answer to #1 lsource-receptor distance i hydraulic gradient k hydraulic conductivity n effective...

Answer to #1Answer to #1

L source-receptor distancei hydraulic gradientK hydraulic conductivityn effective soil porosityBD soil bulk densityfoc fraction organic carbonKoc organic partition coefficient

Porosity and conductivity would likely be positively relatedPorosity and density would likely be inversely relatedThus, conductivity and density should be inversely related tooFraction organic carbon and the partition coefficient would likely be

positively related

Answer to #2Answer to #2L = [80,120] m // source-receptor distancei = [0.0003,0.0008] m per m // hydraulic gradientK = [300,3000] m per yr // hydraulic conductivityn = [0.2,0.35] // effective soil porosityBD = [1500,1750] kg per m3 // soil bulk densityfoc = [0.0001,0.005] // fraction organic carbonKoc = [5 ,20] m3 per kg // organic partition coefficient Kd = foc * KocR = 1 + BD * Kd / n V = K * i / (n * R)T = L/V T [ 20.952380952, 408800] yr

func TT() return L/(K * i / ($1 * (1 + BD * foc * Koc / $1)))

b = left(n)e = TT(b)w = width(n)/100for k=1 to 100 do begin s = [ (k-1), k] * w + b e = env(e, TT(s)) ende [ 31.480562448, 235352] yr

Answer to #3Answer to #3Iwater = [1.5, 2.5] liters per day // water intakeIfish = [0, 8] g per day // dietary ingestion of fish tissueB = [0.9, 2.1] liters per g // bioaccumulation factorW = [60, 90] kg // receptor biomass

D = [0, 6] mg per kg per day // tolerable dose

// forward equation// D = (Iwater * C) / W + (Ifish * B * C) / W

// condense repeated parameters// D = C * (Iwater + Ifish * B) / W

// solve for C by factoring the quotient (Iwater+Ifish*B)/W out of DC = factor((Iwater + Ifish * B) / W, D) C [ 0, 18.652849741] mg liters1

// put C back into original forward equation (and ensure correct units)d = (Iwater * C) / W + (Ifish * C * B) / W + 0 mg per kg per day

d [ 0, 6] mg kg1 day1

105 104 103

Probability of tank rupturing under pumping

Mixed dependencies

Frank to Fréchet

All Fréchet

Mixed dependence with intervals

Vesely et al. (all independent)3.5105

[3.499105, 3.504105]

[3.50105, 1.35104]

[3105, 1.4104]

[2.9105, 4.1105]

[ 2.5105, 1.905 104] All Fréchet

Answer to #5Answer to #5

Plan of attack for #6Plan of attack for #6

• Use the events from the slide “Snakes on a plane” • Write as Boolean expression

– Represent ANDs as conjunctions &– Represent ORs as disjunctions V

(plane V ship V raft V pet V zoo) & (pregnant V (malefemale & repro) V (female & parthenog)) & (dogs & cats & pigs) & (eggs V rodents) & (hides V quick V disperse)

• Specify the temporal context, e.g., a year• Estimate marginal event probabilities

– Maybe as submodels, e.g., plane = flights * stowaway per flight

• Decide which events are independent

plane =randomship=randomraft =randompet =randomzoo=randompregnant =randommalefemale=randomrepro=randomfemale =randomparthenog=randomdogs =randomcats =randompigs=randomeggs =randomrodents=randomhides =randomquick =randomdisperse=random

(plane ||| ship ||| raft ||| pet ||| zoo) |&| (pregnant ||| (malefemale |&| repro) ||| (female |&| parthenog)) |&| (dogs |&| cats |&| pigs) |&| (eggs ||| rodents) |&| (hides ||| quick ||| disperse)

Answer to #7Answer to #7Assuming a = 0.29, b = 0.22, and r = 1.0, yields

a * b + r * sqrt(a*(1-a)*b*(1-b)) Pearson AND0.2517692528

[ max(0, a+b–1), min(a, b) ] Fréchet AND [ 0, 0.22]

The probability assuming Pearson correlation r =1 is larger than the largest possible probability. This happens because the Pearson correlation for two events with these marginals cannot be as large as one (so we are making a false assumption which leads to the error). The extreme Pearson correlations for them are

(-a*b)/sqrt(a*b*(1-a)*(1-b)) // when they don’t overlap -0.33941721344 (b-a*b)/sqrt(a*b*(1-a)*(1-b)) // when b’s inside a 0.83098697084

We can check this with this calculationr = 0.83098697084a * b + r * sqrt(a*(1-a)*b*(1-b))

0.22