 1 / 1 point the circle and the edge of the square are the same length. He asks me to close my eyes and pick a point at random inside the square. He says the probability that my point will also be inside the circle is π 4 Is this correct? Yes No Correct Probabilities can be any real number between 0 and 1. They do not need to be rational numbers – a numerator that is a transcendental number like Pi is acceptable. Note that the correct probability does not depend on the length r of the circle’s radius. For a circle with any radius r to be circumscribed inside a square, the square must have sides each of length 2r. The area of the circle is Pi*r^2 and the area of the square is (2r)^2 = 4*r^2 = The probability of landing in a circle of area Pi*r^2 when it is known that one is in the area of the square is equal to the ratio of the area of the circle to the area of the square in which it is circumscribed, or Pi*r^2/4*r^2, which equals Pi/4. 7. The probability of drawing a straight flush (including a Royal Flush) in a five-card poker hand is 0.8 What is the probability of not drawing a straight flush? .2 .2 .9 .8 Correct p(∼ x) = 1 − p(x) 8. What is the probability that a fair, six-sided die will come up with a prime number? (Recall that prime numbers are positive integers other than 1 that are divisible only by themselves and 1) 1 3 2 3 1 6 1 2 Correct The faces with 2, 3 and 5 satisfy the condition – which makes 3 relevant outcomes out of the “universe” of 6 outcomes = 3 = 1 6 2 1 / 1 point 1 / 1 point 9. The joint probability p (the die will come up 5, the next card will be a heart) Is equal to the joint probability: p (the next card will not come up 5, the next card will be a heart) p (the die will not come up 5, the next card will not be a heart) p (the next card will be a heart, the die will come up 5) p (the next card will be a heart, the die will not come up 5) Correct In joint probabilities, the order does not change the probability: p(A, B) = p(B, A) Practice quiz on Problem Solving TOTAL POINTS 9 1. I am given the following 3 joint probabilities: 1 / 1 point p(I am leaving work early, there is a football game that I want to watch this afternoon) = .1 p(I am leaving work early, there is not a football game that I want to watch this afternoon) = .05 p(I am not leaving work early, there is not a football game that I want to watch this afternoon) = .65 What is the probability that there is a football game that I want to watch this afternoon? .35 .2 .3 .1 Correct Getting the answer is a two-step process. First, recall that the sum of probabilities for a probability distribution must sum to 1. So the “missing” joint distribution p(I am not leaving work early, there is a football game I want to watch this afternoon) must be 1–(0.1 + 0.05 + 0.65) = 0.2 By the sum rule, the marginal probability p(there is a football game that I want to watch this afternoon) = the sum of the joint probabilities P(I am leaving work early, there is a football game that I want to watch this afternoon) + P(I am not leaving work early, there is a football game I want to watch this afternoon) = .1 + .2 = .3 2. The Joint probability of my summiting Mt. Baker in the next two years AND 1 / 1 point publishing a best-selling book in the next two years is .05. If the probability of my publishing a best-selling book in the next two years is 10%, and the probability of my summiting Mt. Baker in the next two years is 30%, are these two events dependent or independent? Independent Dependent Correct We know this because the joint distribution of 5% does not equal the product distribution of (0.1) × (0.3) = 3%. If I summit Mt. Baker, I am more likely to publish a best-selling book, and vice versa. 