A basketball player makes 70% of his free throws the random numbers below represent 20 trials for a simulation let the numbers 0 to 6 represent a made free throw and let 7 to 9 represent a missed free throw use this simulation to estimate the probability that the player makes both free throws. (5,5) (0,6) (9,6) (7,4) (9,1) (7,3) (0,4) (9,3) (2,9) (1,4) (6,8) (5,3) (1,7) (1,7) (1,8) (3,5) (7,2) (2,1) (0,6) (1,8)

40%

Step-by-step explanation:

Given the result of the simulation below:

(5,5) (0,6) (9,6) (7,4) (9,1) (7,3) (0,4) (9,3) (2,9) (1,4)

(6,8) (5,3) (1,7) (1,7) (1,8) (3,5) (7,2) (2,1) (0,6) (1,8)

n(S)=20

Let Event A be the event that the player made both throws

The pairs in which the player made both throws are:

(5,5) (0,6) (0,4) (1,4) (5,3)  (3,5)  (2,1) (0,6)

n(A)=8

Therefore:

The probability that the player made both throws is 40%.

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