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: 134

Answer

$158.3$

Work Step by Step

Here is the formula for the NFL rating: $\text{Rating}=\dfrac{\left(250\cdot \frac{C}{A}\right) + \left(1000\cdot \frac{T}{A}\right) + \left(12.5 \cdot \frac{Y}{A}\right) +6.25-\left(1250\cdot \frac{I}{A}\right)}{3}$ where A = attempted passes, C = completed passes, T= touchdown passes, Y= yards gained passing, and I = interceptions Here are the maximums for each of the weighing factors: $C/A= 0.775\\ T/A = 0.11875\\ Y/A = 12.5\\ I/A = 0.0995\\$ We want the lowest amount subtracted possible, so we will use $0$ for $\frac{I}{A}$ which is acceptable because $0.0995$ is the MAXIMUM, not the minimum. Subsitutute each of these values into the rating formula above to obtain: $\text{Rating}=\dfrac{250\cdot 0.775 +1000\cdot 0.11875 + 12.5 \cdot 12.5-1250\cdot 0}{3}=158.33\approx 158.3$
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