A cable is made up of four wires. The breaking strength of each wire is a normally distributed random variable with mean 10 kN a
1 answer:
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Step-by-step explanation:
<u>Let us assume that:</u>
<u>Therefore, the equation becomes:</u>
<u>Now substitute the value of u. We get:</u>
<u>Therefore:</u>
★ <u>Which is our required answer.</u>
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
a² - b² = (a + b)(a - b)
(a + b)³ = a³ + 3ab(a + b) + b³
(a - b)³ = a³ - 3ab(a - b) - b³
a³ + b³ = (a + b)(a² - ab + b²)
a³ - b³ = (a - b)(a² + ab + b²)
(x + a)(x + b) = x² + (a + b)x + ab
(x + a)(x - b) = x² + (a - b)x - ab
(x - a)(x + b) = x² - (a - b)x - ab
(x - a)(x - b) = x² - (a + b)x + ab
Answer:
Step-by-step explanation:
<em>Step 1: Re-state the problem in an easier way to set up a permutation problem</em>
"The numbers 1,6,8,13,15,20 can be placed in the circle below, each exactly once, so that the sum of each pair of numbers adjacent in the circle is a multiple of seven"
The series above can be represented again, in form of:
7a + 1, 7b - 1, 7c + 1, 7d - 1, 7e + 1, 7f - 1
with a, b, c, d, e, f are non-negative integers.
=> 3 numbers are multiply of 7 plus 1
=> 3 numbers are multiple of 7 minus 1
<em>Step 2: Perform the counting:</em>
For the 1st number, there are 6 ways to select
To satisfy that each pair of numbers creates a multiple of 7, then:
For the 2nd number, there are 3 ways left to select
For the 3rd number, there are 2 ways left to select
For the 4rd number, there are 2 ways left to select
For the 5th number, there are 1 way left to select
For the 6th number, there are 1 way left to select
=> In total, the number of possible ways to select:
N = 6 x 3 x 2 x 2 x 1 x 1 = 72
However, these numbers are located around a circle, each option is counted twice.
=> The final number of possible ways:
N = 72/2 = 36
Hope this helps!
:)