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d. 17.5 marks] A remotely situated fuel cell has an installed cost of $2,000 and will reduce existing surveillance expenses b
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Part (D)

2000=350(\frac{P}{A},10\%,n)\\ (\frac{P}{A},10\%,n)=\frac{2000}{350}=5.7142\\

at\ n=8.889322\\ (\frac{P}{A},10\%,n)=5.71405

correct answer is (D) 8.89

Part (E)

if\ X\ is\ a\ random\ variable\\ expected\ value,E(X)=\sum{x*p(x)}\\ variance,var(X)=E(X^{2})-[E(X)]^{2}=\sum{x^{2}*p(x)}-[\sum{x*p(x)}]^{2}\\

expected\ production\ rate=100*0.4+125*0.6=40+75=115\\ expected\ profit\ per\ unit\ sold=5*0.35+7*0.65=1.75+4.55=6.3\\

variance\ of\ production\ rate=(100^{2}*0.4+125^{2}*0.6)-115^2=4000+9375-13225=150

variance\ of\ profit\ per\ unit=(5^{2}*0.35+7^{2}*0.65)-6.3^{2}=8.75+34.85-39.69=3.91

total\ profit=(production\ rate)*(profit\ per\ unit)\\

E(XY)=E(X)*E(Y)\\ expected\ value\ of\ total\ profit,=115*6.3=724.5\\

var(XY)=(\sigma_{X}^{2}+[E(X)]^{2})*(\sigma_{Y}^{2}+[E(Y)]^{2})-[E(X)]^{2}*[E(Y)]^{2}\\

variance\ of\ total\ profit=(150+115^{2})*(3.91+6.3^{2})-115^{2}*6.3^{2}\\ variance\ of\ total\ profit=18111.40496\\ standard\ deviation=18111.404^{0.5}=134.5786

correct answer is (B)

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