Explanation:
the Ethics of emerging Technology can only make use of speculative data about future products,uses and impacts.
A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. 1) The inspector randomly picks 20 items from a shipment. What is the probability that there will be at least one defective item among these 20? 2) Suppose that the company receives 10 shipments in a month and the inspector randomly tests 20 devices per shipment. What is the probability that there will be exactly 3 Shipments each containing at least one defective device among the 20 that are selected and tested from the shipment?
Answer:
1) The probability of at least 1 defective is approximately 45.621%
2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%
Explanation:
The given parameters are;
The defective rate of the device = 3%
Therefore, the probability that a selected device will be defective, p = 3/100
The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;
The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97
The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927
The probability of at least 1 = 1 - The probability of 0 defective in 20
∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621
The probability of at least 1 defective ≈ 0.45621 = 45.621%
2) The probability of at least 1 defective in a shipment, p ≈ 0.45621
Therefore, the probability of not exactly 1 defective = q = 1 - p
∴ q ≈ 1 - 0.45621 = 0.54379
The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;
P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212
Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%
When my animal is to old to physically carry her baby what process can I use to still produce her young from her
Multiple Select
In the following list, what are criteria that would be important to someone buying a Jersey cow
milk production
health
Speed
age
hoof color
Answer:
Milk Production, Health, and Age.
Explanation:
You want a younger cow when buying cattle so you can have gather more milk from it over it's lifetime. You also want to make sure that it can actually produce milk. Then you want a cow in good health.
differentiate from first principles if y=5
[tex] {?}^{2} [/tex]
Answer:
y'(x) = 7 ;
y'(x) = 15x²
Explanation:
Differentiation from first principle :
y = 7x
Formula to obtain derivative from first principle:
y'(x) = lim(h - - > 0) : [y(x + h) - y(x)] / h
Obtain : y(x + h)
y(x) = 7x
y(x + h) = 7(x + h)
Substitute into formula :
y'(x) = lim(h - - > 0) : [7(x + h) - 7x] / h
Expand :
y'(x) = lim(h - - > 0) : [7x + 7h - 7x] / h
y'(x) = lim(h - - > 0) : 7h/ h
y'(x) = 7
2.)
y = 5x³
y'(x) = lim(h - - > 0) : [y(x + h) - y(x)] / h
Obtain : y(x + h)
y(x) = 5x³
y(x + h) = 5(x + h)³
Substitute into formula :
y'(x) = lim(h - - > 0) : [5(x + h)³ - 5x³] / h
Expand :
y'(x) = lim(h - - > 0) : [5(x³ + 3x²h + 3xh² + h³) - 5x³] / h
y'(x) = lim(h - - > 0) : [5x³ + 15x²h + 15xh² + 5h³ - 5x³] / h
y'(x) = lim(h - - > 0) : [15x²h + 15xh² + 5h³] / h
y'(x) = lim(h - - > 0) : h(15x² + 15xh + 5h²) / h
y'(x) = lim(h - - > 0) : (15x² + 15xh + 5h²)
y'(x) = 15x²
For an Na+—Cl- ion pair, attractive and repulsive energies EA and ER, respectively, depend on the distance between the ions r, according to EA = -1.436/r ER =(7.32 *10-6 )/r8 For these expressions, energies are expressed in electron volts per Na+—Cl- pair, and r is the distance in nanometers. The net energy EN is just the sum of the preceding two expressions
Answer:
Explanation:
[tex]\text{The curve of the plot}[/tex] [tex]\mathbf{E_A,E_R, \ and \ E_N}[/tex] [tex]\text{can be seen in the attached diagram below}[/tex]
[tex]\text{From the plot}[/tex], [tex]\mathbf{r_o = 0.2 4nm \ and \ E_o =-5.3 eV}[/tex]
[tex]\mathbf{We \ knew \ that: E_N = E_A + E_R}[/tex]
[tex]\mathtt{GIven \ E_A = \dfrac{-1.436}{r}\ \ \ , E_R = \dfrac{7.32 \times 10^{-6}}{r^n} \ \ and \ \ n=8 }[/tex]
[tex]\mathtt{Then; E_N = -\dfrac{-1.436}{r}+ \dfrac{7.32\times 10^{-6}}{r^8}}[/tex]
[tex]\mathtt{Also; r_o = \Big( \dfrac{A}{nB} \Big)^{\dfrac{1}{1-n}}} \\ \\ \mathtt{ r_o = \Big( \dfrac{1.986}{8 \times 7.32\times 10^{-6}} \Big)^{\dfrac{1}{1-8}}} \\ \\ \mathbf{r_o = 0.236 nm}[/tex]
[tex]E_o = \dfrac{-1.436}{\Big[\dfrac{1.436}{8(732\times 10^{-6})}\Big]^{\dfrac{1}{1-8}}} + \dfrac{7.32 \times 10^{-6}}{\Big[ \dfrac{1.436}{8\times7.32 \times 10^{-6} } \Big]^{\dfrac{8}{1-8}}}[/tex]
[tex]\mathbf{E_o = -5.32 \ eV}[/tex]
A workpiece in the form of a bar 100 mm in diameter is to be turned down to 70 mm diameter for 50 mm of its length. A roughing cut using maximum power and a depth of cut of 12 mm is to be followed by a finishing cut using a feed of 0.1 mm and a cutting speed of 1.5 m/s. It takes 20 s to load and unload the workpiece and 30 s to set the cutting conditions, set the tool at the beginning of the cut and engage the feed. The specific cutting energy for the material is 2.3 GJ/m3 and the lathe has a 3-kW motor and a 70 percent efficiency. Estimate:
Answer:
Hello your question has some missing part below is the missing part
Estimate The matching time for the Finish cut
answer: 79.588 seconds
Explanation:
Calculate the matching time for the Finish cut
Diameter of workpiece before cutting = 100 mm
hence
Diameter of workpiece before Finishing cut ( same as after rough cut )
D2 = D1 - 2 * d = 100 - (2 * 12) = 76mm
step 1 : determine spindle speed
