Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AMC 10B Problems/Problem 9"

(→‎Solution)
(→‎Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
Since <math>AB = AC</math>, then <math>\triangle ABC</math> is isosceles, so <math>BD = CD</math>. Therefore, the coordinates of <math>C</math> are <math>(-1 - 3, 3 + 6) = \boxed{\textbf{(C) } (-4,9)}</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2017|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:50, 16 February 2017

Problem

A radio program has a quiz consisting of $3$ multiple-choice questions, each with $3$ choices. A contestant wins if he or she gets $2$ or more of the questions right. The contestant answers randomly to each question. What is the probability of winning?

$\textbf{(A)}\ \frac{1}{27}\qquad\textbf{(B)}\ \frac{1}{9}\qquad\textbf{(C)}\ \frac{2}{9}\qquad\textbf{(D)}\ \frac{7}{27}\qquad\textbf{(E)}\ \frac{1}{2}$

Solution

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_9&oldid=83768"