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Gennadij [26K]
4 years ago
15

Assuming the point estimate in Problem 6.36 is the true population parameter, what is the probability that a particular assay, w

hen expressed in the log scale to the base 2, is no more than 1.5 log units off from its true mean value for a particular woman?

Engineering
1 answer:
sp2606 [1]4 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability is 0.958

Explanation:

The explanation is shown on the second and third uploaded image

You might be interested in
2. The moist weight of 0.1 ft3 of soil is 12.2 lb. If the moisture content is 12% and the specific gravity of soil solids is 2.7
adell [148]

The answers to dry unit weight, void ratio, porosity, degree of saturation, volume occupied by water are respectively;

γ_d = 108.93 lb/ft³; e = 0.56; n = 0.36; S = 0.58; V_w = 0.021 ft³

<h3>Calculation of Volume and Weight of soil</h3>

We are given;

Moist weight; W = 12.2 lb

Volume of moist soil; V = 0.1 ft³

moisture content; w = 12% = 0.12

Specific gravity of soil solids; G_s = 2.72

A) Formula for dry unit weight is;

γ_d = γ/(1 + w)

where γ_w is moist unit weight as;

γ_w = W/V

γ_w = 122/0.1 = 122 lb/ft³

Thus;

γ_d = 122/(1 + 0.12)

γ_d = 108.93 lb/ft³

B) Formula for void ratio is;

e = [(G_s * γ_w)/γ_d] - 1

e = [(2.72 * 122)/108.93] - 1

e = 0.56

C) Formula for porosity is;

n = e/(1 + e)

n = 0.56/(1 + 0.56)

n = 0.36

D) Formula for degree of saturation is;

S = (w * G_s)/e

S = (0.12 * 2.72)/0.56

S = 0.58

E) Volume occupied by water is gotten from;

V_w = S*V_v

where;

V_v is volume of voids = nV

V_v = 0.36*0.1

V_v = 0.036 ft³

Thus;

V_w = 0.58 * 0.036

V_w = 0.021 ft³

Read more about Specific Gravity of Soil at; brainly.com/question/14932758

4 0
3 years ago
The hull of a vessel develops a leak and takes on water at a rate of 57.5 gal/min. When the leak is discovered the lower deck is
leva [86]

Answer:

It will be around 146,27 min since the pump is turned on until the deck is clear of the water.

Explanation:

When the leak is discovered and the pump is turned on, the lower deck is already submerged and the leak is not fixed; then, in order to have the deck clear of water, the bilge pump has to remove the <em>accumulated water </em>(V_{0}) and the <em>water that is taking on</em> (r_{in}*t) through the leak. We can represent this mathematically as follow:

V_{0} +r_{in} *t-r_{out}*t=0  <em>Equation 1</em>

Where:

V_{0}: is the accumulated water when the leak was discovered

r_{in}: is the takes on rate through the leak = 57.5 gal/min

r_{out}: is the removing rate of the bilge pump = 73.8 gal/min

t= is the time since the pump is turned on until the deck is clear of water.

To calculate the accumulated water (V_{0}), we will model the lower deck as a flat-bottomed container with a bottom surface area of 510 ft^{2} and straight vertical sides. Knowing that the level submerged is 7.5 inches, and performing the corresponding unit conversions, we obtain:

V_{0}= bottom surface area * lever submerged

V_{0}= 510ft^{2}*7.5 in*\frac{1ft}{12in}=318.75 ft^{3}*7.48\frac{gal}{1ft^{3}}=2384.25 gal <em>Equation 2</em>

Solving equation 1 for time (t), and replacing the value obtained in equation 2, we get:

t=\frac{V_{0}}{(r_{out}-r_{in})} =\frac{2384.25 gal}{(73.8-57.5)gal/min}=146,27 min

8 0
3 years ago
i know this isnt suppose to be here or any of that but this needs to be stop any where...ok now that i have your attention every
AfilCa [17]
I know this isnt suppose to be here or any of that but this needs to be stop any where...ok now that i have your attention everyone needs to know this people all over the world are burning bibles LETS STOP THIS because that is so disrepecful that people are doing that to the bible and thats hurting jesus and gods heart LETS STOP THIS BURNING BIBLE STUFF pls copy and paste this and share the word so we can stop this.
6 0
3 years ago
Read 2 more answers
A balanced three phase load is supplied over a three-phase , 60 hz, transmission line with each line have a series impedance of
posledela

Answer:

Explanation:

Given a three-phase system

Frequency f=60Hz

Line impedance Z= 12.84 + j72.76 Ω

Then,

The resistance is R=12.84Ω

And reactance is X=72.76Ω

Z=√(12.84²+72.76²)

Z=73.88

Angle = arctan(X/R)

Angle = arctan(72.76/12.84)

Angle=80°

Then, Z=73.88 < 80° ohms

Load voltage is 132 kV

Load power P=55 MWA

Power factor =0.8lagging

the relation between the sending and receiving end specifications are given using ABCD parameters by the equations below.

Vs = AVr + BIr

Is = CVr + DIr

Where

Vs is sending Voltage

Vr is receiving Voltage

Is is sending current

Ir is receiving current

A is ratio of source voltage to received voltage A=Vs/Vr when Ir=0

B is short circuit resistance

B= Vs/Ir when Vr=0

C is ratio of source current to received voltage C=Is/Vr when Ir=0

D is ratio of source current to received current D=Is/Ir when Vr=0

Now,

The load at 55MVA at 132kV (line to line)

Therefore, load current is

Ir= P/V√3

Ir=55×10^6/(132×10^3×√3)

Ir=240.56 Amps

It has a power factor 0.8 lagging

PF=Cosθ

0.8=Cosθ

θ=arcCos(0.8)

θ=36.87°

Therefore, Ir=240.56 <-36.87°

Vr=V/√3

Vr=132/√3

Vr=76.21 kV. Phase voltage

Vr= 76210 < 0° V

For series impedance,

Using short line approximation

Vs = Vr + IrZ

Vs = 76210 < 0° + (240.56 <-36.87° × 73.88 < 80°)

Using calculator

Vs=76210<0° + 17772.5728<(-36.87°+80°)

Vs=76210<0° + 17772.5728<43.13°

Vs=89970.67<7.7°

Also

Is = Ir = 240.56 <-36.87° Amps

Therefore, the ABCD parameters is

A=Vs/Vr

A= 89970.67 <7.7° / 76210 <0°

A=1.181 <7.7-0

A=1.18 <7.7° no unit

B = Vs/Ir

B = 89970.67 < 7.7° / 240.56 <-36.87°

B = 347.01 < 7.7+36.87

B= 347.01 < 44.57° Ω

C= Is/Vr = 240.56 <-36.87° / 76210 < 0°

C= 0.003157 <-36.87-0

C= 3.157 ×10^-3 < -36.87° /Ω

C= 3.157 ×10^-3 < -36.87° Ω~¹

D= Is/Ir

Since Is=Ir

Then, D = 1 no unit.

8 0
3 years ago
Saturated water at 25C enters a tank through a 10-cm diameter pipe at a velocity of 3 m/s. At the bottom of the tank, water exi
harina [27]

Answer:

see answer attached

Explanation:

That velocity for part b should be 3 m/s and I have calculated the potential energy with respect to the base of the tank.

6 0
4 years ago
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