To estimate the long-term settlement of a footing, a geotechnical engineer uses the total load on the footing calculated by summing the dead load of the structure and the live load (loads imposed by furniture and occupancy). Because each is the sum of many relatively small weights, the engineer assumes that the dead load (X) and the live load (Y) are normally distributed. He collected data from numerous buildings of a similar type and found that the mean and the standard deviation of X are 100 kips and 10 kips, respectively, and the mean and the standard deviation of Y are 40 kips and 10 kips, respectively. Assuming that the design load is the total load for which the exceeding probability is 5%. Determine the design load.

Q.1 dr. frank needs to choose 3 students from her advanced biology course to attend a conference in hawaii. the course has 40 students. which of the following methods could dr. frank use to choose a simple, random sample of 3 students? select all that apply. 1 dr. frank could have the students line up from shortest to tallest, then use her birthdate, 2/5/39, to select 3 students from the line (the 2nd student, the 5th student, and the 39th student). 2. dr. frank could upload her class roster into a spreadsheet and use a ranom-number generating function to produce 3 random numbers. the students whose names are located in the rows with these 3 numbers on the spreadsheet would be chosen. 3. dr. frank could choose the students with the 3 highest grades in the class. 4. dr. frank could count to three, then the first students to come to the front of the classroom and grab a golden ticket from her hand would be selected. 5. dr. frank could ask all students to write their name on a card, place all of the cards in a hat, shake the hat, and then pull 3 cards out from the hat. the students whose names appear on those 3 cards will be selected. 6. dr. frank could ask for volunteers to go to the conference, then choose the first 3 students who raise their hands. q2. now, choose one of the methods that you did not select and explain why this method would not result in a simple, random sample: