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: 25.0
Step-by-step explanation:
25
Answer:
Step-by-step explanation:
Just multiply the sides. For example, the roses take 8x4=32 feet. if we want to know the area of Zinnias, we multiply the sides 8x12 and remove the roses, which would be 8x12-32=96-32=60 for the Zinnias.
For a)
The roses take 8x4
Zinnias take 8x12 - roses
Snapdragons take 16x12 - roses - zinnias
Petunias take 16x20 - snapdragons - zinnias- roses
b) The entire area of the garden is 20x16=320 square feet
The area of the Zinnias(including roses) is 8x12 = 96 square feet
To get the percent, we divide the smaller number by the bigger number and multiply by 100 or in practice 96/320x100=30%
c) Let's see the progression here. The width progresses in the first rectangle by four 8 to 12, then again by four 12, 16 and again by four 16 to 20, which means that the next rectangle would have a width of 20+4=24. Doing the sale analysis for the height, we get that the increase is with 4 feet at a time or the new rectangle would have a width of 24 and height of 20, which means a total area of 480 feet. We have to subtract all other rectangles or the new area would be 480 - (20x16) = 160 square feet
Answer:
xy = 1
k = 79
Step-by-step explanation:
Question One
The first and third frames look to me to be the same. I'll treat them that way.
y = x^2 Equate y = x^2 to the result of 2y + 6 = 2x + 6
2y + 6 = 2(x + 3) Remove the brackets
2y + 6 = 2x + 6 Subtract 6 from both sides
2y = 2x Divide by 2
y = x
Now solve these two equations.
so x^2 = x
x > 0
1 solution is x = 0 from which y = 0. This won't work. x must be greater than 0. So the other is
x(x) = x Divide both sides by x
x = 1
y = x^2 Put x = 1 into x^2
y = 1^2 Solve
y = 1
The second solution is
(1,1)
xy = 1*1
xy = 1
Answer: A
Question Two
square root(k + 2) - x = 0
Subtract x from both sides
sqrt(k + 2) = x Square both sides
k + 2 = x^2 Let x = 9
k + 2 = 9^2 Square 9
k + 2 = 81
k = 81 - 2
k = 79
$0.25 • 20 = 5•20 = 100 every 40 runs then subtract 18 from 100 and the total would be 82 in total. I hope this helped:)
