Suppose your last chief was slow to respond to fires, or he sent engines and men to the neighboring community to help fight their fires with your tax money. So you have him replaced.
His replacement has never been in a fire engine, but gets the job. When he’s slow to respond to fires he blames the former chief. When he sends even MORE men and engines to help the neighboring town, he blames the former chief because "everyone does it".
How long do you keep your faith in the new chief?
Yea! Doesn’t that "BURN" you up??? Geez, I wouldn’t want this guy as a Chief of my local fire Department….Perhaps DOG CATCHER???
I need to set up a 4 zone panel with 4 detectors and install 4 emergency lights in a small industry unit. We are a charity and I want to avoid being ripped off. How much on average should just installation of the above cost?
Get in contact with one of the bigger companies like ADT they will not rip you off and will give you the best advice. Also contact you local fire station for help and advice.
The volunteer firefighters were frustrated with their efforts at basic fire prevention training. It seemed that no matter how many fire – safety meetings they set up, hardly anybody ever paid attention. They devised a plan. They decided that each of them (there were eight firefighters) would teach two other people the fire-safety basics. At that point, the teacher would retire but each student would then teach two others. Those people, in turn, would teach two others. The whole thing would be mandated by the city council, and each person would have a month to fulfill his or her teaching requirement. The firefighters taught the first group of people in the first month. Under this plan, how many people would know the fire-safety basics after 10 months?
At the beginning (t = 0 months), the number of people N who knew fire-safety basics were the eight firefighters. So,
N(0) = 8
After the first month (t = 1), 16 more people knew fire-safety basics. So,
N(1) = 8 + 16 = 24
After the second month, 32 more people knew fire-safety basics. So,
N(2) = 8 + 16 + 32 = 56
Note that all the terms in the sum are powers of two. That is,
N(2) = 2^3 + 2^4 + 2^5
So, after the third month,
N(3) = 2^3 + 2^4 + 2^5 + 2^6,
… and
N(4) = 2^3 + 2^4 + 2^5 + 2^6 + 2^7
In general,
N(t) = 2^3 + 2^4 + … + 2^(t + 3)
= 8 (2^0 + 2^1 + … + 2^t)
Also,
N(t + 1) – N(t) = [ 2^3 + 2^4 + ... + 2^(t + 3) + 2^(t + 4) ] – [ 2^3 + 2^4 + ... + 2^(t + 3) ]
= 2^(t + 4)
= (16) (2^t)
So,
N(t) = (16) (2^t) – 8
And,
N(10) = (16) (2^10) – 8
= 16,376