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:
1/30
Step-by-step explanation:
P(champion)= 1/30
F(x) is a piecewise function defined as:
f(x) = (3/2)x when 0 <= x <= 2
or
f(x) = -(3/2)x+6 when 2 < x <= 4
g(x) is defined as:
g(x) = -(1/4)x+1 when 0 <= x <= 4
Based on the graph or through the equations we can say:
f(1) = (3/2)*1 = 1.5
g(1) = -(1/4)*1+1 = 0.75
f(3) = (3/2)*3 = 4.5
g(3) = -(1/4)*3+1 = 0.25
And the derivative values are:
f ' (1) = 3/2 = 1.5
f ' (3) = -3/2 = -1.5
g ' (1) = -1/4 = -0.25
g ' (3) = -1/4 = -0.25
which are the slopes of each line
So...
h(x) = f(x)*g(x)
h ' (x) = f ' (x)*g(x) + f(x)*g ' (x) ... product rule
h ' (1) = f ' (1)*g(1) + f(1)*g ' (1)
h ' (1) = 1.5*0.75 + 1.5*(-0.25)
h ' (1) = 0.75
and,
h(x) = f(x)*g(x)
h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)
h ' (3) = f ' (3)*g(3) + f(3)*g ' (3)
h ' (3) = -1.5*0.25 + 4.5*(-0.25)
h ' (3) = -1.5
5(x+y)= 20+ 3x
5x+5y=20+3x
5x+5y-3x=20+3x-3x
2x+5y=20
2x+5y-2x=20-2x
5y=-2x+20
5y/5=-2x/5+20/5
y=-2/5x+4
ANSWER: y=-2/5x+4
