A machine that produces ball bearings has initially been set so that the mean diameter of the bearings it produces is 0.500 inches. A bearing is acceptable if its diameter is within 0.004 inches of this target value. Suppose, however, that the setting has changed during the course of production, so that the distribution of the diameters produced is now approximately normal with mean 0.499 inch and standard deviation 0.002 inch. What percentage of the bearings produced will not be acceptable

7.3% of the bearings produced will not be acceptable.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 0.499 inch

Standard Deviation, σ = 0.002 inch

We are given that the distribution of the diameters is a bell shaped distribution that is a normal distribution.

Formula:

A bearing is acceptable if its diameter is within 0.004 inches that is the range is from 0.496 inch to 0.504 inches.

P(diameter between 0.496 inch and 0.504 inch)

Percentage of the bearings produced will not be acceptable =

Thus, 7.3% of the bearings produced will not be acceptable.

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