Geometry For Enjoyment And Challenge
91st Edition
ISBN: 9780866099653
Author: Richard Rhoad, George Milauskas, Robert Whipple
Publisher: McDougal Littell
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 1.9, Problem 4PSA
To determine
To find the probability of the angle to be straight when one of the five
Expert Solution & Answer
Answer to Problem 4PSA
Explanation of Solution
Given information: The angles can be classified as follows:
Formula used:
Probability means possibility.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
Probability of event to happen is
Calculation:
A straight angle is an angle that is exactly 180 degrees.As you can see, a straight angle forms a straight line.
In our example, there is no favorable outcome.
There are five total number of outcomes, i.e.
Therefore, the probability will be
Chapter 1 Solutions
Geometry For Enjoyment And Challenge
Ch. 1.1 - Prob. 1PSACh. 1.1 - Prob. 2PSACh. 1.1 - Prob. 3PSACh. 1.1 - Prob. 4PSACh. 1.1 - Prob. 5PSACh. 1.1 - Prob. 6PSACh. 1.1 - Prob. 7PSACh. 1.1 - Prob. 8PSACh. 1.1 - Prob. 9PSACh. 1.1 - Prob. 10PSA
Ch. 1.1 - Prob. 11PSBCh. 1.1 - Prob. 12PSBCh. 1.1 - Prob. 13PSCCh. 1.1 - Prob. 14PSCCh. 1.2 - Prob. 1PSACh. 1.2 - Prob. 2PSACh. 1.2 - Prob. 3PSACh. 1.2 - Prob. 4PSACh. 1.2 - Prob. 5PSACh. 1.2 - Prob. 6PSACh. 1.2 - Prob. 7PSACh. 1.2 - Prob. 8PSACh. 1.2 - Prob. 9PSACh. 1.2 - Prob. 10PSACh. 1.2 - Prob. 11PSACh. 1.2 - Prob. 12PSBCh. 1.2 - Prob. 13PSBCh. 1.2 - Prob. 14PSBCh. 1.2 - Prob. 15PSBCh. 1.2 - Prob. 16PSBCh. 1.2 - Prob. 17PSBCh. 1.2 - Prob. 18PSBCh. 1.2 - Prob. 19PSCCh. 1.2 - Prob. 20PSCCh. 1.2 - Prob. 21PSCCh. 1.2 - Prob. 22PSCCh. 1.2 - Prob. 23PSCCh. 1.3 - Prob. 1PSACh. 1.3 - Prob. 2PSACh. 1.3 - Prob. 3PSACh. 1.3 - Prob. 4PSACh. 1.3 - Prob. 5PSACh. 1.3 - Prob. 6PSACh. 1.3 - Prob. 7PSACh. 1.3 - Prob. 8PSACh. 1.3 - Prob. 9PSACh. 1.3 - Prob. 10PSBCh. 1.3 - Prob. 11PSBCh. 1.3 - Prob. 12PSBCh. 1.3 - Prob. 13PSBCh. 1.3 - Prob. 14PSBCh. 1.3 - Prob. 15PSCCh. 1.3 - Prob. 16PSCCh. 1.3 - Prob. 17PSCCh. 1.4 - Prob. 1PSACh. 1.4 - Prob. 2PSACh. 1.4 - Prob. 3PSACh. 1.4 - Prob. 4PSACh. 1.4 - Prob. 5PSACh. 1.4 - Prob. 6PSACh. 1.4 - Prob. 7PSACh. 1.4 - Prob. 8PSACh. 1.4 - Prob. 9PSACh. 1.4 - Prob. 10PSACh. 1.4 - Prob. 11PSBCh. 1.4 - Prob. 12PSBCh. 1.4 - Prob. 13PSBCh. 1.4 - Prob. 14PSCCh. 1.4 - Prob. 15PSCCh. 1.5 - Prob. 1PSACh. 1.5 - Prob. 2PSACh. 1.5 - Prob. 3PSACh. 1.5 - Prob. 4PSACh. 1.5 - Prob. 5PSACh. 1.5 - Prob. 6PSACh. 1.5 - Prob. 7PSACh. 1.5 - Prob. 8PSACh. 1.5 - Prob. 9PSACh. 1.5 - Prob. 10PSACh. 1.5 - Prob. 11PSACh. 1.5 - Prob. 12PSACh. 1.5 - Prob. 13PSACh. 1.5 - Prob. 14PSACh. 1.5 - Prob. 15PSACh. 1.5 - Prob. 16PSACh. 1.5 - Prob. 17PSACh. 1.5 - Prob. 18PSBCh. 1.5 - Prob. 19PSBCh. 1.5 - Prob. 20PSBCh. 1.5 - Prob. 21PSBCh. 1.5 - Prob. 22PSCCh. 1.5 - Prob. 23PSCCh. 1.5 - Prob. 24PSCCh. 1.6 - Prob. 1PSACh. 1.6 - Prob. 2PSACh. 1.6 - Prob. 3PSACh. 1.6 - Prob. 4PSACh. 1.6 - Prob. 5PSACh. 1.6 - Prob. 6PSACh. 1.6 - Prob. 7PSBCh. 1.6 - Prob. 8PSBCh. 1.6 - Prob. 9PSBCh. 1.6 - Prob. 10PSCCh. 1.6 - Prob. 11PSCCh. 1.6 - Prob. 12PSDCh. 1.7 - Prob. 1PSACh. 1.7 - Prob. 2PSACh. 1.7 - Prob. 3PSACh. 1.7 - Prob. 4PSACh. 1.7 - Prob. 5PSACh. 1.7 - Prob. 6PSACh. 1.7 - Prob. 7PSACh. 1.7 - Prob. 8PSBCh. 1.7 - Prob. 9PSBCh. 1.7 - Prob. 10PSBCh. 1.7 - Prob. 11PSBCh. 1.7 - Prob. 12PSBCh. 1.7 - Prob. 13PSCCh. 1.7 - Prob. 14PSCCh. 1.8 - Prob. 1PSACh. 1.8 - Prob. 2PSACh. 1.8 - Prob. 3PSACh. 1.8 - Prob. 4PSACh. 1.8 - Prob. 5PSACh. 1.8 - Prob. 6PSBCh. 1.8 - Prob. 7PSBCh. 1.8 - Prob. 8PSBCh. 1.8 - Prob. 9PSBCh. 1.8 - Prob. 10PSCCh. 1.9 - Prob. 1PSACh. 1.9 - Prob. 2PSACh. 1.9 - Prob. 3PSACh. 1.9 - Prob. 4PSACh. 1.9 - Prob. 5PSACh. 1.9 - Prob. 6PSBCh. 1.9 - Prob. 7PSBCh. 1.9 - Prob. 8PSBCh. 1.9 - Prob. 9PSBCh. 1.9 - Prob. 10PSBCh. 1.9 - Prob. 11PSBCh. 1.9 - Prob. 12PSCCh. 1.9 - Prob. 13PSCCh. 1.9 - Prob. 14PSCCh. 1.9 - Prob. 15PSCCh. 