Raw materials are studied for contamination. suppose that the number of particles of contamination per pound of the material is a poisson random varable with a mean of 0.01 paterical per pounda)what is the expected number of pounds of material requried to obtain 15 particles of contamination?b) what is the standard deviation of the per pounds pf material required to obtain 15 particales of contamination?

A= 1500 ; B=387.3

Step-by-step explanation:

A. the expected number of pounds of material requried to obtain 15 particles of contamination

We represent:

Y as the poisson random variable with λ to be 0.01

X as Erlang random variable with r to be 15 particles

Therefore

E(X) = r/λ

= 15/0.01

=1500

B. standard deviation of the per pounds pf material required to obtain 15 particales of contamination

Standard deviation X = √ r/λ^2

√ 15/0.0001

=387.3

a

step-by-step explanation:

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12