3.5 A lake has the following dissolved oxygen readings Month DO (mg/L) 12 10 9 10...
the equilibrium concentration of dissolved oxygen jn mg/L in Lake Titicaca. The elevation is 3.850 m (atmospheric pressure = 0.62 atm)? Assume T-20 °C and K 0.62 atm)? Assume T-20 °C and Kn730 atm-L/mol. What would the concentration be with the same temperature at sea level? Explain the significance of this equation for a) developing and b) developed countries: 1 - (P)(A)(T)
2. (30 points) If the dissolved oxygen (DO) concentration measured during a BOD test is 9 mg/L initially, 6 mg/L after 5 days and 3 mg/L after an indefinitely long period of time, calculate the BOD remaining after 10 days (i.e., BOD(10)).
20 points 1) Capnophillc baaequre ovgento grow. Based on 11 samples of these bacteria,the average oxygen needed was found to be 15 mg/L. The lab technician knows the standard deviation from experience as 3. a) Can we claim the mean oxygen of the bacteria is less than 16 at 5% significance? What is the p- value? b) How many more samples do you need for the 95% confidence interval of the mean pH to be within 1? 20 points 1)...
Oxygen demand is a term biologists use to describe the oxygen needed by fish and other aquatic organisms for survival. The Environmental Protection Agency conducted a study of a wetland area. In this wetland environment, the mean oxygen demand was μ = 9.7 mg/L with 95% of the data ranging from 5.9 mg/L to 13.5 mg/L. Let x be a random variable that represents oxygen demand in this wetland environment. Assume x has a probability distribution that is approximately normal....
QUESTION 3 a) After a point of wastewater discharge, the dissolved oxygen (DO) concentration is found as 83.33% of the saturation DO. The remaining BOD after 5 days at the point of discharge is 10.13 mg/L. The BOD rate constant determined in the laboratory under standard conditions is 0.115/day. The saturation DO concentration in the river water is 9 mg/L. The deoxygenation and the reaeration rate coefficients are 0.25/day and 0.45/day, respectively. The river travels at a velocity of 0.12...
Oxygen demand is a term biologists use to describe the oxygen needed by fish and other aquatic organisms for survival. The Environmental Protection Agency conducted a study of a wetland area. In this wetland environment, the mean oxygen demand was μ = 9.8 mg/L with 95% of the data ranging from 6.0 mg/L to 13.6 mg/L. Let x be a random variable that represents oxygen demand in this wetland environment. Assume x has a probability distribution that is approximately normal....
Oxygen demand is term biologists use to describe the oxygen needed by fish and other aquatic organisms for survival. The Environmental Protection Agency wetland area. In this wetland environment, the mean oxygen demand was u=9.8 mg/L with 95% of the data ranging from 7.0 mg/L to conducted a study of 12.6 mg/L. Let x be a random variable that represents oxygen demand in this wetland environment. Assume x has a probability distribution that is approximately normal. (a) Use the 95%...
9. ?Using the following figure, describe how BOD effects oxygen concentration and fish.? 10.What is the difference between an oligotrophic and eutrophic lake? What is anthropogenic eutrophication? 11.How can groundwater become contaminated? 12.What can you do at home to reduce water pollution?
Is (a) and (b) correct? Am i doing (c) correctly so far and how do i finish (c) ? Answer the following questions based on this data: The cholesterol content of large chicken eggs is normally distributed with a mean of 200mg and standard deviation 15 mg. a. (10 pts)What is the hypotheses for the mean cholesterol content of a random sample of 25 1. of these eggs is less than 205mg? (10 pts)To test if the claimed 200mg value...
The average amount of nicotine in a cigarette is about 12 mg (milligrams), but for your information, when you smoke a cigarette, you only consume about 0.95 mg. We assume that nicotine distribution in cigarettes is normal with standard deviation 1.5 mg. A cigarette manufacturer claims that it has reduced the amount of nicotine in its cigarettes. You the curious cat, actually go out and take a random sample of 50 cigarettes; you find that average amount is 11.7 mg....