From Omilili

### "Queue Correlation Values not enabled" error even **after** adding "WaitForIdle...

Hi, I'm getting the "Queue Correlation Values not enabled" error, even **after**adding the "WaitForIdle"

**argument**. Following is the extract of the code, WorkItemService work = new WorkItemService on using the .NET WF can be taken only

**after**solving the issue. If the .NET WF is unstable, i need

From Yahoo Answers

**Question:**a weak acid has a pH of 5.31 and a concentration of 5.35 x 10^-3 mol dm-3, I have to work out the Ka and pKa values for this. please help (: thanks

**Answers:**Here's the technique: http://www.chemteam.info/AcidBase/Calc-Ka-from-data.html

**Question:**There are many organic acids and bases in our cells and their presence modifies the ph of fluids inside them. One of these is lactic acid, CH3CHOHCOOH. A solution of equal concentrations of lactic acid and sodium lactate has a ph = 3.08 Using these values determine the pKa and Ka values of lactic acid.. What would the Ph be if the acid had twice the concentration of the salt sodium lactate? Thanks

**Answers:**The relationship between pKa and pH is: pH = pKa + log [HA]/[A-] With the concentration equal, the pKa is just the pH of the system. pKa = 3.08 Ka = 10^-3.08 = 8.31E-4 [EDIT]: If the acid was twice the concentration of the sodium salt, then [HA] = 2x, [A-] = x. pH = pKa + log [HA]/[A-] pKa = pH - log [HA]/[A-] pKa = 3.08 - log [2x/1x] pKa = 3.08 - log[2] pH = 3.08 - .301 pH = 2.78

**Question:**An airline claims that the no-show rate for passengers booked on its flights is less than 6%. Of 380 randomly selected reservations, 18 were no-shows. Find the P-value for a test of the airline's claim. I am not worried so much about the answer as to HOW to get the answer.

**Answers:**Let P be the proportion of no-show rate for passengers booked on its flights. Ho: P =0.06 Ha: P < 0.06 Estimated p = 18 / 380 = 0.0474 Variance of proportion = p*(1-p)/n = 0.06(0.94)/380 =0.0001484 S.D. of p is sqrt[0.000148] = 0.0122 Z = ( 0.0474 - 0.06 ) / 0.0122 = -1.0368 From Normal probability table , P( z < -1.0368) = 0.1492 p-value = 0.1492

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