Answer:
attached below
Step-by-step explanation:
a) Diagram of state transition representing failure and repair process of component
attached below
b) expression for the system to be completely available
attached below
c) Expression for system to be operating in a degraded mode
attached below
One way to solve this system of equations would be by substitution. It would helpful to write these equation in the form y=mx + b in order to substitute one equation to the other. We do as follows:
<span>1.) x - 2y = 6
y = x/2 -3
2.) 2x - 4y = 10
y = x/2 - 5/2
As we look at the rearranged equations, we see that their slopes are equal which means they are parallel lines so they do not intersect at one point. There is no solution for this system of equations.</span>
Solution :
The objective of the study is to test the claim that the loaded die behaves at a different way than a fair die.
Null hypothesis, 
That is the loaded die behaves as a fair die.
Alternative hypothesis,
: loaded die behave differently than the fair die.
Number of attempts , n = 200
Expected frequency, 

Test statistics, 


≈ 5.8
Degrees of freedom, df = n - 1
= 6 - 1
= 5
Level of significance, α = 0.10
At α = 0.10 with df = 5, the critical value from the chi square table

= 9.236
Thus the critical value is 
![$P \text{ value} = P[x^2_{df} \geq x^2]$](https://tex.z-dn.net/?f=%24P%20%5Ctext%7B%20value%7D%20%3D%20P%5Bx%5E2_%7Bdf%7D%20%5Cgeq%20x%5E2%5D%24)
![$=P[x^2_5\geq 5.80]$](https://tex.z-dn.net/?f=%24%3DP%5Bx%5E2_5%5Cgeq%205.80%5D%24)
= chi dist (5.80, 5)
= 0.3262
Decision : The value of test statistics 5.80 is not greater than the critical value 9.236, thus fail to reject
at 10% LOS.
Conclusion : There is no enough evidence to support the claim that the loaded die behave in a different than a fair die.
Answer: Simplifying a fraction means that the unsimplified and simplified fractions both give the same quotient and remainder. So 14:2 can both be divided equally to yield 7:1.
Answer:
Hello,
Answer <u>162</u>
Step-by-step explanation:
Let say n the searched number.
