Answer: OPTION B.
Step-by-step explanation:
Given the following System of equations:
You can use the Elimination Method to solve it. The steps are:
1. You can mutliply the second equation by -3.
2. Then you must add the equations.
3. Solve for the variable "y".
Then:
4. Now that you know the value of the variable "y", you must substitute it into any original equation.
5. The final step is to solve for "x" in order to find its value.
Then:
Therefore, the solution is:
Answer:
(c) Probability that a failure is due to loose keys = 0.2376
(d) Probability that a failure is due to improperly connected or poorly welded wires = 0.078
Step-by-step explanation:
The Whole probability scenario is given for Computer Keyboard failures.
(a) Let F be the event of failure due to faulty electrical connects, P(F) = 0.12
M be the event of failure due to mechanical defects, P(M) = 0.88
LK be the event of mechanical defect due to loose keys, P(LK/M) = 0.27
IA be the event of mechanical defect due to improper assembly, P(IA/M) =0.73
DW be the event of electrical connects due to defective wires,P(DW/F) = 0.35
IC be the event of electrical connects due to improper connections,
P(IC/F) = 0.13 .
PWW be the event of electrical connects due to poorly welded wires,
P(PWW/F) = 0.52
(b) <u> </u><u>Keyboard failures</u>
<h2> / \</h2><u> </u><u> Faulty electrical connects </u> <u>Mechanical Defects </u>
P(F) = 0.12 P(M) = 0.88
<h2> / | \ / \</h2><u><em>Defective wires</em></u> <u><em>Improper</em></u> <u><em>Poorly</em></u> <u><em>Loose Keys</em></u> <u><em>Improper</em></u><em> </em>
P(DW/F)=0.35 <u><em>Connections</em></u> <u><em>Welded wires</em></u> P(LK/M)=0.27 <em> </em><u><em>Assembly</em></u>
P(IC/F)=0.13 P(PWW/F)=0.52 P(IA/M)=0.73
This is the required tree diagram.
(c) Probability that a failure is due to loose keys is given by:
P(LK) =P(LK/M) * P(M) {This means mechanical failure is due to loose
keys}
P(LK) = 0.27 * 0.88 = 0.2376 .
(d) Probability that a failure is due to improperly connected or poorly welded
wires is given by P(IC PWW) ;
P(IC PWW) = P(IC) + P(PWW) - P(IC
PWW) { Here P(IC
PWW) = 0 }
P(IC) = P(IC/F) * P(F) = 0.13 * 0.12 = 0.0156
P(PWW) = P(PWW/F) * P(F) = 0.52 * 0.13 = 0.0676
Therefore, P(IC PWW) = 0.0156 + 0.0676 - 0 = 0.078 .