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.
Step-by-step explanation:
The parent function of the functions of the form f(x)=√x−a+b is f(x)=√x . Note that the domain of f(x)=√x is x≥0 and the range is y≥0 . The graph of f(x)=√x−a+b can be obtained by translating the graph of f(x)=√x to a units to the right and then b units up.
I hope it's helpful!
It would be in a state of loss because the units manufactured neither reaches nor exceeds the break even point.