Answer:
The z score for bolt of diameter 18.12 mm is 1.20.
Step-by-step explanation:
Let <em>X</em> = diameter of bolts.
It is provided that the random variable <em>X</em> follows a Normal distribution with mean, <em>μ</em> = 18 mm and standard deviation, <em>σ</em> = 0.10 mm.
A <em>z</em>-score is a standardized score, a numerical, that defines how far a data value from the mean.
The distribution of <em>z</em>-scores is defined by the Standard Normal distribution.

The formula to compute the <em>z</em>-score is:

The value of the diameter of a bolt is, <em>x</em> = 18.12 mm.
Compute the <em>z</em>-score for this value as follows:

Thus, the z score for bolt of diameter 18.12 mm is 1.20.
1/7 x 9/4 = 9/28
This process to solving division involving fractions is called “keep, change, flip”
Given triangle ABC with coordinates A(−6, 4), B(−6, 1), and C(−8, 0), and its image A′B′C′ with A′(−2, 0), B′(−5, 0), and C′(−6,
Zinaida [17]
Answer:
The line of reflection is at y = x+6.
Step-by-step explanation:
The perpendicular bisector of AA' is a line with slope 1 through the midpoint of AA', which is (-4, 2). In point-slope form, the equation is ...
y = 1(x +4) +2
y = x + 6 . . . . . . . line of reflection
It would be 2769.44 still since it would technically be 2769.440 if you were rounding to the TENTHS place. The Zero(0) doesn't change anything so the answer is
2769.44