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tia_tia [17]
2 years ago
14

Could -free, automatic faucets actually be housing more bacteris than the old fashioned, manual kind The concern is that decreas

ed water flow may increase the chance that bacteria grows, because the automatic faucets are not being thoroughly flushed through. It is known that 15% of cultures from older faucets in hospital patient care areas test positive for Legionella bacteria. A recent study at Johns Hopkins Hospital found Legionella bacteria growing in 10 of cultured water samples from 20 electronic faucets aIf the probability of Legionella bacteria growing in a faucet is 0.15, what is the prob ability that in a sample of 20 faucets. 10 or more have the bacteria growing (the Johns Hopkins study provide sufficient evidence that the probability of Legionella bacteria growing in electronic faucets is greater than 15%? Explain.
Mathematics
1 answer:
Volgvan2 years ago
3 0

Answer:

Not enough evidence to reject Null hypothesis

Step-by-step explanation:

Solution:-

- A comparative study for bacterial growth in manual and electronic faucets is made.

- It is observed that there is a higher growth in electronic faucets due to slower flow rates, i.e electronic faucets are not thoroughly flushed; hence, giving more resident time for the scaled bacteria to grow.

- It is known that 15% of cultures from older faucets were tested positive for the Legionella bacteria.

- A study at John Hopkins was conducted on a sample n = 20 electronic faucets with the probability of bacteria growing in a faucet is 0.15.

- We will conduct a hypothesis for at-least half proportion of electronic faucets have cultured bacteria.

- State the hypothesis for the proportion of electronic faucets culturing Legionella bacteria:

        Null Hypothesis:  P = 0.15

        Alternate hypothesis: P > 0.15    

- To determine the test statistics for the study conducted at John hopkins. We had a sample size of n = 20, and the probability for a bacteria to grow in a faucet is 0.15.

- Denote random variable, X: The number of electronic faucets culturing Legionella bacteria.

- Since, the probability for a bacteria to grow in a faucet is independent for each new faucet. We will assume the RV " X " to follow binomial distribution with probability of success 0.15:

                      X ~ Bin ( 20 , 0.15 )                  

- We are to determine that at-least half of the sample is subjected to the said bacteria. This is the probability of P ( X ≥ 10 ).

- The pmf for a binomially distributed random variable X is given below:

                     P ( X = r ) = n_C__r * ( p(success) )^r * ( p (fail) )^(^n^-^r^)

Where,

            p ( success ) = 0.15

            p ( fail ) = 1 - p ( success ) = 1 - 0.15 = 0.85

- Use the pmf to determine the required test statistics:

P ( X \geq 10 ) = 1 - P ( X \leq  9 )\\\\P ( X \geq 10 ) = 1 - [ (0.85)^2^0 + 20*(0.15)*(0.85)^1^9 + 20_C_2 (0.15)^2*(0.85)^1^8 +\\\\ 20_C_3 (0.15)^3*(0.85)^1^7 + 20_C_4 (0.15)^4*(0.85)^1^6 + 20_C_5 (0.15)^5*(0.85)^1^5+\\\\ 20_C_6 (0.15)^6*(0.85)^1^4 + 20_C_7 (0.15)^7*(0.85)^1^3 + 20_C_8 (0.15)^8*(0.85)^1^2 + \\\\ 20_C_9 (0.15)^9*(0.85)^1^1\\\\\\P ( X \geq 10 ) = 1 - [  0.03875 + 0.13679 + 0.22933 + 0.24282 + 0.18212 + 0.10284 + \\\\ 0.04537 + 0.01601 + 0.00459 + 0.00108 ]\\\\

P ( X \geq 10 ) = 1 - [ 0.997 ] = 0.003

- The probability that 10 or more electronic faucets is found to have Legionella bacteria growing is 0.003              

- The test proportion of 10 and more electronic faucets have culturing bacteria is p = 0.003.

- Assuming normality of the population, the Z-statistics would be:

                  Z-test = \frac{ (p - P) \sqrt{n} }{\sqrt{P*(1 - P )} } \\\\Z-test = \frac{ (0.003 - 0.15) \sqrt{20} }{\sqrt{0.15*(0.85)} } \\\\Z-test = -1.84109

- If we were to test the claim to 90% level of confidence:

                  significance level (α) = 1 - CI = 1 - 0.9 = 0.1

- The rejection region Z-critical is defined by a right-tail:

                 Z-critical \geq Z_\alpha \geq Z_0_._2\\\\Z-critical \geq 1.28    

- Compare the test statistics with the rejection criteria defined by the Z-critical:

                Z-test < Z-critical

                -1.84 < 1.28

Conclusion:

There is not enough evidence that the probability of Legionella bacteria growing in electronic faucets is greater than 15%.

