BIG IDEAS MATH Algebra 1: Common Core Student Edition 2015
BIG IDEAS MATH Algebra 1: Common Core Student Edition 2015
1st Edition
ISBN: 9781608408382
Author: HOUGHTON MIFFLIN HARCOURT
Publisher: Houghton Mifflin Harcourt
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 11.1, Problem 31E

(a)

To determine

To find which Team win the match.

(a)

Expert Solution
Check Mark

Answer to Problem 31E

Team A wins the match because it has greater mean than Team B.

If Team with greater median will win, then Team B wins.

Explanation of Solution

Given information: Team A Score: 172,130,173,212 and Team B Score: 136,184,168,192Formula Used: Mean = SumofallobservationsTotalobservations and Median = (n2)th+(n2+1)th2term

Calculation:

A) Team A:

Team A Score: 172,130,173,212

So, In Ascending Order

Team A Score: 130,172,173,212

  • Mean

Mean = SumofallobservationsTotalobservations

So,

Mean = 130+172+173+2124

Now,

Mean = 6874

Hence,

Mean = 171.75

  • Median

Median for even numbers = (n2)th+(n2+1)th2term

So,

Median = (42)th+(42+1)th2term

Now,

Median = (2)nd+(3)rd2term

As, (2)ndterm=172 and (3)rdterm=173

So,

Median = 172+1732

Hence,

Median = 172.5

B) Team B:

Team B Score: 136,184,168,192

So, In Ascending Order

Team B Score: 136,168,184,192

  • Mean

Mean = SumofallobservationsTotalobservations

So,

Mean = 136+168+184+1924

Now,

Mean = 6804

Hence,

Mean = 170

  • Median

Median for even numbers = (n2)th+(n2+1)th2term

So,

Median = (42)th+(42+1)th2term

Now,

Median = (2)nd+(3)rd2term

As, (2)ndterm=168 and (3)rdterm=184

So,

Median = 168+1842

Hence,

Median = 176

So,

Team A wins the match because it has greater mean than Team B.

Now,

If Team with greater median will win, then Team B wins.

(b)

To determine

To find which Team is more consistent.

(b)

Expert Solution
Check Mark

Answer to Problem 31E

Team A.

Explanation of Solution

Given information: Team A: Mean = 171.75 , Median = 172.5 and Team B: Mean = 170 , Median = 176

Calculation:

Team A: Mean = 171.75 , Median = 172.5

And

Team B: Mean = 170 , Median = 176

So,

Mean and Median values of Team A are much similar than Team B.

Hence,

Team A is more consistent.

(c)

To determine

To find which Team win the match.

(c)

Expert Solution
Check Mark

Answer to Problem 31E

Team B wins the match because it has greater mean than Team A.

Explanation of Solution

Given information: Team A Score: 172,130,173,212 and Team B Score: 136,184,168,192

Team A score increase by 15 and Team B score increase by 12.5%

Formula Used: Mean = SumofallobservationsTotalobservations

Calculation:

A) Team A:

Team A Score: 172,130,173,212

As,

Team A score increase by 15

Now,

Team A Score: 187,145,188,227

  • Mean

Mean = SumofallobservationsTotalobservations

So,

Mean = 187+145+188+2274

Now,

Mean = 7474

Hence,

Mean = 186.75

B) Team B:

Team B Score: 136,184,168,192

As,

Team B score increase by 12.5%

Now,

Team B Score: 153,207,189,216

  • Mean

Mean = SumofallobservationsTotalobservations

So,

Mean = 153+207+189+2164

Now,

Mean = 7654

Hence,

Mean = 191.25

So,

Team B wins the match because it has greater mean than Team A.

