Geophysicists determine the age of a zircon by counting the number ofuranium fission tracks on a polished surface. A particular zircon is of such anage that the average number of tracks per square centimeter is five. What is the probability that a 2cm^2 sample of this zircon will reveal at most three tracks, thus leading to an underestimation of the age of the material?

p(x = 3, λ = 5) = 0.14044

Step-by-step explanation:

Given

λ = 5 (the average number of tracks per square centimeter)

ε = 2.718 (constant value)

x = 3 (the variable that denotes the number of successes that we want to occur)

p(x,λ) = probability of x successes, when the average number of occurrences of them is λ

We can use the equation

p(x,λ) = λˣ*ε∧(-λ)/x!

⇒ p(x = 3, λ = 5) = (5)³*(2.718)⁻⁵/3!

⇒ p(x = 3, λ = 5) = 0.14044

0.0108

Step-by-step explanation:

Let X denote the number of uranium fission tracks occurring on the average 5 per square centimetre.We need to find the probability that a 2cm² sample of this zircon will reveal at most three tracks. X follows Poisson distribution, λ = 5 and s = 2.

k = λs = 5×2 = 10

Since we need to reveal at most three tracks the required probability is:

P (X≤3) = P (X =0) + P (X =1) + P (X =2) + P (X =3)

P (X≤3)  = (((e^-10) × (10)⁰)/0!) +  (((e^-10) × (10)¹)/1! +  (((e^-10) × (10)²)/2! + (((e^-10) × (10)3)/3!

P (X≤3)  = 0.0004 + 0.0005 +0.0023 +0.0076

P (X≤3)  = 0.0108

Therefore, the probability that a 2cm² sample of this zircon will reveal at most three tracks is 0.0108

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.