V = [tex]\frac{\pi D_{2}N_{s} }{60}[/tex] -------- ( 1 ). where : D2 = 76 mm , Ns = ? , V = 1.5 ( input values into equation 1 )
therefore Ns = 376.94 rpm
Finally : Determine the matching time
T = [tex]\frac{L + A}{Ft * Ns}[/tex] ---- ( 2 ) . where : Ft = 0.1 mm , Ns = 376.94 rpm, L = 50 mm , A = 0 ( input values into equation 2 )
note : A = allowance length ( not given ) , ft = feed rate
T = 50 / ( 0.1 * 376.94 )
= 1.3265 minutes ≈ 79.588 seconds
The dry weather average flow rate for a river is 8.7 m3/s. During dry weather flow, the average COD concentration in the river is 32 mg/L. An industrial source continuously discharges 18,000 m3/d of wastewater contains an average 342 mg/L COD concentration into the river. What is the COD mass loading in the river upstream of the industrial source discharge
Answer:
6156 kg /day
Explanation:
Determine the COD mass loading in the river upstream of the industrial source discharge
Given data:
Flow rate of river = 8.7 m^3/s
Average COD concentration in river = 32 mg/L
Industrial source continuous discharge ( Qw )= 18,000 m^3/d
Yw = 342 mg/l
since :
1 m^3 = 1000 liters
Qw = 18 * 10^6 liters = ( 18 million per day )
Hence the COD mass loading
= Yw * Qw
= 342 * 18 liters
= 6156 kg /day
For heat transfer purposes, a standing man can be mod-eled as a 30-cm-diameter, 170-cm-long vertical cylinderwith both the top and bottom surfaces insulated and with theside surface at an average temperature of 34°C. For a con-vection heat transfer coefficient of 15 W/m2·K, determinethe rate of heat loss from this man by convection in still airat 20°C. What would your answer
Answer:
Rate of Heat Loss = 336 W
Explanation:
First, we will find the surface area of the cylinder that is modelled as the man:
[tex]Area = A = (2\pi r)(l)[/tex]
where,
r = radius of cylinder = 30 cm/2 = 15 cm = 0.15 m
l = length of cylinder = 170 cm = 1.7 m
Therefore,
[tex]A = 2\pi(0.15\ m)(1.7\ m)\\A = 1.6\ m^2[/tex]
Now, we will calculate the rate of heat loss:
[tex]Rate\ of\ Heat\ Loss = hA\Delta T[/tex]
where,
h = convective heat tranfer coefficient = 15 W/m²K
ΔT = Temperature difference = 34°C - 20°C = 14°C
Therefore,
[tex]Rate\ of\ Heat\ Loss = (15\ W/m^2K)(1.6\ m^2)(14\ K)\\[/tex]
Rate of Heat Loss = 336 W
The stagnation chamber of a wind tunnel is connected to a high-pressure airbottle farm which is outside the laboratory building. The two are connectedby a long pipe of 4-in inside diameter. If the static pressure ratio between thebottle farm and the stagnation chamber is 10, and the bottle-farm staticpressure is 100 atm, how long can the pipe be without choking? Assumeadiabatic, subsonic, one-dimensional flow
This question is not complete, the complete question is;
The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005
Answer:
the length of the pipe is 11583 in or 965.25 ft
Explanation:
Given the data in the question;
Static pressure ratio; p1/p2 = 10
friction coefficient f = 0.005
diameter of pipe, D =4 inch
first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so
4fL[tex]_{max}[/tex] / D = 57.915
we substitute
(4×0.005×L[tex]_{max}[/tex]) / 4 = 57.915
0.005L[tex]_{max}[/tex] = 57.915
L[tex]_{max}[/tex] = 57.915 / 0.005
L[tex]_{max}[/tex] = 11583 in
Therefore, the length of the pipe is 11583 in or 965.25 ft
6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The temperature inside the refrigerator is 38°F. Room temperature is 72°F. [answer in BTUs]
Answer:
The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU
Explanation:
The given parameters are;
The duration of the heat transfer, t = 24 hours = 86,400 seconds
The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²
The material of the insulator in the door = Cellulose fiber
The thickness of the insulator in the door, d = 2.0" = 0.0508 m
The temperature inside the fridge = 38° F = 276.4833 K
The temperature of the room = 78°F = 298.7056 K
The thermal conductivity of cellulose fiber = 0.040 W/(m·K)
By Fourier's law, the heat flow through a by conduction material is given by the following formula;
[tex]\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}[/tex]
[tex]Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t[/tex]
Therefore, we have;
[tex]Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522[/tex]
The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU
A storage tank is purged continuously with a stream of nitrogen. The purgestream leaving the tank is saturated with the product stored in the tank. A majorpart of the product lost in the purge could be recovered by installing a scrubbingtower to absorb the product in a solvent. The solution from the tower could befed to a stage in the production process, and the product and solvent recoveredwithout significant additional cost. A preliminary design of the purge recoverysystem has been made. It would consist of:1. A small tower 0.5 m diameter, 4 m high, packed with 25 mm ceramic saddles,packed height 3 m.2. A small storage tank for the solution, 5 m3 capacity.3. The necessary pipe work, pump, and instrumentation.All materials of construction, carbon steel.Using the following data, evaluate whether it would be economical to install therecovery system:1. cost of product £5 per kg,2. cost of solvent 20 p/kg,3. additional solvent make-up 10 kg/d,4. current loss of product 0.7 kg/h,5. anticipated recovery of product 80 per cent,6. additional service(utility) costs, negligible.Other operating costs will be insignificant.