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Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one. Steam enters the first turbi
miss Akunina [59]

Answer:

Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one, and reheat. Steam enters the first turbine stage at 12 MPa, 480∘C, and expands to 2 MPa. Some steam is extracted at 2 MPa and fed to the closed feedwater heater. The remainder is reheated at 2 MPa to 440∘C and then expands through the second-stage turbine to 0.3 MPa, where an additional amount is extracted and fed into the open feedwater heater operating at 0.3 MPa. The steam expanding through the third-stage turbine exits at the condenser pressure of 6 kPa. Feedwater leaves the closed heater at 210∘C, 12 MPa, and condensate exiting as saturated liquid at 2 MPa is trapped into the open feedwater heater. Saturated liquid at 0.3 MPa leaves the open feedwater heater. Assume all pumps and turbine stages operate isentropically. Determine for the cycle

a. Draw the cycle on a T-S diagram using the same numbering in the schematic

b. Determine the thermal efficiency of the cycle.

c. Determine the mass flow rate of steam entering the first turbine of the cycle.

(i) Thermal efficiency of the cycle = 43.185 %

(ii) The mass flow rate of steam =93.66 kg/h

Step-by-step explanation:

So we have at

For Point 1 on the T-S diagram we have

p₁ = 80 bar,  

t₁ = 480 °C,

From the super-heated steam tables we have

h₁ = 3349.6 kJ/kg, s₁ = 6.6613 kJ/kg·K

Point 2

p₂ = 20 bar

s₁ = s₂  =with x₂ = (6.6613 -6.6409)/(6.6849-6.6409) = 0.464

therefore h₂ =2953.1 + 0.464×(2977.1 - 2953.1) = 2964.22 kJ/kg

Point 3 on the T-S diagram we have

p₃ = 3 bar again s₁ = s₃  so we go to 3 bar on the steam tables and look up s = 6.6613 kJ/kg·K which is on the saturated steam tables

and x₃ is given as (6.6613 -1.6717)/(6.9916-1.6717) = 0.9379 and

h₃ = 561.43 + x₃×2163.5 = 2590.6 kJ/kg

Point 4

p₄ = 0.08 bar, s₁ = s₄, x₄ = 0.7949 and h₄ = 2083.45 kJ/kg

Point 5  

p₅ = 0.08 bar, h_{f5}= 173.84 kJ/kg

Point 6

Here h₆ is given by  h_{f5} plus the work done to move the water to the open heater therefore h₆ =

= 173.84 kJ/kg + 0.00100848×(3 - 0.08) × 100

= 173.84 kJ/kg + 0.29447616 kJ/kg = 174.13 kJ/kg

Point 7

p₇ = 3 bar, and h_{f7} = 561.43 kJ/kg

Point 8

Here again work is done to convey the fluid t constant pressure thus

h₈ = h_{f7} + v_{f7}× (p₈ - p₇)

561.43 kJ/kg + 0.00107317×(80 - 3)×100 = 569.69 kJ/kg

Point 9

p₉ = 80 bar  and T₉  = 205°C

By interpolating the values on the subcooled teperature tables we get

x₉ = 0.5 and h₉ =  854.94 + 0.5× (899.79 - 854.94) = 877.365 kJ/kg

Point 10

p₁₀ =  20 bar, h₁₀   = h_{f10} = 908.50 kJ/kg

point 11

Here h₁₁ = h₁₀ = 908.50 KJ/kg

For the closed feed water heater, energy and mass flow rate balance gives

m₁ × (h₂ - h₁₀) + (h₈ - h₉) = 0

Therefore m₁ = \frac{ (h_{9}  - h_8)}{(h_{2} - h_{10})}  = 0.14967

while the open water heater we get

m₂×h₃+(1-m₁-m₂)×h₆+m₁×h₁₁ - h₇ = 0

from where m₂ = 0.11479

W_{T} = (h₁-h₂) + (1 - m₁)(h₂ - h₃) +(1 - m₁ - m₂)(h₃ - h₄)

= 1076.11 kJ/kg

W_{p} = (h₈ - h₇) + (1 - m₁ - m₂)×(h₆ - h₅)

= 8.4733 kJ/kg

Q = h₁ -h₉ = 2472.235 kJ/kg

Efficiency = η = \frac{W_{T} - W_{P} }{Q} = 43.185 %

(ii)W_{cycle} = m_1*(W_T -W_P)

m'₁ = 100×10³/1066.63 = 93.66 kg/h

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2 years ago
Madelyn buys 3 bottles of orange juice at the corner store for a total cost of $2.31.
Tom [10]

Answer:

$0.77

Step-by-step explanation:

To find how much each bottle of juice costs, divide 2.31 by 3:

2.31/3

= 0.77

So, each bottle of juice is $0.77

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Step-by-step explanation:

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Step-by-step explanation:

Multiply exponents

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What is the slope of the line through ( − 5 , − 10 ) (−5,−10)left parenthesis, minus, 5, comma, minus, 10, right parenthesis and
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3.75

Step-by-step explanation:

We have to find the slope of a straight line through points (-5,-10) and (-1,5).

Now. we know the slope of a straight line through the points (x_{1},y_{1}) and (x_{2},y_{2}) is given by [\frac{y_{2} - y_{1} }{x_{2} - x_{1}} ].

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