For section 1.....Problem a is 7.2 Problem B is 7.6 Problem C is 18.8 Problem E is 32.5 Problem F is 28.8 Problem G is 10 Problem I is 61.5 Problem J is 170.8 Problem K is 81.6.
For Section 2 Problem A is 9.36 Problem B is 10.35 Problem C is 25.85 Problem E is 1.82 Problem F is 3.28 Problem G is 2.8 Problem I is 42.21 Problem J is 56.16 Problem K is 7.65
It 36 hope i helped!!
Assume answer as x;
x÷4=9 or x4=9
Multiply 9 with 4;
x=9×4
x=36
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:
(Square, Right Triangle)
18 meters²
Step-by-step explanation:
You can divide the shape into a square and right triangle, see the attachment :)
The area of the square is the length times the width. So,
3 · 3 = 9
The area of the square is 9 m².
The triangle's area is the base times the height divided by 2. So,
6 · 3 = 18
18 / 2 = 9
The triangle has an area of 9 m².
Add these together.
9 + 9 = 18
The area is 18 meters².