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:






= 0.00334
therefore,
3,872.80 > 0.00334
Step-by-step explanation:
Which statement is correct?
(2.06x102x1.88 x 10-1) <7.69x 10
2.3x10-5
(2.06 x 10-2)(1.88 x 10-1) 7.69x 10-
2.3x10-5
7.69 x 10"
(2.06 x 10-2x1.88x10-)>
2.3x10-5
(2.06x 10-2X(1.88x 10-1)-7.69x104
2.3x10-5
Given that;






= 0.00334
therefore,
3,872.80 > 0.00334
That’s the same thing I’m on
Answer:
27,824
Step-by-step explanation:
i just calculated it.