This causes increased alveolar collapse on expiration, air trapping and hyperinflation, impaired ventilation, and increased dead space, leading to hypoxemia and hypercapnia. E- Emphysema is a component of chronic obstructive pulmonary disease characterized by loss/damage of alveolar elastin and lung parenchyma due to a protease/antiprotease imbalance.

This patient is at risk for the development of this condition based on her history of type 2 diabetes mellitus. A vaginal smear reveals the presence of normal vaginal epithelium and pseudohyphae, confirming the diagnosis of vaginal candidiasis. Physical exam reveals edematous, erythematous vaginal mucosa with white discharge, suggestive of a candidal infection.

A- This patient has vaginal pain, pruritus, and dyspareunia.

Based on this, 50% of individuals will have leukocyte counts greater than 7500 and 50% will have leukocyte counts less than 7500. In this example, the mean leukocyte count is 7500 cells/mm3.

D- For a normal distribution of data, mean (average) is the most reliable measure of central tendency.

Severe acute kidney injury can be lethal if not properly managed, although the question states that this patient’s death was secondary to a cardiac arrhythmia, likely due to damage sustained from ischemia during cardiac arrest. This is not uncommon after several weeks. Most patients with ATN will experience tubular re-epithelialization and spontaneous full recovery.

Two weeks after his initial cardiac arrest, the oliguria and azotemia resolved, representing the recovery of acute tubular necrosis. The diagnosis of ATN is supported by his oliguria, azotemia, and urinary granular casts. This was most likely due to severe renal ischemia due to impaired cardiac output. E- Six weeks ago, this patient developed acute renal failure and ischemia-induced acute tubular necrosis (ATN) in the setting of cardiac arrest due to underlying coronary artery disease.The lower end of the confidence interval is thus 130 – (1.96 x 5), which equals 120.2, and the higher end of the confidence interval is 130 + (1.96 x 5) which equals 139.8. B- The formula for a 95% confidence interval for data that follow a normal distribution is x̅ +/- (1.96 x Standard Error), where x̅ is the mean.

How to decrease dead space ventilation