Algebra 2
Algebra 2
1st Edition
ISBN: 9780078884825
Author: McGraw-Hill/Glencoe
Publisher: Glencoe/McGraw-Hill School Pub Co
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Question
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Chapter 10, Problem 15STP

(a)

To determine

To find: The difference in the sales of yearbook and frames.

(a)

Expert Solution
Check Mark

Answer to Problem 15STP

Explanation of Solution

Given information:

The given table is:

  Algebra 2, Chapter 10, Problem 15STP , additional homework tip  1

Calculation:

Calculate the table.

The sales of yearbook made to the 11th grade classes and 10th classes.

Difference in sales of yearbook made to the 11th grade classes and 10th classes.

  546464=82

Difference in sales of frames made to the 11th grade classes and 10th classes.

  344278=66

Therefore the difference in the sales of yearbook and frames 66.

(b)

To determine

To find: the total number of yearbook and frames.

(b)

Expert Solution
Check Mark

Answer to Problem 15STP

The total number of yearbook and frames is 1375.

Explanation of Solution

Given information:

The given table is.

  Algebra 2, Chapter 10, Problem 15STP , additional homework tip  2

Calculation:

Calculate the table.

Total number of yearbook sold is equal to.

  423+464+546+575=2008

Total number of frames sold is equal to.

  256+278+344+497=1375

Therefore the total number of yearbook and frames is 1375.

(C)

To determine

To find: the sales of yearbook and frames for each class.

(C)

Expert Solution
Check Mark

Answer to Problem 15STP

The total number of sales of yearbook and frames is each class 36546$

.

Explanation of Solution

Given information:

The given table is.

  Algebra 2, Chapter 10, Problem 15STP , additional homework tip  3

Calculation:

Calculate the table.

Cost of a yearbook is equal to $48 and cost of a frame is equal to $18.

The amount of sales of yearbook and frames for each class.

  $48*423+$18*256=20304+4608=24912$$48*464+$18*278=22272+5004=27276$$48*546+$18*497=27600+8946=36546$@48*546+$18*344=26208+6192=32400$

Therefore the total number of sales of yearbook and frames is each class 36546$.

Chapter 10 Solutions

Algebra 2

Ch. 10.1 - Prob. 7CYUCh. 10.1 - Prob. 8CYUCh. 10.1 - Prob. 9CYUCh. 10.1 - Prob. 10PPSCh. 10.1 - Prob. 11PPSCh. 10.1 - Prob. 12PPSCh. 10.1 - Prob. 13PPSCh. 10.1 - Prob. 14PPSCh. 10.1 - Prob. 15PPSCh. 10.1 - Prob. 16PPSCh. 10.1 - Prob. 17PPSCh. 10.1 - Prob. 18PPSCh. 10.1 - Prob. 19PPSCh. 10.1 - Prob. 20PPSCh. 10.1 - Prob. 21PPSCh. 10.1 - Prob. 22PPSCh. 10.1 - Prob. 23PPSCh. 10.1 - Prob. 24PPSCh. 10.1 - Prob. 25PPSCh. 10.1 - Prob. 26PPSCh. 10.1 - Prob. 27PPSCh. 10.1 - Prob. 28PPSCh. 10.1 - Prob. 29PPSCh. 10.1 - Prob. 30PPSCh. 10.1 - Prob. 31PPSCh. 10.1 - Prob. 32PPSCh. 10.1 - Prob. 33PPSCh. 10.1 - Prob. 34PPSCh. 10.1 - Prob. 35PPSCh. 10.1 - Prob. 36PPSCh. 10.1 - Prob. 37PPSCh. 10.1 - Prob. 38PPSCh. 10.1 - Prob. 39PPSCh. 10.1 - Prob. 40PPSCh. 10.1 - Prob. 41PPSCh. 10.1 - Prob. 42HPCh. 10.1 - Prob. 43HPCh. 10.1 - Prob. 44HPCh. 10.1 - Prob. 45HPCh. 10.1 - Prob. 46HPCh. 10.1 - Prob. 47STPCh. 10.1 - Prob. 48STPCh. 10.1 - Prob. 49STPCh. 10.1 - Prob. 50STPCh. 10.1 - Prob. 51SRCh. 10.1 - Prob. 52SRCh. 10.1 - 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Use of ALGEBRA in REAL LIFE; Author: Fast and Easy Maths !;https://www.youtube.com/watch?v=9_PbWFpvkDc;License: Standard YouTube License, CC-BY
Compound Interest Formula Explained, Investment, Monthly & Continuously, Word Problems, Algebra; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=P182Abv3fOk;License: Standard YouTube License, CC-BY
Applications of Algebra (Digit, Age, Work, Clock, Mixture and Rate Problems); Author: EngineerProf PH;https://www.youtube.com/watch?v=Y8aJ_wYCS2g;License: Standard YouTube License, CC-BY