3. The Joint probability of my summiting Mt. Baker in the next two years AND my publishing a 1 / 1 point best-selling book in the next two years is .05. If the probability of my publishing a best-selling book in the next two years is 10%, and the probability of my summiting Mt. Baker in the next two years is 30%, what is the probability that (sadly) in the next two years I will neither summit Mt. Baker nor publish a best-selling book? .95 .9 .65 .25 Correct Set A = I will summit Mt. Baker in the next two years Set B = I will publish a best-selling book in the next two years. Since p(A) = 0.3 and p(A, B) = 0.05, by the SUM RULE we know that p(A, ∼ B) = (0.3 − 0.05) = 0.25 Since p(B) = 0.1,p(∼ B) = 0.9 Since p(∼ B) = 0.9 and p(A, ∼ B) = 0.25 and again by the SUM RULE, p(∼ A, ∼ B) = 0.9 − 0.25 = .65 4. I have two coins. One is fair, and has a probability of coming up heads of .5. The 1 / 1 point second is bent, and has a probability of coming up heads of .75. If I toss each coin once, what is the probability that at least one of the coins will come up heads? .375 .875 1.0 .625 Correct We apply the rule p(A or B or both) = 1 - (p(~A)p(~B)) = 1 ((1 5)(1 75)) =1-.125 =.875 5. What is 11! 9! ? 110, 000 110 554, 400 4, 435, 200 Correct 11! 9! = 11 × 10 = 110 6. What is the probability that, in six throws of a die, there will be exactly one each of “1” “2” “3” “4” “5” and “6” ? . . . . Correct There are 6! = 720 permutations where each face occurs exactly once. There are 6 × 6 × 6 × 6 × 6 × 6 = 46656 total permutations of 6 throws. 1 / 1 point 1 / 1 point The probability is therefore 46656 720 = 0. 7. On 1 day in 1000, there is a fire and the fire alarm rings. 1 / 1 point On 1 day in 100, there is no fire and the fire alarm rings (false alarm) On 1 day in 10, 000, there is a fire and the fire alarm does not ring (defective alarm). On 9, 889 days out of 10, 000, there is no fire and the fire alarm does not ring. If the fire alarm rings, what is the (conditional) probability that there is a fire? Written p(there is a fire | fire alarm rings) 1.12% 90.9% 1.1% 9.09% Correct 10 days out of every 10, 000 there is fire and the fire alarm rings. 100 days out of every 10, 000 there is no fire and the fire alarm rings. 110 days out of every 10, 000 the fire alarm rings. The probability that there is a fire, given that the fire alarm rings, is 10 = 9.09% 110 8. On 1 day in 1000, there is a fire and the fire alarm rings. 1 / 1 point On 1 day in 100, there is no fire and the fire alarm rings (false alarm) On 1 day in 10, 000, there is a fire and the fire alarm does not ring (defective alarm). On 9, 889 days out of 10, 000, there is no fire and the fire alarm does not ring. If the fire alarm does not ring, what is the (conditional) probability that there is a fire? p(there is a fire | fire alarm does not ring) .01000% .10011% 1.0001% 0.01011% Correct On (1 + 9, 889) = 9, 890 days out of every 10, 000 the fire alarm does not ring. On 1 of those 10, 000 days there is a fire. 1 = 0.01011% 9890 9. A group of 45 civil servants at the State Department are newly qualified to serve as Ambassadors to foreign governments. There are 22 countries that currently need Ambassadors. How many distinct groups of 22 people can the President promote to fill these jobs? $$4.1167 times (10^12) =1.06*(10^35) 8.2334 times (10^12) =2.429*(10^-13) Correct ( 45 22) = 45!/(23!)(22!) 45! = 23! × 22! Practice quiz on Bayes Theorem and the Binomial Theorem TOTAL POINTS 9 1. A jewelry store that serves just one customer at a time is concerned about the 1 / 1 point safety of its isolated customers. The store does some research and learns that: 10% of the times that a jewelry store is robbed, a customer is in the store. A jewelry store has a customer on average 20% of each 24-hour day. The probability that a jewelry store is being robbed (anywhere in the world) is 1 in 2 million. What is the probability that a robbery will occur while a customer is in the store? 1 1 1 1 Correct What is known is: A: "a customer is in the store," P (A) = 0.2 B: "a robbery is occurring," P (B) = 1 2,000,000 P (a customer is in the store ∣ a robbery occurs) = P (A ∣ B) P(A ∣ B) = 10% What is wanted: P (a robbery occurs ∣ a customer is in the store) = P (B ∣ A) By the product rule: P(B∣A)= P(A, B) P(A) and P (A, B) = P (A ∣ B)P (B) Therefore: P(A ∣ B)P(B) (0.1)( 1 ) 1 P(B∣A)= = = P(A) 0.2 2. If I flip a fair coin, with heads and tails, ten times in a row, what is the probability 1 / 1 point that I will get exactly six heads? 0.021 0.187 0.2051 0.305 Correct By Binomial Theorem, equals ( 10 )(0.510 ) 6 = ( 10! )( 1 ) 4!×6! 1024 = 0.2051 3. If a coin is bent so that it has a 40% probability of coming up heads, what is the 1 / 1 point probability of getting exactly 6 heads in 10 throws? 