Algebra 1
Algebra 1
1st Edition
ISBN: 9780078884801
Author: McGraw-Hill/Glencoe
Publisher: Glencoe/McGraw-Hill School Pub Co
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Chapter 0.1, Problem 2E
To determine

To calculate: The final cost of boarding the dog at a kennel for four days by Ms. Hernandez.

Expert Solution & Answer
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Answer to Problem 2E

The final cost of boarding the dog at a kennel for four days by Ms. Hernandez is $111 .

Explanation of Solution

Given information:

The per day cost to board dog at a kennel is $28.90 . Ms. Hernandez boarded for four days and she had a coupon for $5 as a discount.

Formula used:

Four step plan is used to solve situations with help ofmathematical tools.

First understand the situation. Plan the solution according to it then solve it using mathematical tools and at last verify your solution.

Calculation:

It is provided that the per day cost to board dog at a kennel is $28.90 . Ms. Hernandez boarded for four days and she had a coupon for $5 as a discount.

An estimated answer is required for the problem as cost in dollars is to be evaluated and money can be easily returned back.

Total cost is per day cost for boarding the dog multiplied by number of days and discount coupon amount is subtracted from it.

Denote total cost by C ,

Therefore,

  C=28.90×45=110.6

Round off the result to nearest tenth, to obtain cost, C=111 .

Since, the total cost is positive number, the answer is verified.

Thus, the final cost of boarding the dog at a kennel for four days by Ms. Hernandez is $111 .

Chapter 0 Solutions

Algebra 1

Ch. 0.2 - Prob. 5ECh. 0.2 - Prob. 6ECh. 0.2 - Prob. 7ECh. 0.2 - Prob. 8ECh. 0.2 - Prob. 9ECh. 0.2 - Prob. 10ECh. 0.2 - Prob. 11ECh. 0.2 - Prob. 12ECh. 0.2 - Prob. 13ECh. 0.2 - Prob. 14ECh. 0.2 - Prob. 15ECh. 0.2 - Prob. 16ECh. 0.2 - Prob. 17ECh. 0.2 - Prob. 18ECh. 0.2 - Prob. 19ECh. 0.2 - Prob. 20ECh. 0.2 - Prob. 21ECh. 0.2 - Prob. 22ECh. 0.2 - Prob. 23ECh. 0.2 - Prob. 24ECh. 0.2 - Prob. 25ECh. 0.2 - Prob. 26ECh. 0.2 - Prob. 27ECh. 0.2 - Prob. 28ECh. 0.2 - Prob. 29ECh. 0.2 - Prob. 30ECh. 0.2 - Prob. 31ECh. 0.2 - Prob. 32ECh. 0.2 - Prob. 33ECh. 0.2 - Prob. 34ECh. 0.2 - Prob. 35ECh. 0.3 - Prob. 1ECh. 0.3 - Prob. 2ECh. 0.3 - Prob. 3ECh. 0.3 - Prob. 4ECh. 0.3 - Prob. 5ECh. 0.3 - Prob. 6ECh. 0.3 - Prob. 7ECh. 0.3 - Prob. 8ECh. 0.3 - Prob. 9ECh. 0.3 - Prob. 10ECh. 0.3 - Prob. 11ECh. 0.3 - Prob. 12ECh. 0.3 - Prob. 13ECh. 0.3 - Prob. 14ECh. 0.3 - Prob. 15ECh. 0.3 - Prob. 16ECh. 0.3 - Prob. 17ECh. 0.3 - Prob. 18ECh. 0.3 - Prob. 19ECh. 0.3 - Prob. 20ECh. 0.3 - Prob. 21ECh. 0.3 - Prob. 22ECh. 0.3 - 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Prob. 24POCh. 0 - Prob. 25POCh. 0 - Prob. 26POCh. 0 - Prob. 27POCh. 0 - Prob. 28POCh. 0 - Prob. 29POCh. 0 - Prob. 30POCh. 0 - Prob. 31POCh. 0 - Prob. 32POCh. 0 - Prob. 33POCh. 0 - Prob. 34POCh. 0 - Prob. 35POCh. 0 - Prob. 36POCh. 0 - Prob. 37POCh. 0 - Prob. 38POCh. 0 - Prob. 39POCh. 0 - Prob. 40POCh. 0 - Prob. 41POCh. 0 - Prob. 42POCh. 0 - Prob. 43POCh. 0 - Prob. 44POCh. 0 - Prob. 45POCh. 0 - Prob. 46POCh. 0 - Prob. 47POCh. 0 - Prob. 48POCh. 0 - Prob. 49POCh. 0 - Prob. 50POCh. 0 - Prob. 51POCh. 0 - Prob. 52POCh. 0 - Prob. 53POCh. 0 - Prob. 54POCh. 0 - Prob. 55POCh. 0 - Prob. 56POCh. 0 - Prob. 57POCh. 0 - Prob. 58POCh. 0 - Prob. 59POCh. 0 - Prob. 60POCh. 0 - Prob. 61POCh. 0 - Prob. 62POCh. 0 - Prob. 63POCh. 0 - Prob. 64POCh. 0 - Prob. 53PRCh. 0 - Prob. 54PRCh. 0 - Prob. 55PRCh. 0 - Prob. 56PRCh. 0 - Prob. 57PRCh. 0 - Prob. 58PRCh. 0 - Prob. 59PRCh. 0 - Prob. 65POCh. 0 - Prob. 66POCh. 0 - Prob. 67POCh. 0 - Prob. 68POCh. 0 - Prob. 69POCh. 0 - Prob. 70POCh. 0 - Prob. 71PO
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