Craig measured a length as 0.034 millimeters. what is the margin of error? 0.03 mm to 0.04 mm 0.033 mm to 0.034 mm 0.0345 mm to 0.0350 mm 0.0335 mm to 0.0345 mm

  0.0335 mm to 0.0345 mm

Step-by-step explanation:

The least significant digit of Craig's measurement is in the 0.001 mm place. The error in the measurement is presumed to fall within half that value either side of Craig's measurement. So, the actual value is expected to be in the range ...

  0.034 -0.0005 = 0.0335

to

  0.034 +0.0005 = 0.0345

The measurement is presumed to be in the range 0.0335 mm to 0.0345 mm.

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

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