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	<title>Comments on: Calculus of Blogging: The Whys and Hows of Blogging Success</title>
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	<link>http://www.dailyblogtips.com/calculus-of-blogging-the-whys-and-hows-of-blogging-success/</link>
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	<lastBuildDate>Thu, 09 Feb 2012 22:50:25 +0000</lastBuildDate>
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	<item>
		<title>By: Oguibe</title>
		<link>http://www.dailyblogtips.com/calculus-of-blogging-the-whys-and-hows-of-blogging-success/#comment-901854</link>
		<dc:creator>Oguibe</dc:creator>
		<pubDate>Mon, 22 Jun 2009 20:12:38 +0000</pubDate>
		<guid isPermaLink="false">http://www.dailyblogtips.com/?p=3648#comment-901854</guid>
		<description>Hi, thanks everyone, your comments and contributions has been helpful. Realy an eye openner. You all are great</description>
		<content:encoded><![CDATA[<p>Hi, thanks everyone, your comments and contributions has been helpful. Realy an eye openner. You all are great</p>
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	<item>
		<title>By: new blogger</title>
		<link>http://www.dailyblogtips.com/calculus-of-blogging-the-whys-and-hows-of-blogging-success/#comment-879869</link>
		<dc:creator>new blogger</dc:creator>
		<pubDate>Thu, 28 May 2009 08:41:22 +0000</pubDate>
		<guid isPermaLink="false">http://www.dailyblogtips.com/?p=3648#comment-879869</guid>
		<description>very scientific.....</description>
		<content:encoded><![CDATA[<p>very scientific&#8230;..</p>
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		<title>By: Tyrone – Internet Business Path</title>
		<link>http://www.dailyblogtips.com/calculus-of-blogging-the-whys-and-hows-of-blogging-success/#comment-848730</link>
		<dc:creator>Tyrone – Internet Business Path</dc:creator>
		<pubDate>Wed, 22 Apr 2009 06:20:29 +0000</pubDate>
		<guid isPermaLink="false">http://www.dailyblogtips.com/?p=3648#comment-848730</guid>
		<description>This is a niceinterpretation of blogging in mathematical areas. 

True enough, as bloggers, we must focus in our &quot;D&quot;s = the difference you believe between your chances of success and everyone else. 

Be original, and the rest will follow, i guess.</description>
		<content:encoded><![CDATA[<p>This is a niceinterpretation of blogging in mathematical areas. </p>
<p>True enough, as bloggers, we must focus in our &#8220;D&#8221;s = the difference you believe between your chances of success and everyone else. </p>
<p>Be original, and the rest will follow, i guess.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: AskYourPreacher</title>
		<link>http://www.dailyblogtips.com/calculus-of-blogging-the-whys-and-hows-of-blogging-success/#comment-843875</link>
		<dc:creator>AskYourPreacher</dc:creator>
		<pubDate>Fri, 17 Apr 2009 08:54:06 +0000</pubDate>
		<guid isPermaLink="false">http://www.dailyblogtips.com/?p=3648#comment-843875</guid>
		<description>D was such a HUGE factor in why our group started our blog.  I remember sitting down and looking for a blog that would fill the particular need that our blog now does.  I couldn&#039;t find anything in that niche.  D became the driving force behind our start.  Great post!</description>
		<content:encoded><![CDATA[<p>D was such a HUGE factor in why our group started our blog.  I remember sitting down and looking for a blog that would fill the particular need that our blog now does.  I couldn&#8217;t find anything in that niche.  D became the driving force behind our start.  Great post!</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Daniel Lucraft</title>
		<link>http://www.dailyblogtips.com/calculus-of-blogging-the-whys-and-hows-of-blogging-success/#comment-843195</link>
		<dc:creator>Daniel Lucraft</dc:creator>
		<pubDate>Thu, 16 Apr 2009 19:36:39 +0000</pubDate>
		<guid isPermaLink="false">http://www.dailyblogtips.com/?p=3648#comment-843195</guid>
		<description>This is a really good way to take a hardheaded look at your blogs potential. May I nitpick (did maths at college) and point out that you probably mean:

  U = pB – C + D

In this equation p is a factor, not a function, so the parentheses are unnecessary.

Also, and perhaps more importantly, the implied units of the equation are not consistent: you are subtracting a probability (D) from a cost (pB-C, which is probably in dollars, or at least dollar equivalents). 

The term D needs to be included in the calculation of p, which is your chance of success. Let&#039;s suppose that your chance of success is:

p = K + D

where K is the chance of success that absolutely everyone has, and D is as before. Then

U = (K+D)B + C

which is a consistent equation and a perfect illustration of your - very good - point.</description>
		<content:encoded><![CDATA[<p>This is a really good way to take a hardheaded look at your blogs potential. May I nitpick (did maths at college) and point out that you probably mean:</p>
<p>  U = pB – C + D</p>
<p>In this equation p is a factor, not a function, so the parentheses are unnecessary.</p>
<p>Also, and perhaps more importantly, the implied units of the equation are not consistent: you are subtracting a probability (D) from a cost (pB-C, which is probably in dollars, or at least dollar equivalents). </p>
<p>The term D needs to be included in the calculation of p, which is your chance of success. Let&#8217;s suppose that your chance of success is:</p>
<p>p = K + D</p>
<p>where K is the chance of success that absolutely everyone has, and D is as before. Then</p>
<p>U = (K+D)B + C</p>
<p>which is a consistent equation and a perfect illustration of your &#8211; very good &#8211; point.</p>
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