Question

Rick​ Wing, salesperson for Wave Soldering​ Systems, Inc.​ (WSSI), has provided you with a proposal for improving the temperature control on your present machine. The machine uses a​ hot-air knife to cleanly remove excess solder from printed circuit​ boards; this is a great​ concept, but the​ hot-air temperature control lacks reliability. According to​ Wing, engineers at WSSI have improved the reliability of the critical temperature controls. The new system still has the four sensitive integrated circuits controlling the​ temperature, but the new machine has a backup for each. The four integrated circuits have reliabilities of 0.88​, 0.90​, 0.92​, and 0.94. The four backup circuits all have a reliability of 0.88. ​
The overall reliability of the new temperature controller​ = nothing​% ​(enter your response as a percentage rounded to two decimal​ places).

Answer 1

Answer:

The answer is "A, B , C, D is now considered a single one".

Explanation:

$\to (A = 0.84) \\\\ \to (B=0.86)\\\\ \to (C = 0.88)\\\\ \to (D =0.90)$

In the question, the components will be in series throughout or they must both work while the machine operates. The inability of one part fails. ABCD    In each adjacent component is assisted and if those modules stop running the system works.  

This device for one central ingredient and then one back up becomes perpendicular from each circuit throughout the given problem. For example, Is the device with two adjacent components  

Parallel sequence method  

Some modules were replicated in parallel here and in sequence. The whole four-circuit welding system is one.  

Consistency in parallel systems  

The probability each component would fail

The formula for calculating the Probability:

$\text{Probability (single failed component) = 1-(component reliability)}$

Probability (A-key IC component failure)=1-(component reliability)=1-0.84  

Probability  (B-key IC component failure) = 1-(component reliability) = 1-0.86  

Probability (C-key IC component failure) = 1-(component reliability) = 1-0.88  

Probability (D-key IC component failure) = 1-(component reliability) = 1-0.90  

Probability (backup fails)=1-0.84  

Device stable = probability (at least one part works)  

Using the approach as a supplement,  

P (The function of at least one component)  

= 1-Probability (Failure in all parts)  

= 1- The result of single potential error probabilities  

= 1-[Probability (Main IC-Fails) \times  Probability  \times (Backup-Fails)]  

The Durability of A$=1- [ (1-0.84)(1-0.84)]$

                               $= 1- 0.256\\\\=0.9744$

The Durability of B $=1- [ (1-0.86)(1-0.84)]$

                                $=1- [ (0.14)(0.16)]\\\\=1- 0.0224\\\\= 0.9776$

The Durability of  C $=1- [ (1-0.88)(1-0.84)]$

                                 $=1- [ (0.12)(0.16)]\\\\=1- [0.0192]\\\\= 0.9808$

The Durability of  D $=1- [ (1-0.90)(1-0.84)]$

                                 $=1- [ (0.10)(1-0.16)]\\\\= 1- 0.016\\\\=0.984$

Durability of the system sequence  

Durability of Device = consumer durability of series components  

$=0.9744\times 0.9766 \times 0.9808 \times 0.9840 \\\\=0.9184 \ \ or\ \ 91.84 \% \ \ reliability$

Note:

Any single part A, B , C, D is now considered a single one