# HNC Maths Help

• Jan 22nd 2010, 05:55 AM
JP103
HNC Maths Help
Not sure if I have posted this in the right place, so I aplogise if I havn't.

I have started doing a HNC course in Engineering and the first topic is Analytical Methods for Engineers. It is a self learning course, I don;t go to any lectures or get to see people for guidance, the only help I get is a booklet of lessons. I have worked through this but I am still struggling.

Maths has never been my strongest subject so it would be much appreciated if anyone could point me in the right direction with any of this question:

The age of a machine, x, in years is related to the probablity of breakdown, y, by the formula:

..................y
x = 3 + ln [ ___ ]
................1 - y

Determine the probablity of breakdown for 1, 3 and 10 years.
• Jan 22nd 2010, 07:07 AM
tonio
Quote:

Originally Posted by JP103
Not sure if I have posted this in the right place, so I aplogise if I havn't.

I have started doing a HNC course in Engineering and the first topic is Analytical Methods for Engineers. It is a self learning course, I don;t go to any lectures or get to see people for guidance, the only help I get is a booklet of lessons. I have worked through this but I am still struggling.

Maths has never been my strongest subject so it would be much appreciated if anyone could point me in the right direction with any of this question:

The age of a machine, x, in years is related to the probablity of breakdown, y, by the formula:

..................y
x = 3 + ln [ ___ ]
................1 - y

Determine the probablity of breakdown for 1, 3 and 10 years.

So it seems to be that you've to calculate y in each of the cases x =1,3,or 10. Let's do the first one, and you do the following ones:

$\displaystyle 1=3+\ln\left(\frac{y}{1-y}\right)\Longrightarrow \ln\left(\frac{y}{1-y}\right)=-2\Longrightarrow \frac{y}{1-y}=e^{-2}$ $\displaystyle \Longrightarrow y\left(1+e^{-2}\right)=e^{-2}\Longrightarrow y=\frac{e^{-2}}{\left(1+e^{-2}\right)}\cong 0.1192$

Tonio
• Jan 23rd 2010, 08:55 AM
JP103
Thanks alot for that. I've done the other two as well now.