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:
y = 40
Step-by-step explanation:
__y_ x 8 = 8 x 5
8
y = 40
Sum of the terms of the series is
Sn = n/2 ( a1+an )
we have n= 6 , a1= 17, an = 57
so Sn = 6/2 ( 17+57) = 3(74) = 222
Answer:
24 / x = 14 - 2x
Step-by-step explanation:
Quotient means division (÷ or /)
Quotient of 24 and x = 24 ÷ x
24 ÷ x is equivalent to 24 / x
14 minus 2 times x = 14 - 2x
So, the equation is
24 / x = 14 - 2x
Answer:
<h3>
1) 7x - 5
</h3><h3>
2) 9y - 18
</h3><h3>
3) 0.5n + 4n
</h3><h3>
4) 2(w³+23)</h3><h3>
Step-by-step explanation:</h3>
1)
The product of seven and a number x: 7·x = 7x
<u>Five less than the product of seven and a number x:</u>
<h3>
7x - 5
</h3>
2)
nine times a number y: 9·y = 9y
<u>The difference of nine times a number y and eighteen:</u>
<h3>
9y - 18
</h3>
3)
half a number n: 0.5n
four times the number: 4·n = 4n
<u>Half a number n increased by four times the number:</u>
<h3>
0.5n + 4n
</h3>
4)
a number w cubed: w³
the sum of a number w cubed and twenty-three: w³+23
<u>Twice the sum of a number w cubed and twenty-three:</u>
<h3>2(
w³+23)</h3>
