Find the value of the standard normal random variable z, called z0 such that: (a) p(z≤z0)=0.7004 z0= .525 (b) p(−z0≤z≤z0)=0.7952 z0= (c) p(−z0≤z≤z0)=0.7206 z0= (d) p(z≥z0)=0.00677 z0= (e) p(−z0≤z≤0)=0.4634 z0= (f) p(−1.57≤z≤z0)=0.5302 z0=

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

yes

step-by-step explanation: