SOLUTION: Hello here is a word problem:
Due to melting, an ice sculpture loses one-half its weight every hour. After 8 hours, it weights 5/16 of a pound. How much did it weigh in the beginn
Algebra ->
Percentage-and-ratio-word-problems
-> SOLUTION: Hello here is a word problem:
Due to melting, an ice sculpture loses one-half its weight every hour. After 8 hours, it weights 5/16 of a pound. How much did it weigh in the beginn
Log On
Question 167511: Hello here is a word problem:
Due to melting, an ice sculpture loses one-half its weight every hour. After 8 hours, it weights 5/16 of a pound. How much did it weigh in the beginning?
I know the answer is 80lb but don't understand how to set the problem up.
Thank you for helping. Found 2 solutions by ankor@dixie-net.com, Mathtut:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Due to melting, an ice sculpture loses one-half its weight every hour. After 8 hours, it weights 5/16 of a pound. How much did it weigh in the beginning?
;
Use the decay formula: A = Ao(2^(-t/h))
where
Ao is initial amt (what we are solving for)
A = resulting amt (5/16)
t = time in hrs (8)
h = time required for substance to lose half it's weight (1hr)
:
Ao(2^(-8/1)) = .3125; (decimal value for 5/16)
;
Using a calc find 2^-8
Ao * .0039 = .3125
Ao =
Ao = 80.1 ~ 80 lbs
You can put this solution on YOUR website!
the formula I came up with was where x=original weight and n is the number of hours and r equals the remaining weight after n hours.
it evolved from this (x-(1/2)x)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)
if I plug given numbers into the equation I get ----> now cross multiply
(