0.0974 0.1045 0.1115 0.1219 Correct ( 106) × 0.46 × 0.64 = 0.1115 4. A bent coin has 40% probability of coming up heads on each independent toss. 1 / 1 point If I toss the coin ten times, what is the probability that I get at least 8 heads? 0.0312 0.0123 0.0213 0.0132 Correct The answer is the sum of three binomial probabilities: ((10 8) × (0.48 ) × (.62 )) + ((10 9) × (0.49 ) × (0.61 ))+ ((10 10)) × (0.410) × (0.60 )) 5. Suppose I have a bent coin with a 60% probability of coming up heads. I throw 1 / 1 point the coin ten times and it comes up heads 8 times. What is the value of the “likelihood” term in Bayes’ Theorem -- the conditional probability of the data given the parameter. 0. 0. 0. 0. Correct Bayesian “likelihood” --- the p(observed data | parameter) is p(8 of 10 heads | coin has p = .6 of coming up heads) ( 108) × (0.68 ) × (0.42 ) = 0. 6. We have the following information about a new medical test for diagnosing 1 / 1 point cancer. Before any data are observed, we know that 5% of the population to be tested actually have Cancer. Of those tested who do have cancer, 90% of them get an accurate test result of “Positive” for cancer. The other 10% get a false test result of “Negative” for Cancer. Of the people who do not have cancer, 90% of them get an accurate test result of “Negative” for cancer. The other 10% get a false test result of “Positive” for cancer. What is the conditional probability that I have Cancer, if I get a “Positive” test result for Cancer? **Formulas in the feedback section are very long, and do not fit within the standard viewing window. Therefore, the font is a bit smaller and the word "positive test" has been abbreviated as PT. 67.9% 4.5% 9.5% 32.1% probability that I have cancer Correct I still have a more than 2 3 probability of not having cancer Posterior probability: p(I actually have cancer | receive a “positive” Test) By Bayes Theorem: = (chance of observing a PT if I have cancer)(prior probability of having cancer) (marginal likelihood of the observation of a PT) = p(receiving positive test| has cancer)p(has cancer [before data is observed]) p(positive | has cancer)p(has cancer)+p(positive | no cancer )p(no cancer) = (90%)(5%) / ((90%)(5%) + (10%)(95%) =32.1% 7. We have the following information about a new medical test for diagnosing 1 / 1 point cancer. Before any data are observed, we know that 8% of the population to be tested actually have Cancer. Of those tested who do have cancer, 90% of them get an accurate test result of "Positive'' for cancer. The other 10% get a false test result of "Negative'' for Cancer. Of the people who do not have cancer, 95% of them get an accurate test result of "Negative'' for cancer. The other 5% get a false test result of "Positive'' for cancer. What is the conditional probability that I have cancer, if I get a "Negative'' test result for Cancer? 99.1% 88.2% 0.9% .80% Correct p(cancer ∣ negative test) = p(negative test ∣ Cancer) p(Cancer) p(negative test ∣ cancer) p(cancer) + p(negative test ∣no cancer) p(no cancer) (10%)(8%) (10%)(8%)+(95%)(92%) 0.8% 0.8%+87.4% 0.8% 88.2% = 0.9% 8. An urn contains 50 marbles – 40 blue and 10 white. After 50 draws, exactly 40 1 / 1 point blue and 10 white are observed. You are not told whether the draw was done “with replacement” or “without replacement.” What is the probability that the draw was done with replacement? 13.98% 1 87.73% 12.27% Correct p(40 blue and 10 white | draws without replacement) = 1 [this is the only possible outcome when 50 draws are made without replacement] p(40 blue and 10 white | draws with replacement) S=40 N=50 P = .8 [for draws with replacement] because 40 blue of 50 total means p(blue) = 40/50 = .8 (( 5040))(0.840)(0.210) = 13.98% By Bayes’ Theorem: p(draws with replacement | observed data) = 13.98%(.5) (13.98%)(.5)+(1)(.5) = 0 1 . . = 12.27% 9. According to Department of Customs Enforcement Research: 99% of people crossing into the United States are not smugglers. The majority of all Smugglers at the border (65%) appear nervous and sweaty. Only 8% of innocent people at the border appear nervous and sweaty. If someone at the border appears nervous and sweaty, what is the probability that they are a Smuggler? 92.42% 8.57% 7.58% 7.92% Correct By Bayes’ Theorem, the answer is (.65)(.01) ((.65)(.01) + (.08)(.99)) = 7.58%