In exercise 5.10, we proved that 1, 0≤y1 ≤2,0≤y2 ≤1,2y2 ≤y1, f (y1, y2) = 0, elsewhere 0, elsewhere is a valid joint probability density function for y1, the amount of pollutant per sample collected above the stack without the cleaning device, and for y2, the amount collected above the stack with the cleaner. a if we consider the stack with the cleaner installed, find the probability that the amount of pollutant in a given sample will exceed .5. b given that the amount of pollutant in a sample taken above the stack with the cleaner is observed to be 0.5, find the probability that the amount of pollutant exceeds 1.5 above the other stack (without the cleaner).

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3

both the histogram and the stem and leaf plot will visually show how many shoppers were of various ages within each range of data.   the histogram does not give us the specific ages of these shoppers unlike the stem and leaf plot.a histogram does not give us a true range of the data because we do not know exactly how old the shoppers were. a stem and leaf plot will give exact ages, so we can determine the exact range along with all measures of center and the others measures of center.   with the histogram, we cannot find any of these.

step-by-step explanation: