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shepuryov [24]
3 years ago
9

I know you have to use the half life formula but I don’t know how to use it could someone show how to find the solution step by

step? Thank you

Mathematics
1 answer:
n200080 [17]3 years ago
7 0

Answer:

A = Pe ^{-0.03108 * t}

Step-by-step explanation:

Note: these are typically two part questions...

you have the model and have to FIRST find a constant value "k"

once you have the "k" value you plug that into the model to finish the problem

A = Pe ^{-kt} is a typical version of the continuous decay...

e is "e 2.717" not a variable "e"

so... step 1.. fink k

you are told that the 1/2 life is 22.3 years...

so lets assume that we started with 2 units of lead, and waited for it to decrease to 1 (1 is 1/2 of 2)

you would have

1 = 2e^{k * 22.3}

.5 = e^{k * 22.3}

ln(.5) = 22.3k ln(e)

ln(e) = 1

ln(.5) = 22.3k

ln(.5) = 22.3k ln(e)

\frac{ln(.5) }{22.3} = k

\frac{\ln \left(0.5\right)}{22.3}=-0.03108 = k

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(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 \bigcup PWW) ;

 P(IC \bigcup PWW) = P(IC) + P(PWW) - P(IC \bigcap PWW) { Here P(IC \bigcap 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 \bigcup PWW) = 0.0156 + 0.0676 - 0 = 0.078 .

8 0
3 years ago
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