Comments on: How To Analyze Data Using the Average http://betterexplained.com/articles/how-to-analyze-data-using-the-average/ Learn Right, Not Rote. Fri, 03 Feb 2012 19:38:49 +0000 hourly 1 http://wordpress.org/?v=3.2.1 By: Muhammad Atif Naseem http://betterexplained.com/articles/how-to-analyze-data-using-the-average/#comment-7015 Muhammad Atif Naseem Sat, 29 Oct 2011 21:55:58 +0000 http://betterexplained.com/articles/how-to-analyze-data-using-the-average/#comment-7015 Just amazing. such a great ideas u shared with us. thank u very much. Just amazing. such a great ideas u shared with us. thank u very much.

]]> By: SomeShmuck http://betterexplained.com/articles/how-to-analyze-data-using-the-average/#comment-4919 SomeShmuck Sat, 08 Oct 2011 15:26:42 +0000 http://betterexplained.com/articles/how-to-analyze-data-using-the-average/#comment-4919 The data transmission is incorrect for GB/$. If you're only paying one side it's exactly the amount you pay there's no average to be had. But if you're paying both client and server, it's not each doing a half the work, it's each doing a whole part each. If you're paying for a client and server and you get 10GB/$ on both of them, when you transmit 10GB from the client to the server, both have consumed 10GB worth of transmitted data, you pay 2 dollars. So in effect two 10GB/$ machines averages to 5GB/$. That's half the estimate if both had 10GB/$, which would be 2/(1/10+1/10)=10GB/$, which is clearly wrong. A 20GB/$ server and a 10GB/$ client transmitting the data would average 6.66GB/$ since you need to set the same ratio for GB. Take this to the extreme with 1000GB/$ or 10000000GB/$ and and 10GB/$ and the cost should move towards being significantly towards 10GB/$ because the servers cost is almost nothing. The data transmission is incorrect for GB/$.
If you’re only paying one side it’s exactly the amount you pay there’s no average to be had.
But if you’re paying both client and server, it’s not each doing a half the work, it’s each doing a whole part each. If you’re paying for a client and server and you get 10GB/$ on both of them, when you transmit 10GB from the client to the server, both have consumed 10GB worth of transmitted data, you pay 2 dollars. So in effect two 10GB/$ machines averages to 5GB/$.
That’s half the estimate if both had 10GB/$, which would be 2/(1/10+1/10)=10GB/$, which is clearly wrong.
A 20GB/$ server and a 10GB/$ client transmitting the data would average 6.66GB/$ since you need to set the same ratio for GB.
Take this to the extreme with 1000GB/$ or 10000000GB/$ and and 10GB/$ and the cost should move towards being significantly towards 10GB/$ because the servers cost is almost nothing.

]]> By: NANDEESH http://betterexplained.com/articles/how-to-analyze-data-using-the-average/#comment-4918 NANDEESH Sun, 14 Aug 2011 08:22:51 +0000 http://betterexplained.com/articles/how-to-analyze-data-using-the-average/#comment-4918 In problems involving tanks and taps or problems such as follows I find solving the problem without reciprocals easier. Ex:If 12 men and 16 boys can do a piece of work in 5 days; 13men and 24 boys can do it in 4 days, then blah.... Soln:From the first condition 1 job=12*5 mandays+16*5 boydays From the second condition 1 job=13*4 mandays+24*4 boydays. Equating the two we get 60mandays+80boydays=52mandays+96boydays. So, 1manday=2boydays. We can solve any question from here onward. ==== Ex:If a tap takes 12 hours to fill a tank and another tap takes 10 hrs to fill the same tank how long will both the taps together take to fill the tank? Multiply12*10=120 ; In 120hrs first tap fills 10 tanks and second tap fills 12 tanks, a total of 22 tanks. So both taps together take 120/22 hrs. In problems involving tanks and taps or problems such as follows I find solving the problem without reciprocals easier.
Ex:If 12 men and 16 boys can do a piece of work in 5 days; 13men and 24 boys can do it in 4 days, then blah….
Soln:From the first condition 1 job=12*5 mandays+16*5 boydays
From the second condition 1 job=13*4 mandays+24*4 boydays.
Equating the two we get 60mandays+80boydays=52mandays+96boydays.
So, 1manday=2boydays.
We can solve any question from here onward.
====
Ex:If a tap takes 12 hours to fill a tank and another tap takes 10 hrs to fill the same tank how long will both the taps together take to fill the tank?
Multiply12*10=120 ;
In 120hrs first tap fills 10 tanks and second tap fills 12 tanks, a total of 22 tanks.
So both taps together take 120/22 hrs.

]]> By: Anonymous http://betterexplained.com/articles/how-to-analyze-data-using-the-average/#comment-4917 Anonymous Mon, 28 Feb 2011 10:53:00 +0000 http://betterexplained.com/articles/how-to-analyze-data-using-the-average/#comment-4917 A boat takes t1 time to travel a certain distance when it travels in the direction of flow of water, and takes t2 time to travel the same distance against the stream. How much time will it take to travel the same distance in still water ? Is it a case of Harmonic Mean ? Please Explain, how ? A boat takes t1 time to travel a certain distance when it travels in the direction of flow of water, and takes t2 time to travel the same distance against the stream. How much time will it take to travel the same distance in still water ?

Is it a case of Harmonic Mean ?

Please Explain, how ?

]]>