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November 5th, 2009, 02:10 AM
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| | can you assist please? hi guys can someone assist with this one please before i pull any more hair out.
the age of a machine,x, in years is related to the probability of breakdown,y, by the formula
x=3 + ln( y/1-y)
determine the probability of breakdown for 1,3,10 years!
i am fairly sure 1st i take away the 3 from both sides thus leaving x-3=ln(y/1-y).
then inverse of ln makes y/1-y=e^x-3?
what do you think any help appreciated,logs,powers etc not my strong point!
ps. hope this is the correct section to be in.
thankyou | 
November 5th, 2009, 02:52 AM
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| | Quote:
the age of a machine,x, in years is related to the probability of breakdown,y, by the formula
x=3 + ln( y/1-y)
determine the probability of breakdown for 1,3,10 years!
i am fairly sure 1st i take away the 3 from both sides thus leaving x-3=ln(y/1-y).
then inverse of ln makes y/1-y=e^x-3?
| yes, now multiply both sides by 1-y, and move the ye^x-3 to the left and factorise out the y and then divide both sides by e^x-3
y= (1-y) (e^x-3)
y=e^x-3 - y e^x-3
y + y e^x-3=e^x-3
y(1 + e^x-3)=e^x-3
y=e^x-3/(1 + e^x-3)
now you got y by itself.
sub in those values of x to get probabilty.
it seems correct but correct me if im wrong
Last edited by mr fantastic; November 5th, 2009 at 05:11 AM.
Reason: Added quote tags.
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November 5th, 2009, 03:04 AM
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| | ah of course! how stupid of me.
i was actually trying to cross multiply which didnt leave y on its own ,many thanks for the very quick reply. | 
November 5th, 2009, 03:11 AM
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| | Quote:
Originally Posted by ads114 ah of course! how stupid of me.
i was actually trying to cross multiply which didnt leave y on its own ,many thanks for the very quick reply.  | np glad i could help.
btw theres a thanks button some where | 
November 5th, 2009, 04:31 AM
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| | Oddly enough, posts that I, myself, made, don't have a "thank you" button by them! That cuts down on my "thank you"s a lot! | | The following users thank HallsofIvy for this useful post: | |  | | Thread Tools | | | | Display Modes | Linear Mode |
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