A Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of O and a standard deviation of 1. Draw
a graph and find the bone density test scores that can be used as cutoff values separating the lowest 16% and highest 16%, indicating levels that are too low or too
high, respectively.
Sketch the region. Choose the correct graph below.
ОА.
B.
Ос.
D.
Q
Q
Q
a
Q
22
The bone density scores are
(Use a comma to separate answers as needed. Round to two decimal places as needed.)
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step-by-step explanation:

107 units

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form