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: 121

Answer

$\text{Rating }\approx 111.0 $

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 $A= 492 ,$ $C= 324 ,$ $T= 36 ,$ $Y= 3900 ,$ and $I= 4 ,$ then \begin{array}{l}\require{cancel} \text{Rating = } \dfrac{\left(250\cdot\dfrac{ 324 }{ 492 } \right)+\left(1000\cdot\dfrac{ 36 }{ 492 } \right)+\left(12.5\cdot\dfrac{ 3900 }{ 492 } \right)+6.25-\left(1250\cdot\dfrac{ 4 }{ 492 } \right)}{3} \\\\ \text{Rating = } 110.9925 \\\\ \text{Rating }\approx 111.0 .\end{array}
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