A frequenter of a pub had observed that the new barman poured in average 0.47 liters of beer into the glass with a standard deviation equal to 0.09 liters instead of a half a liter with the same standard deviation. The frequenter had used a random sample of 47 glasses of beer in his experiment. Consider the one-sided hypothesis test for volume of beer in a glass: H0: u=0.5 against H1: u<0.5. Determine the P-value of this test.
Round your answer to four decimal places (e.g. 98.7654).
Answer:
P-value = 0.0011
Explanation:
Formula for the test statistic is;
z = (x¯ - μ)/(σ/√n)
We have;
Sample mean;x¯ = 0.47
Population mean; μ = 0.5
Standard deviation; σ = 0.09
Sample size; n = 47
Thus;
z = (0.46 - 0.5)/(0.09/√47)
z = -3.05
From z-distribution table attached, the p-value corresponding to z = -3.05 is;
P = 0.00114
To four decimal places gives;
P-value = 0.0011
The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 5.5 x 3^106 Btu/h. The combustion efficiency of the boiler is measured to be 0.7 by a hand-held flue gas analyzer. After tuning up the boiler, the combustion efficiency rises to 0.8. The boiler operates 4200 hours a year intermittently. Taking the unit cost of energy to be $4.35/10^6 Btu, determine the annual energy and cost savings as a result of tuning up the boiler.
Answer:
Energy Saved = 6.93 x 10⁹ Btu
Cost Saved = $ 30145.5
Explanation:
The energy generated by each boiler can be given by the following formula:
[tex]Annual\ Energy = (Heat\ In)(Combustion\ Efficiency)(Operating\ Hours)[/tex]
Now, the energy saved by the increase of efficiency through tuning will be the difference between the energy produced before and after tuning:
[tex]Energy\ Saved = (Heat\ In)(Efficiency\ After\ Tune - Efficiency\ Before\ Tune)(Hours)[/tex][tex]Energy\ Saved = (5.5\ x\ 3\ x\ 10^{6}\ Btu/h)(0.8-0.7)(4200\ h)[/tex]
Energy Saved = 6.93 x 10⁹ Btu
Now, for the saved cost:
[tex]Cost\ Saved = (Energy\ Saved)(Unit\ Cost)\\Cost\ Saved = (6.93\ x\ 10^{9}\ Btu)(\$4.35/10^{6}Btu)\\[/tex]
Cost Saved = $ 30145.5
QUESTION 4:
4.1
Name FOUR principles of kinetic friction
Answer:
The force of friction always acts in a direction, opposite to that in which the body is moving.
The magnitude of kinetic friction bears a constant ratio to the normal reaction between the two surfaces. ...
For moderate speeds, the force of friction remains constant.
Answer:
Explanation:Kinetic friction is a force that acts between moving surfaces. An object that is being moved over a surface will experience a force in the opposite direction as its movement. The magnitude of the force depends on the coefficient of kinetic friction between the two kinds of material.
Another name for a load-center distribution system is a A. primary radial system. B. complex radial system. C. split-radial system. D. dual-radial system.
Answer:
A
Explanation:
Primary radial system
The alternator must be operated with the battery disconnected or with the terminals at the back of the alternator
disconnected.
True or false
Answer:
true
Explanation:
Build a 32-bit accumulator circuit. The circuit features a control signal inc and enable input en. If en is 1 and inc is 1, the circuit increments the stored value by an amount specified by an input A[31:0] on the next clock cycle. If en is 1 and inc is 0 the circuit decrements the stored value by the amount specified in the input A on the next clock cycle. If en is 0, the circuit simply stores its current value without modification. The circuit has the following interface:______.
Input clock governs the state transitions in the circuit upon each rising edge.
Input clear is used as a synchronous reset for the stored value.
Input inc controls whether the value stored is to be incremented or decremented.
Input en is a control signal that activates the values increment/decrement
Input A determines how much to increment or decrement by
Output value is a 32-bit signal that can be used to read the stored value at any time.
* Note: Use any combination of combinational or sequential logic. It may be helpful to look into D Flip Flops and Registers.
Sorry need.points I'm new
You have available three blocks of different material, at various temperatures. They are, respectively, a 2 kg block of iron at 600 K, a 3 kg block of copper at 800 K and a 10 kg block of granite at 300 K. The heat capacities for the three materials are 0.460 (iron), 0.385 (copper), and 0.790 (granite), in kj/(kg*K), all independent of temperature. For solids, the heat capacities at constant pressure and constant volume can be assumed to be equal, Cp=Cv. what is the minimum temperature that could be obtained in any one of the block? what is the maximum temperature that could be obtained? no heat or work interactions with the enviroment are allowed.
