Answer:
Azithromycin will be in your system for <u>around 15.5 days</u>, after the last dose.
Explanation:
Azithromycin has an elimination half-life of 68 hours. The prolonged terminal half-life is thought to be due to extensive uptake and subsequent release of drug from tissues. It takes around 5.5 x elimination half life's for a medicine to be out of your system. Therefore it would take 374 hours about 15.5 days (5.5 x 68 hours) for it to be eliminated from the system. So it'll be in your system for that period of time, after the last dose.
Answer:
cognitive dissonance
Explanation:
his actions are not matching his emotions, so he is trying to make them match each other
I love statistics So I will use The principles of it
![\begin{cases}\\ \dag \: \underline{\Large\bf Formulas\:of\:Statistics} \\ \\ \bigstar \: \underline{\rm Mean:} \\ \\ \bullet\sf M=\dfrac {\Sigma x}{n} \\ \bullet\sf M=a+\dfrac {\Sigma fy}{\Sigma f} \\ \\ \bullet\sf M=A +\dfrac {\Sigma fy^i}{\Sigma f}\times c \\ \\ \bigstar \: \underline{\rm Median :} \\ \\ \bullet\sf M_d=\dfrac {n+1}{2} \:\left[\because n\:is\:odd\:number\right] \\ \bullet\sf M_d=\dfrac {1}{2}\left (\dfrac {n}{2}+\dfrac {n}{2}+1\right)\:\left[\because n\:is\:even\:number\right] \\ \\ \bullet\sf M_d=l+\dfrac {m-c}{f}\times i \\ \\ \bigstar \: {\boxed{\sf M_0=3M_d-2M}}\end {cases}](https://tex.z-dn.net/?f=%20%5Cbegin%7Bcases%7D%5C%5C%20%20%5Cdag%20%5C%3A%20%5Cunderline%7B%5CLarge%5Cbf%20Formulas%5C%3Aof%5C%3AStatistics%7D%20%5C%5C%20%5C%5C%20%5Cbigstar%20%5C%3A%20%5Cunderline%7B%5Crm%20Mean%3A%7D%20%5C%5C%20%5C%5C%20%5Cbullet%5Csf%20M%3D%5Cdfrac%20%7B%5CSigma%20x%7D%7Bn%7D%20%5C%5C%20%5Cbullet%5Csf%20M%3Da%2B%5Cdfrac%20%7B%5CSigma%20fy%7D%7B%5CSigma%20f%7D%20%5C%5C%20%5C%5C%20%5Cbullet%5Csf%20M%3DA%20%2B%5Cdfrac%20%7B%5CSigma%20fy%5Ei%7D%7B%5CSigma%20f%7D%5Ctimes%20c%20%5C%5C%20%5C%5C%20%5Cbigstar%20%5C%3A%20%5Cunderline%7B%5Crm%20Median%20%3A%7D%20%5C%5C%20%5C%5C%20%5Cbullet%5Csf%20M_d%3D%5Cdfrac%20%7Bn%2B1%7D%7B2%7D%20%5C%3A%5Cleft%5B%5Cbecause%20n%5C%3Ais%5C%3Aodd%5C%3Anumber%5Cright%5D%20%5C%5C%20%5Cbullet%5Csf%20M_d%3D%5Cdfrac%20%7B1%7D%7B2%7D%5Cleft%20%28%5Cdfrac%20%7Bn%7D%7B2%7D%2B%5Cdfrac%20%7Bn%7D%7B2%7D%2B1%5Cright%29%5C%3A%5Cleft%5B%5Cbecause%20n%5C%3Ais%5C%3Aeven%5C%3Anumber%5Cright%5D%20%5C%5C%20%5C%5C%20%5Cbullet%5Csf%20M_d%3Dl%2B%5Cdfrac%20%7Bm-c%7D%7Bf%7D%5Ctimes%20i%20%5C%5C%20%5C%5C%20%5Cbigstar%20%5C%3A%20%7B%5Cboxed%7B%5Csf%20M_0%3D3M_d-2M%7D%7D%5Cend%20%7Bcases%7D)
The answer is False. It can’t be harmful
Answer:
False.
Explanation:
Zwitter ion may be defined as the molecule that contains the net neutral charge. The change in pH can change the charge on the zwitter ions. Amino acids can exist in the form of zwitter ions.
The basic environment, causes the deprotonation of the ammonia group. The carboxyl end of a protein release the hydrogen ions in the acidic environment , not in the basic environment.
Thus, the answer is false.
