Know how to calculate sample size: A restaurant would like to estimate the average $ amount of a take-out order. The standard deviation is $15. If a 99% confidence level and an interval (error) of $5 is desired, how many customers should be in the sample? Which of the following examples would use the z-distribution:.
a. x = 100; µ = 99; s = 5; n = 70
b. x = 50; µ = 53; σ = 7; n = 15
c. x = 7; µ = 4; s = 1; n = 20
d. x = 15; µ = 17; σ = 3; n = 45

yo answer should be 6 if im not mistaken

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

it would be c. 25 units