The probability that a wafer contains a large particle of contaminat.pdf

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The probability that a wafer contains a large particle of contamination is .02. If it is assumed that the wafers are independent, what is the probability that exactly 120 wafers need to be analyzed before a large particle is detected? Solution your complete answer is in this link staff.najah.edu/sites/default/files/Chapter%203.pdf.

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The population of a city is increasing such that a new water treatment plant is planned. Potential source water for the plant is considered to have high concentrations of multivalent cations, high suspended solids concentration (turbidity), and high pathogens concentration. What treatment operations/processes would you recommend and why Solution Water treatment plants are kept for purification of water for drinking purpose So for removal high concentration of multivalent ions , high suspended solids concentration and highpathogens in water are removed by 1) Coagulation - alum is added to raw water to remove dirt and tiny particles 2) Sedimentation - removal of turbidity in water using sand filters 3) Filtration - To remove suspended particles in water 4) Disinfection = Removal of high pathogen concentration in water 5) Sludge drying - Solids collected in sedimentation and filtration are removed using dry lagoons 6) Flouridation = removal of flourine substance in water 7) Chlorination = Chlorine is added to remove high concentration of multivalent ions.

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The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial population P0 has doubled in 5 years. Suppose it is known that the population is 11,000 after 3 years. What was the initial population P0? (Round your answer to one decimal place.) P0 = 1 What will be the population in 10 years? (Round your answer to the nearest person.) 2 persons How fast is the population growing at t = 10? (Round your answer to the nearest person.) 3 persons/year Solution dP/dt = P(t) P = ? P(t) dt P = e^(kt) So P(t) = P0*e^(kt) If P(5) = 2P0: P(5) = P0*e^(k5) 2P0 = P0*e^(5k) 2 = e^(5k) ln(2) = 5k k = ln(2)/5 P(t) = P0*e^(ln(2)t/5) P(t) = P0*2^(t/5) 2) P(t) = 11,000 when t = 3 P(3) = 11,000 P0*2^(3/5) = 11,000 P0 = 11000/(2^(3/5)) P0 = 7257.29, or rounded to 7257 P(10) = 7257*2^(10/5) P(10) = 7257*2.

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The previous help did not produce a correct answer. Archimedes Levers is financed by a mixture of debt and equity. You have the following information about its cost of capital: Calculate the rE, rA, E, D, and A. I received the following response for rE &rA, which are not correct: Re=Riskfree Rate+Beta*(Market Rate-Riskfree Rate)=10+1.6*(15-10)=18% Ra=0.35*11+0.65*18=12.91%rE = ? %rD = 11%rA = ? %E = ?D = .2A = ?rf = 10 %rm = 15%D/V = .35 Solution rE = rf + E * (rm - rf) = 10% + 2.4 * (15% - 10%) = 22% D = (rD - rf) / (rm - rf) = (12% - 10%) / (15% - 10%) = 0.4 A = E/V * (E + D * D/E) = 0.45 * (2.4 + 0.4 * 0.45/0.55) = 1.23 rA = rf + A * (rm - rf) = 10% +1.23 * (15% - 10%) = 16.14% Archimedes repurchases debt and issues equity rE = rf + E * (rm - rf) = 10% + 2.4 * (15% - 10%) = 22% D = (rD - rf) / (rm - rf) = (11% - 10%) / (15% - 10%) = 0.2 A = E/V * (E + D * D/E) = 0.65 * (2.4 + 0.2 * 0.35/0.65) = 1.63 rA = rf + A * (rm - rf) = 10% +1.63 * (15% - 10%) = 18.15%.

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