Answer:
max temp = 711.32 k
mini temp = 331.29 k
Explanation:
Given data:
2kg block of Iron : temperature = 600k , C = 0.460 kJ/kgk
3 kg block of copper : temp = 800k , C = 0.385 KJ /kgk
10 kg block of granite : temp = 300k , C = 0.790 KJ/kgk
Cp = Cv at constant pressure and constant volume
Determine the minimum temperature that is obtained in any one of the block
considering the heat transfer equation
Q = mC ( T2 - T1 )
attached below is a detailed solution of the problem
A contractor excavates 10,000 m3 soil at moist unit weight of 17.5 kN/m3 and moisture content of 10% from a borrow pit and transports it to a project site. The project has an area of 20,000 m2 to be filled with this compacted soil. If the required dry unit weight and moisture content of the compacted soil are 18.3 kN/m3 and 12.5% (assume there is no soil loss during transportation and compaction), what is the thickness of the compacted soil and how much water needs to be added?
Answer:
Part A
The thickness of the compacted soil is approximately 4.3467 × 10⁻¹ m
Part B
The weight of water to be added is approximately 19886.[tex]\overline{36}[/tex] kN, the volume of the water added is approximately 2,027.77 m³
Explanation:
The parameters of the soil are;
The volume of sol the excavator excavates, [tex]V_T[/tex] = 10,000 m³
The moist unit weight, W = 17.5 kN/m³
The moisture content = 10%
The area of the project, A = 20,000 m²
The required dry unit weight = 18.3 kN/m³
The required moisture content = 12.5%
Part A
Therefore, we have;
The moist unit weight = Unit weight = ([tex]W_s[/tex] + [tex]W_w[/tex])/[tex]V_T[/tex]
The moisture content, MC = 10% = ([tex]W_w[/tex]/[tex]W_s[/tex]) × 100
∴ [tex]W_w[/tex] = 0.1·[tex]W_s[/tex]
∴ The moist unit weight = 17.5 kN/m³ = ([tex]W_s[/tex] + 0.1·[tex]W_s[/tex])/(10,000 m³)
1.1·[tex]W_s[/tex] = 10,000 m³ × 17.5 kN/m³ = 175,000 kN
[tex]W_s[/tex] = 175,000 kN/1.1 = 159,090.[tex]\overline{09}[/tex] kN
For the required soil, we have;
The required dry unit weight = 18.3 kN/m³ = [tex]W_s[/tex]/[tex]V_T[/tex] = 159,090.[tex]\overline{09}[/tex] kN/[tex]V_T[/tex]
[tex]V_T[/tex] = 159,090.[tex]\overline{09}[/tex] kN/(18.3 kN/m³) ≈ 8,693.4923 m³
The total volume of the required soil ≈ 8,693.4923 m³
Volume [tex]V_T[/tex] = Area, A × Thickness, d
∴ d = [tex]V_T[/tex]/A
d = 8,693.4923 m³/(20,000 m²) ≈ 4.3467 × 10⁻¹ m
The thickness of the compacted soil ≈ 4.3467 × 10⁻¹ m
Part A
The moisture content, MC = 12.5% = ([tex]W_w[/tex]/[tex]W_s[/tex]) × 100
[tex]W_w[/tex] = [tex]W_s[/tex] × MC/100 = 159,090.[tex]\overline{09}[/tex] kN × 12.5/100 = 19886.[tex]\overline{36}[/tex] kN
The weight of water to be added, [tex]W_w[/tex] = 19886.[tex]\overline{36}[/tex] kN
Where the density of water, ρ = 9.807 kN/m³
Therefore, we have;
The volume of water, V = [tex]W_w[/tex]/ρ
∴ V = 19886.[tex]\overline{36}[/tex] kN/(9.807 kN/m³) ≈ 2027.77 m³
The volume of water, V ≈ 2027.77 m³
pls help me it’s due today
Answer:
C. 14.55
Explanation:
12 x 10 = 120
120 divded by 10 is 12
so now we do the left side
7 x 3 = 21 divded by 10 is 2
so now we have 14
and the remaning area is 0.55
so 14.55