how would exponents be used in a math (runs per game )

by Dan ONeill
(Boston, MA. USA)

how is this solved ??

(5^1.95)/(5^1.95)+ (3^1.95)= % ??

also:
(5^1.805)/ (5^1.805) + (3^1.805)= % ??

Could you please explain how each is solved

Comments for how would exponents be used in a math (runs per game )

 Aug 25, 2011 Exponents and Runs per Game by: Staff The question: by Dan ONeill (Boston, MA. USA) (5^1.95)/(5^1.95)+ (3^1.95)= % ?? also: (5^1.805)/ (5^1.805) + (3^1.805)= % ?? Could you please explain how each is solved The answer: I’m unclear if you meant what you wrote, or if you actually meant this: The Pythagorean expectation: (used to estimate the proportion of how many games a baseball team should win): Win ≈ (runs scored)²/[(runs scored)² + (runs allowed)²] The calculations you asked for help on are: a. (5^1.95)/(5^1.95)+ (3^1.95)= % ?? (5^1.95)/(5^1.95)+ (3^1.95) = 1+ (3^1.95) = 1 + 8.51896 = 9.51896 (this is not a percent) b. (5^1.805)/ (5^1.805) + (3^1.805)= = % ?? (5^1.805)/ (5^1.805) + (3^1.805) = 1+ (3^1.805) = 1+ 7.26447 = 8.26447 (this is not a percent) The calculations for the Pythagorean expectation are: a1. (5^1.95)/[(5^1.95)+ (3^1.95)]= % ?? (5^1.95)/[(5^1.95)+ (3^1.95)] = 23.067/(23.067 + 8.51896) = 23.067/(31.586) = 0.730292 = 73.0292% b1. (5^1.805)/ (5^1.805) + (3^1.805) = % ?? (5^1.805)/ [(5^1.805) + (3^1.805)] = 18.2659/(18.2659 + 7.26447) = 18.2659/(25.5304) = 0.715457 = 71.5457% Final Note: Any number can be converted to a %. Simply multiply the number by 100. A % is used to compare a number (any number) to 100 using division. For example, the number 8.26447 in (part b of your question) can be converted to 826.447 percent (because 826.447 ÷ 100 = 8.26447). However, this is a bit misleading. A % sign should only used to represent a proportion. A percent tells you that you are dealing with a ratio. Thanks for writing. Staff www.solving-math-problems.com

Join in and write your own page! It's easy to do. How? Simply click here to return to Math Questions & Comments - 01.

Copyright © 2008-2015. All rights reserved. Solving-Math-Problems.com