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nataly862011 [7]
3 years ago
6

Certain pieces made by an acoustic lathe are subject to three kinds of defects X,Y,Z. A sample of 100 pieces was inspected with

the following results:
2.1% had type X defect.
2.4% had type Y .
2.8 had typre Z
0.3 had both X and Y
0.4 had both X and Z
0.6 had both Y and Z
Then Find:
A. What percent had none of these defects.
B. What percent had at least one of these defects.
C. What percent were free of type X and/ or type Y defects?
D. What percent had not more than one of these defects?
Engineering
1 answer:
Sonja [21]3 years ago
8 0

Explanation:

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An engineer is working with archeologists to create a realistic Roman village in a museum. The plan for a balance in a marketpla
NeTakaya

Answer:

The minimum volume requirement for the granite stones is 1543.64 cm³

Explanation:

1 granite stone weighs 10 denarium

100 granted stones will weigh 1000 denarium

1 denarium = 3.396g

1000 denarium = 3396g.

But we're told that 20% of material is lost during the making of these stones.

This means the mass calculated represents 80% of the original mass requirement, m.

80% of m = 3396

m = 3396/0.8 = 4425 g

This mass represents the minimum mass requirement for making the stones.

To now obtain the corresponding minimum volume requirement

Density = mass/volume

Volume = mass/density = 4425/2.75 = 1543.64 cm³

Hope this helps!!!

3 0
3 years ago
A certain piece of property is assessed at $150,000. If the tax rate is $2.50 per $100, what is the tax on this property?
stiks02 [169]

Answer:

The tax on this property is 3750 dollars

Explanation:

Given

Tax on per $100 is $2.50

Tax on every $1 is \frac{2.5}{100} = 0.025 dollars

Tax on property of value $150,000 is

150,000 * 0.025 = 3750 dollars

The tax on this property is 3750 dollars

7 0
3 years ago
what is the method of the slope stabilization, and how many type of method of the stabilization, please help explain the types o
Alik [6]

Answer:

1.Plant Grass and Shrubs. Grass and shrubs are very effective at stopping soil erosion. ...

2.Use Erosion Control Blankets to Add 3.Vegetation to Slopes. ...

4.Build Terraces. ...

5.Create Diversions to Help Drainage

5 0
3 years ago
Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is
Leona [35]

Answer:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

dQ = dU - dW

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) \int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW

As per work definition:

dW = F*dr

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

dW = PA*dx

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

dW = - P*dV

So the third integral in equation (1) is:

\int\limits^1_2 {- P} \, dV

Considering the gas as ideal, the pressure can be calculated as P = \frac{n*R*T}{V}, so:

\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Replacion this and solving equation (1) between state 1 and 2:

\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}

The internal energy depends only on the temperature of the gas, so there is no internal energy change U_{2} - U_{1} = 0, so the heat exchanged to the system equals the work done by the system:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

4 0
3 years ago
10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de
Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span  20.83\E ft

at mid span = 1.23\E ft

shear stress  7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

\delta = \frac{wa^2b^2}{3EI}

= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}

         =\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}

         = 20.83\E ft

at mid span

\delta = \frac{wl^3}{48EI}

= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}

\delta = 1.23\E ft

shear stress

\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi

6 0
3 years ago
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