1 - Prob. 1RPCh. 1 - Prob. 2RPCh. 1 - Prob. 3RPCh. 1 - Prob. 4RPCh. 1 - Prob. 5RPCh. 1 - Prob. 6RPCh. 1 - Prob. 7RPCh. 1 - Prob. 8RPCh. 1 - Prob. 9RPCh. 1 - Prob. 10RPCh. 1 - Prob. 11RPCh. 1 - Prob. 12RPCh. 1 - Prob. 13RPCh. 1 - Prob. 14RPCh. 1 - Prob. 15RPCh. 1 - Prob. 16RPCh. 1 - Prob. 17RPCh. 1 - Prob. 18RPCh. 1 - Prob. 19RPCh. 1 - Prob. 20RPCh. 1 - Prob. 21RPCh. 1 - Prob. 22RPCh. 1 - Prob. 23RPCh. 1 - Prob. 24RPCh. 1 - Prob. 25RPCh. 1 - Prob. 26RPCh. 1 - Prob. 27RPCh. 1 - Prob. 28RPCh. 1 - Prob. 29RPCh. 1 - Prob. 30RPCh. 1 - Prob. 31RPCh. 1 - Prob. 32RPCh. 1 - Prob. 33RPCh. 1 - Prob. 34RPCh. 1 - Prob. 35RPCh. 1 - Prob. 36RPCh. 1 - Prob. 37RPCh. 1 - Prob. 38RPCh. 1 - Prob. 39RPCh. 1 - Prob. 40RPCh. 1 - Prob. 41RPCh. 1 - Prob. 42RPCh. 1 - Prob. 43RP
Additional Math Textbook Solutions
Find more solutions based on key concepts
In Exercises 17-20, refer to the accompanying table showing results from a Chembio test for hepatitis C among H...
Elementary Statistics (13th Edition)
4. You construct a 95% confidence interval for a population mean using a random sample. The confidence interval...
Elementary Statistics: Picturing the World (7th Edition)
In Exercises 1–8, use the Ratio Test to determine whether each series converges absolutely or diverges.
5.
University Calculus: Early Transcendentals (4th Edition)
Version 2 of the Chain Rule Use Version 2 of the Chain Rule to calculate the derivatives of the following funct...
Calculus: Early Transcendentals (2nd Edition)
CHECK POINT 1 Write a word description of the set L = {a, b, c, d, e, f}.
Thinking Mathematically (6th Edition)
A pair of fair dice is rolled. What is the probability that the second die lands on a higher value than does th...
A First Course in Probability (10th Edition)
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, geometry and related others by exploring similar questions and additional content below.Similar questions
- Find the area of a square whose diagonal is 10arrow_forwardDecomposition geometry: Mary is making a decorative yard space with dimensions as shaded in green (ΔOAB).Mary would like to cover the yard space with artificial turf (plastic grass-like rug). Mary reasoned that she could draw a rectangle around the figure so that the point O was at a vertex of the rectangle and that points A and B were on sides of the rectangle. Then she reasoned that the three smaller triangles resulting could be subtracted from the area of the rectangle. Mary determined that she would need 28 square meters of artificial turf to cover the green shaded yard space pictured exactly.arrow_forward7. 11 m 12.7 m 14 m S V=B₁+ B2(h) 9.5 m 16 m h+s 2 na 62-19 = 37 +, M h² = Bu-29arrow_forward
- 39 Two sides of one triangle are congruent to two sides of a second triangle, and the included angles are supplementary. The area of one triangle is 41. Can the area of the second triangle be found?arrow_forwardA parallelogram with an area of 211.41 m^2 hast a base Thatcher measures 24.3m. Find ist height.arrow_forwardBH is tangent to circle A and DF is a diameter. I don't know where to go from here. May you help please?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Elementary Geometry For College Students, 7eGeometryISBN:9781337614085Author:Alexander, Daniel C.; Koeberlein, Geralyn M.Publisher:Cengage,
Elementary Geometry for College StudentsGeometryISBN:9781285195698Author:Daniel C. Alexander, Geralyn M. KoeberleinPublisher:Cengage Learning
Elementary Geometry For College Students, 7e
Geometry
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Cengage,
Elementary Geometry for College Students
Geometry
ISBN:9781285195698
Author:Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:Cengage Learning
Surface Area Of A Sphere | Geometry | Math | Letstute; Author: Let'stute;https://www.youtube.com/watch?v=T_DBkFnr4NM;License: Standard YouTube License, CC-BY