Chapter 11 Solutions

BIG IDEAS MATH Algebra 1: Common Core Student Edition 2015

Ch. 11.1 - Prob. 11ECh. 11.1 - Prob. 12ECh. 11.1 - Prob. 13ECh. 11.1 - Prob. 14ECh. 11.1 - Prob. 15ECh. 11.1 - Prob. 16ECh. 11.1 - Prob. 17ECh. 11.1 - Prob. 18ECh. 11.1 - Prob. 19ECh. 11.1 - Prob. 20ECh. 11.1 - Prob. 21ECh. 11.1 - Prob. 22ECh. 11.1 - Prob. 23ECh. 11.1 - Prob. 24ECh. 11.1 - Prob. 25ECh. 11.1 - Prob. 26ECh. 11.1 - Prob. 27ECh. 11.1 - Prob. 28ECh. 11.1 - Prob. 29ECh. 11.1 - Prob. 30ECh. 11.1 - Prob. 31ECh. 11.1 - Prob. 32ECh. 11.1 - Prob. 33ECh. 11.1 - Prob. 34ECh. 11.1 - Prob. 35ECh. 11.1 - Prob. 36ECh. 11.1 - Prob. 37ECh. 11.1 - Prob. 38ECh. 11.1 - Prob. 39ECh. 11.1 - Prob. 40ECh. 11.1 - Prob. 41ECh. 11.1 - Prob. 42ECh. 11.1 - Prob. 43ECh. 11.1 - Prob. 44ECh. 11.1 - Prob. 45ECh. 11.1 - Prob. 46ECh. 11.1 - Prob. 47ECh. 11.2 - Prob. 1ECh. 11.2 - Prob. 2ECh. 11.2 - Prob. 3ECh. 11.2 - Prob. 4ECh. 11.2 - Prob. 5ECh. 11.2 - Prob. 6ECh. 11.2 - Prob. 7ECh. 11.2 - Prob. 8ECh. 11.2 - Prob. 9ECh. 11.2 - Prob. 10ECh. 11.2 - Prob. 11ECh. 11.2 - Prob. 12ECh. 11.2 - Prob. 13ECh. 11.2 - Prob. 14ECh. 11.2 - Prob. 15ECh. 11.2 - Prob. 16ECh. 11.2 - Prob. 17ECh. 11.2 - Prob. 18ECh. 11.2 - Prob. 19ECh. 11.2 - Prob. 20ECh. 11.2 - Prob. 21ECh. 11.2 - Prob. 22ECh. 11.2 - Prob. 23ECh. 11.2 - Prob. 24ECh. 11.2 - Prob. 25ECh. 11.2 - Prob. 26ECh. 11.2 - Prob. 27ECh. 11.3 - Prob. 1ECh. 11.3 - Prob. 2ECh. 11.3 - Prob. 3ECh. 11.3 - Prob. 4ECh. 11.3 - Prob. 5ECh. 11.3 - Prob. 6ECh. 11.3 - Prob. 7ECh. 11.3 - Prob. 8ECh. 11.3 - Prob. 9ECh. 11.3 - Prob. 10ECh. 11.3 - Prob. 11ECh. 11.3 - Prob. 12ECh. 11.3 - Prob. 13ECh. 11.3 - Prob. 14ECh. 11.3 - Prob. 15ECh. 11.3 - Prob. 16ECh. 11.3 - Prob. 17ECh. 11.3 - Prob. 18ECh. 11.3 - Prob. 19ECh. 11.3 - Prob. 20ECh. 11.3 - Prob. 21ECh. 11.3 - Prob. 22ECh. 11.3 - Prob. 23ECh. 11.3 - Prob. 24ECh. 11.3 - Prob. 25ECh. 11.3 - Prob. 26ECh. 11.3 - Prob. 27ECh. 11.3 - Prob. 1QCh. 11.3 - Prob. 2QCh. 11.3 - Prob. 3QCh. 11.3 - Prob. 4QCh. 11.3 - Prob. 5QCh. 11.3 - Prob. 6QCh. 11.3 - Prob. 7QCh. 11.3 - Prob. 8QCh. 11.3 - Prob. 9QCh. 11.4 - Prob. 1ECh. 11.4 - Prob. 2ECh. 11.4 - Prob. 3ECh. 11.4 - Prob. 4ECh. 11.4 - Prob. 5ECh. 11.4 - Prob. 6ECh. 11.4 - Prob. 7ECh. 11.4 - Prob. 8ECh. 11.4 - Prob. 9ECh. 11.4 - Prob. 10ECh. 11.4 - Prob. 11ECh. 11.4 - Prob. 12ECh. 11.4 - Prob. 13ECh. 11.4 - Prob. 14ECh. 11.4 - Prob. 15ECh. 11.4 - Prob. 16ECh. 11.4 - Prob. 17ECh. 11.4 - Prob. 18ECh. 11.4 - Prob. 19ECh. 11.4 - Prob. 20ECh. 11.4 - Prob. 21ECh. 11.4 - Prob. 22ECh. 11.4 - Prob. 23ECh. 11.4 - Prob. 24ECh. 11.4 - Prob. 25ECh. 11.4 - Prob. 26ECh. 11.4 - Prob. 27ECh. 11.4 - Prob. 28ECh. 