College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.2 - Real Numbers and Their Properties - R.2 Exercises - Page 20: 127

Answer

$\text{Rating }\approx 93.6 $

Work Step by Step

Using the given formula, $ \text{Rating = }\dfrac{\left(250\cdot\dfrac{C}{A} \right)+\left(1000\cdot\dfrac{T}{A} \right)+\left(12.5\cdot\dfrac{Y}{A} \right)+6.25-\left(1250\cdot\dfrac{I}{A} \right)}{3} ,$ where $A= 489 ,$ $C= 306 ,$ $T= 25 ,$ $Y= 3622 ,$ and $I= 10 ,$ then \begin{array}{l}\require{cancel} \text{Rating = } \dfrac{\left(250\cdot\dfrac{ 306 }{ 489 } \right)+\left(1000\cdot\dfrac{ 25 }{ 489 } \right)+\left(12.5\cdot\dfrac{ 3622 }{ 489 } \right)+6.25-\left(1250\cdot\dfrac{ 10 }{ 489 } \right)}{3} \\\\ \text{Rating = } 93.61367 \\\\ \text{Rating }\approx 93.6 .\end{array}
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