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2 years ago

Find E[x] when x is sum of two fair dice?

Engineering

1 answer:

Ksenya-84 [330]2 years ago

4 0

Answer:

When two fair dice are rolled, 6×6=36 observations are obtained.

P(X=2)=P(1,1)=

P(X=3)=P(1,2)+P(2,1)=

P(X=4)=P(1,3)+P(2,2)+P(3,1)=

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)=

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)=

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=

P(X=10)=P(4,6)+P(5,5)+P(6,4)=

P(X=11)=P(5,6)+P(6,5)=

P(X=12)=P(6,6)=

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

⋅P(X

)

=2×

+3×

+4×

+5×

+6×

+7×

+8×

+9×

+10×

+11×

+12×

+1+

E(X

)=∑X

⋅P(X

)

=4×

+9×

+16×

+25×

+36×

+49×

+64×

+81×

+100×

+121×

+144×

+5+

+9+

121

987

329

=54.833

Then, Var(X)=E(X

)−[E(X)]

=54.833−(7)

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

5.833

=2.415

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Answer:

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At resonance, spring constant, k = mw² ( where w = 2πf), when spring constant, k = centripetal force ( F = mw²r).

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3 years ago

A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the ta

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Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

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as satellite is acting parallel to the earth therefore $\theta angle = 0$

and radial component of given velocity is zero

we have $h = r_o v_r_o = 6378+600 =6.97*10^6 m$

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

$\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)$

$GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s$

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solvingt for $\epsilon)$

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3 years ago

A system consists of N very weakly interacting particles at a temperature T sufficiently high so that classical statistical mech

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Answer:

the restoring force is = 3/4NKT

Explanation:

check the attached files for answer.

7 0

3 years ago

A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir

Olenka [21]

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

= 500 - 2(20)

= 460 mm

So, internal radius, r = 230 mm

= 0.23 m

Density of molten metal, ρ = $7.2 \ g/cm^3$

= $7200 \ kg/m^3$

The height of pouring cavity above parting surface is h = 300 mm

= 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

Area, A $=2 \pi r^2$

$=2 \pi (0.23)^2$

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$F=\rho g A h$

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3 0

2 years ago

A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the

ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0

2 years ago