Subtracting the functions
2,0-1,-2-----(1,-2)
The line segment which represents 2 7/16 inches should first be drawn with the use of a ruler.
<h3>What is a Ruler?</h3>
This is referred to an instrument which is used to measure distances between points or draw straight lines and also usually graduated in inches.
On the inches graduation, the zero mark should be located and the 1/16 inch which is found before the half-inch mark between 2 and 3 inches. When this is done, a line should be drawn to meet both points which will therefore measure 2 7/16 inches long.
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The 5 numbers are 105, 210, 315, 420, 630
Answer:
The only thing you will do is to plug the x in the equation.
x = -2 ---> f(x) = -6
x = 0 ---> f(x) = 0
x = 3 ---> f(x) = 9
x = 7 ---> f(x) = 21
f(x) = 3x
f(-2) = <u>3(-2)</u>
f(-2) = -6
f(0) = <u>3(0)</u>
f(0) = 0
f(3) = <u>3(3)</u>
f(3) = 9
21 = 3x
<u>21 / </u><u>3</u> = <u>3x / 3</u>
7 = x
Hope this helps, thank you :) !!
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.