
Concept explainers
To calculate: The centers and variations and to find which airline had shorter flight time.

Answer to Problem 6IP
Air jet Express has a median of 4 hours with an inter quartile range 4 hours.
Cross Country Airlines has a median of 5 hours with an inter quartile range 2 hours.
Air jet Express has a shorter time of flight.
Explanation of Solution
Given information: The double dot plots shows the flight times of two different airlines, Air jet Express and Cross Country Airlines.
Calculation:
Both the distributions are not symmetric. Use the median to compare the centers and the inter quartile range to compare the variations.
Air jet express:
First finding the median of,
The median is the middle number.
Now finding the quartiles.
The first quartile
The third quartile
The inter quartile range is,
Air jet Express has a median of 4 hours with an inter quartile range 4 hours.
Cross Country Airlines:
First finding the median of,
The median is the middle number.
Now finding the quartiles.
The first quartile
The third quartile
The inter quartile range is,
Cross Country Airlines has a median of 5 hours with an inter quartile range 2 hours.
The inter quartile range of Air jet Express is greater than that of Cross Country Airlines. However, the median of Air jet Express is less than that of Cross Country Airlines.
Therefore Air jet Express has a shorter time of flight.
Chapter 10 Solutions
Glencoe Math Accelerated, Student Edition
Additional Math Textbook Solutions
Elementary Statistics: Picturing the World (7th Edition)
A First Course in Probability (10th Edition)
Algebra and Trigonometry (6th Edition)
A Problem Solving Approach To Mathematics For Elementary School Teachers (13th Edition)
Calculus: Early Transcendentals (2nd Edition)
- 5. For the function y-x³-3x²-1, use derivatives to: (a) determine the intervals of increase and decrease. (b) determine the local (relative) maxima and minima. (e) determine the intervals of concavity. (d) determine the points of inflection. (e) sketch the graph with the above information indicated on the graph.arrow_forwardCan you solve this 2 question numerical methodarrow_forward1. Estimate the area under the graph of f(x)-25-x from x=0 to x=5 using 5 approximating rectangles Using: (A) right endpoints. (B) left endpoints.arrow_forward
- 9. Use fundamental theorem of calculus to find the derivative d a) *dt sin(x) b)(x)√1-2 dtarrow_forward3. Evaluate the definite integral: a) √66x²+8dx b) x dx c) f*(2e* - 2)dx d) √√9-x² e) (2-5x)dx f) cos(x)dx 8)²₁₂√4-x2 h) f7dx i) f² 6xdx j) ²₂(4x+3)dxarrow_forward2. Consider the integral √(2x+1)dx (a) Find the Riemann sum for this integral using right endpoints and n-4. (b) Find the Riemann sum for this same integral, using left endpoints and n=4arrow_forward
- Problem 11 (a) A tank is discharging water through an orifice at a depth of T meter below the surface of the water whose area is A m². The following are the values of a for the corresponding values of A: A 1.257 1.390 x 1.50 1.65 1.520 1.650 1.809 1.962 2.123 2.295 2.462|2.650 1.80 1.95 2.10 2.25 2.40 2.55 2.70 2.85 Using the formula -3.0 (0.018)T = dx. calculate T, the time in seconds for the level of the water to drop from 3.0 m to 1.5 m above the orifice. (b) The velocity of a train which starts from rest is given by the fol- lowing table, the time being reckoned in minutes from the start and the speed in km/hour: | † (minutes) |2|4 6 8 10 12 14 16 18 20 v (km/hr) 16 28.8 40 46.4 51.2 32.0 17.6 8 3.2 0 Estimate approximately the total distance ran in 20 minutes.arrow_forwardX Solve numerically: = 0,95 In xarrow_forwardX Solve numerically: = 0,95 In xarrow_forward
- Please as many detarrow_forward8–23. Sketching vector fields Sketch the following vector fieldsarrow_forward25-30. Normal and tangential components For the vector field F and curve C, complete the following: a. Determine the points (if any) along the curve C at which the vector field F is tangent to C. b. Determine the points (if any) along the curve C at which the vector field F is normal to C. c. Sketch C and a few representative vectors of F on C. 25. F = (2½³, 0); c = {(x, y); y − x² = 1} 26. F = x (23 - 212) ; C = {(x, y); y = x² = 1}) , 2 27. F(x, y); C = {(x, y): x² + y² = 4} 28. F = (y, x); C = {(x, y): x² + y² = 1} 29. F = (x, y); C = 30. F = (y, x); C = {(x, y): x = 1} {(x, y): x² + y² = 1}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