11.4 - Prob. 29ECh. 11.4 - Prob. 30ECh. 11.4 - Prob. 31ECh. 11.4 - Prob. 32ECh. 11.4 - Prob. 33ECh. 11.4 - Prob. 34ECh. 11.5 - Prob. 1ECh. 11.5 - Prob. 2ECh. 11.5 - Prob. 3ECh. 11.5 - Prob. 4ECh. 11.5 - Prob. 5ECh. 11.5 - Prob. 6ECh. 11.5 - Prob. 7ECh. 11.5 - Prob. 8ECh. 11.5 - Prob. 9ECh. 11.5 - Prob. 10ECh. 11.5 - Prob. 11ECh. 11.5 - Prob. 12ECh. 11.5 - Prob. 13ECh. 11.5 - Prob. 14ECh. 11.5 - Prob. 15ECh. 11.5 - Prob. 16ECh. 11.5 - Prob. 17ECh. 11.5 - Prob. 18ECh. 11.5 - Prob. 19ECh. 11.5 - Prob. 20ECh. 11.5 - Prob. 21ECh. 11.5 - Prob. 22ECh. 11.5 - Prob. 23ECh. 11.5 - Prob. 24ECh. 11.5 - Prob. 25ECh. 11.5 - Prob. 26ECh. 11.5 - Prob. 27ECh. 11.5 - Prob. 28ECh. 11.5 - Prob. 29ECh. 11.5 - Prob. 30ECh. 11.5 - Prob. 31ECh. 11.5 - Prob. 32ECh. 11.5 - Prob. 33ECh. 11 - Prob. 1CRCh. 11 - Prob. 2CRCh. 11 - Prob. 3CRCh. 11 - Prob. 4CRCh. 11 - Prob. 5CRCh. 11 - Prob. 6CRCh. 11 - Prob. 7CRCh. 11 - Prob. 8CRCh. 11 - Prob. 9CRCh. 11 - Prob. 10CRCh. 11 - Prob. 11CRCh. 11 - Prob. 12CRCh. 11 - Prob. 13CRCh. 11 - Prob. 14CRCh. 11 - Prob. 15CRCh. 11 - Prob. 1CTCh. 11 - Prob. 2CTCh. 11 - Prob. 3CTCh. 11 - Prob. 4CTCh. 11 - Prob. 5CTCh. 11 - Prob. 6CTCh. 11 - Prob. 7CTCh. 11 - Prob. 8CTCh. 11 - Prob. 9CTCh. 11 - Prob. 10CTCh. 11 - Prob. 1CACh. 11 - Prob. 2CACh. 11 - Prob. 3CACh. 11 - Prob. 4CACh. 11 - Prob. 5CACh. 11 - Prob. 6CACh. 11 - Prob. 7CACh. 11 - Prob. 8CACh. 11 - Prob. 9CA
Knowledge Booster
Background pattern image
Algebra
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:PEARSON
Text book image
Contemporary Abstract Algebra
Algebra
ISBN:9781305657960
Author:Joseph Gallian
Publisher:Cengage Learning
Text book image
Linear Algebra: A Modern Introduction
Algebra
ISBN:9781285463247
Author:David Poole
Publisher:Cengage Learning
Text book image
Algebra And Trigonometry (11th Edition)
Algebra
ISBN:9780135163078
Author:Michael Sullivan
Publisher:PEARSON
Text book image
Introduction to Linear Algebra, Fifth Edition
Algebra
ISBN:9780980232776
Author:Gilbert Strang
Publisher:Wellesley-Cambridge Press
Text book image
College Algebra (Collegiate Math)
Algebra
ISBN:9780077836344
Author:Julie Miller, Donna Gerken
Publisher:McGraw-Hill Education
The Shape of Data: Distributions: Crash Course Statistics #7; Author: CrashCourse;https://www.youtube.com/watch?v=bPFNxD3Yg6U;License: Standard YouTube License, CC-BY
Shape, Center, and Spread - Module 20.2 (Part 1); Author: Mrmathblog;https://www.youtube.com/watch?v=COaid7O_Gag;License: Standard YouTube License, CC-BY
Shape, Center and Spread; Author: Emily Murdock;https://www.youtube.com/watch?v=_YyW0DSCzpM;License: Standard Youtube License