Answer:
Probability that the diameter of a selected bearing is greater than 111 millimeters is 0.1056.
Step-by-step explanation:
We are given that the diameters of ball bearings are distributed normally. The mean diameter is 106 millimeters and the standard deviation is 4 millimeters.
<em>Firstly, Let X = diameters of ball bearings</em>
The z score probability distribution for is given by;
Z = 
~ N(0,1)
where, 
= mean diameter = 106 millimeters
= standard deviation = 4 millimeter
Probability that the diameter of a selected bearing is greater than 111 millimeters is given by = P(X > 111 millimeters)
P(X > 111) = P( > 
) = P(Z > 1.25) = 1 - P(Z 
1.25)
= 1 - 0.89435 = 0.1056
Therefore, probability that the diameter of a selected bearing is greater than 111 millimeters is 0.1056.
Answer:
PLZ MARK BRAINLIEST!
The answer is below...Can you farther explain on the part "Find at least one ratio where one of the quan
a fraction or decimal."
Step-by-step explanation:
LEMONS . CUPS
3 : 2 . 3/2
6 : 4 . 6/4
9 : 6 9/6
12 : 8 . 12/8
15 : 10 . 15/10
Y = 1
Using y = mx + c.
Compare to y = 1, y = 0x + 1 ,
We can see that the slope m = 0 and the vertical intercept, c = 1.
For the line perpendicular to y = 1
Condition for perpendicularity m₁m₂ = -1
m₁ = 0, m₂ = ?
0*m₂ = -1
m₂ = -1/0 = Negative Infinite or Infinite
Slope of line perpendicular to y = 1, is = Infinite.
Write "st div. by 6r" as
st
-------
6r
45s 15s 3s
If r =5 and t = 45, but s is unknown, then we get --------- = --------- = -------
6(r) 2(5) 2
