There are 4 accidents, on average, at an intersection. Assume the variable follows a Poisson distribution. Find the probability that there will be less than 2 accidents at this intersection.

P [ X < 2 ]   =  0,091         or       P [ X < 2 ]   =  9,1 %

Step-by-step explanation:

The random variable follows a Poisson Distribution then:

P [ X = x ] =  λˣ *e₋∧-λ / x!

λ = 4

The probability of X less than  2 is:

P [ X < 2 ]   =    P[ X = 0 ]  + P[ X = 1 ]

P[ X = 0 ]  = 4⁰ * e ∧ -4 / 0!

P[ X = 0 ]  = e⁻⁴ / 0!

P[ X = 0 ]  = e⁻⁴ /1

P[ X = 0 ]  = 0,01831

P[ X = 0 ]  = 0,01831    or     P[ X = 0 ]  = 1,8 %

Now

P[ X = 1 ] =  4¹ * e⁻⁴ / 1!

P[ X = 1 ] = 4 * 0,01831

P[ X = 1 ] = 0,073         or    P[ X = 1 ] = 7,3 %

Then

P [ X < 2 ]   =   0,01831  +  0,073

P [ X < 2 ]   =  0,091         or       P [ X < 2 ]   =  9,1 %

step-by-step explanation:

what are the statements? and what do you mean by "216 2 1 6"?

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)