Carolyn usually only misses about 10% of her free throws. if she misses fewer than 2 of her next 4 free throws, then she will hold the school record for free throws. to estimate the probability of this happening, carolyn uses a computer to randomly select 4 numbers from 0 to 9. she lets 0 represent a missed free throw and 1 through 9 represent a made free throw. she repeats this for a total of 18 trials and records the results in the table below. 8100 1347 0693 9869 3399 1376 1273 7905 3471 7987 7604 4587 2968 5064 2697 4248 4545 3308 what is the best estimate, based on this simulation, of the probability that she will miss fewer than 2 of her next 4 free throws? 6% 33% 67% 94%

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