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3 years ago

13

Given the following Grammar and the right sentential form draw a parse tree and show the phrases, simple phrases and handle.S →

AbB | bAcA → Ab | aBBB → Ac | cBb | ca) baabccBbc b) ccBbcbbbB c)baBcBbbc

2 answers:

IRISSAK [1]3 years ago

7 0

Answer:

<u>Answer is written in attached document.</u>

Shtirlitz [24]3 years ago

4 0

Answer:

Find the answer in the attached in the order

Option a, Option c and Option b

Explanation:

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A company, studying the relation between job satisfaction and length of service of employees, classified employees into three le

Wewaii [24]

Answer:

Below see details

Explanation:

A) It is attached. Please see the picture

B) First to calculate the overall mean,

μ=65∗25/75+80∗25/75+95∗25/75

μ=65∗25/75+80∗25/75+95∗25/75 = 80

Next to calculate E(MSTR) = σ2+(1/r−1) ∑ni(μi−μ)^2 = 5634

And E(MSE) = σ^2= 9

C) Yes, it is substantially large than E(MSE) in this case.

D) If we sampled 25 employees from each group, we are likely to get a F statistics to indicate differences of job satisfactions among three types of length of service of employees.

3 0

3 years ago

1: asha started abusness with 30.000

svetoff [14.1K]

Answer:

Explanation:adrive with visual acutity of 20/30 can just decipher asing adistance 20ft from asing determine the maximum destance from the sing which drivers with the flowing visual acuities will able to see the same sing 20/15 20/50

4 0

3 years ago

A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque

Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

τ = 16*T / pi*d^3

T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

T_c = τ_c*pi*d_c^3 / 16

T_c = 12*1000*pi*1^3 / 16

T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

T_s = τ_s*pi*d_s^3 / 16

T_s = 15*1000*pi*0.25^3 / 16

T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

T = min ( 2356.2 , 46.02 )

T = 46.02 lb-in

6 0

3 years ago

A concentrated load P is applied to the upper end of a 1.47-m-long pipe. The outside diameter of the pipe is D = 112 mm and the

myrzilka [38]

Answer:

Pmax = 38251.73 N

Explanation:

Given info

L = 1.47 m

D = 112 mm ⇒ R = D/2 = 112/2 mm = 56 mm

d = 101 mm ⇒ r = D/2 = 101/2 mm = 50.5 mm

a) We can apply the following equation in order to get Q (First Moment of Area):

Q = 2*(A₁*y₁-A₂*y₂)

where

A₁ = π*R² = π*(56 mm)² = 3136 π mm²

y₁ = 4*R/(3*π) = 4*56/(3*π) mm = 224/(3*π) mm

A₂ = π*r² = π*(50.5 mm)² = 2550.25 π mm²

y₂ = 4*r/(3*π) = 4*50.5/(3*π) mm = 202/(3*π) mm

then

Q = 2*(3136 π mm²*224/(3*π) mm-2550.25 π mm²*202/(3*π) mm)

⇒ Q = 62437.833 mm³

b) If τallow = 83 MPa = 83 N/mm²

P = ?

We can use the equation

τ = V*Q / (t*I) ⇒ V = τ*t*I / Q

where

t = D - d = 112 mm - 101 mm = 11 mm

I = (π/64)*(D⁴-d⁴) = (π/64)*((112 mm)⁴-(101 mm)⁴) = 2615942.11 mm⁴

Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3)*(R³-r³)

⇒ Q = (4/3)*((56 mm)³-(50.5 mm)³) = 62437.833 mm³

then we have

V = (83 N/mm²)*(11 mm)*(2615942.11 mm⁴) / (62437.833 mm³)

⇒ V = 2942.255 N

Finally Pmax = V = 38251.73 N

6 0

3 years ago

Socket Programming: (30 points) Use Python TCP socket to implement an application with client-server architecture. In this appli

ololo11 [35]

Answer:

Explanation:

Run the code given in text file following instructions.

8 0

3 years ago