Answer:
It is economical
Explanation:
Given data for determining
cost of product = £5 per kg
cost of solvent = £20 per /kg
additional solvent makeup = 10 kg/d
current loss of product = 0.7 kg/h
anticipated recovery of product = 80%
additional service ( utility cost ) = negligible
Determine if it is economical to install the recovery system
cost of product = £5 per kg
cost of solvent = £20 per /kg
additional solvent required = 10 kg / d
cost of additional solvent in a day = £200 / day ( £20 per kg * 10 kg / d )
product lost = 0.7 kg/h
product recovered = 0.7 * 80% = 0.56 kg/h
therefore product recovered in a day = 0.56 * 24 = 13.44 kg / day
Next :
calculate income generated from product recovery
13.44 kg/day * £20 /kg = £268.88 / day
Determine if it is economical
Income - cost
= £268.88 - £200
= £68.88 ( hence it is economical )
A completely mixed high rate-activated sludge plant with recycletreats 17,500 m3/day of industrial wastewater. The influent to theactivated sludge reactor has a BOD5 of 1000 mg/L. It is desired toreduce the influent BOD5 to 120 mg/L, prior to discharge to a municipalsewer. Pilot plant analysis gave the following results: mean cellresidence time = 6 d, MLSS concentration in reactor = 5500 mg/L,Y = 0.6 kg/kg, kd = 0.03 d–1. Determine the following:________.
a. The hydraulic retention time and volume of the activated sludge
b. The wolumetric loading rate in kg BOD,/m- day to the reactor.
c. The F/M ratio in the reactor
The mass and valume of solids wasted each day, at an underflow solids concentration, Xu = 10,000 mg/L.
e. The sludge recirculation ratio
Answer:
a) 0.488 days ≈ 11.72 hour , 8542.38 m^3
b) 2.05 kg/m^3.day
c) 0.37 day^-1
d)
i) volume of wasted solid = 783 m^3/day
ii) mass of wasted solid = 7830 kg/day
e) 1.12
Explanation:
Given data :
Flow ( Q ) = 17500 m^3/day
BOD 5 ( influent ) = 1000 mg/L
BOD 5 ( efficient ) = 120 mg/L
mean cell residence time ( ∅ c ) = 6-day
MLSS concentration ( X ) = 5500 mg/L
a) Determine the hydraulic retention time and volume of the activated sludge
i) Hydraulic retention time ( OH ) we will use the relation below
X = [tex]\frac{OcY(sin-s)}{OH(1 + Kdθc)}[/tex]
5500 = [tex]\frac{6*0.6(1000-120)}{OH(1 + 0.03*6)}[/tex] . therefore OH = 0.488 days ≈ 11.72 hour
ii) volume of the activated sludge using the relation below
OH = V / Q
V = OH * Q
= 0.488 * 17500
= 8542.38 m^3
b) Determine the volumetric loading rate in kg BOD,/m- day to the reactor
loading rate = ( Q * Sin ) / V
= ( 17500 * 1000 ) / 8542.38
= 2.05 kg/m^3.day
c) Determine The F/M ratio in the reactor
F/M ratio : F/M = ( Q .sin ) / VX
= ( 17500 * 1000 ) / ( 8542.37 * 5500)
= 0.37 day^-1
d) Determine The mass and volume of solids wasted each day, at an underflow solids concentration
i) volume of wasted solid = 783 m^3/day
ii) mass of wasted solid = 7830 kg/day
attached below is the remaining part of the solution
e) Sludge recirculation ratio
= 1.12
attached below is the remaining part of the solution