Which of the following color schemes is composed of hues next to eachother on the color wheel ?

In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length

masya89 [10]

Answer:

Time period = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration $9.4 \times 10^{-5} kg/s$

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration $= 9.4\times 10^{-5} \times 62 = 582.8\times 10{-5} kg/s$

total volume of air infiltration $= \frac{582.8\times 10{-5}}{1.156} = 5.04\times 10^{-3} m^3/s$

Time period $= \frac{210}{5.04\times 10^{-3}} = 41654.08 s$

3 0

3 years ago

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

Ksenya-84 [330]

Answer:

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

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

36

1

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

36

2

=

18

1

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

36

3

=

12

1

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

36

4

=

9

1

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

36

5

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

36

6

=

6

1

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

36

5

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

36

4

=

9

1

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

36

3

=

12

1

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

36

2

=

18

1

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

36

1

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

i

⋅P(X

i

)

=2×

36

1

+3×

18

1

+4×

12

1

+5×

9

1

+6×

36

5

+7×

6

1

+8×

36

5

+9×

9

1

+10×

12

1

+11×

18

1

+12×

36

1

=

18

1

+

6

1

+

3

1

+

9

5

+

6

5

+

6

7

+

9

10

+1+

6

5

+

18

11

+

3

1

=7

E(X

2

)=∑X

i

2

⋅P(X

i

)

=4×

36

1

+9×

18

1

+16×

12

1

+25×

9

1

+36×

36

5

+49×

6

1

+64×

36

5

+81×

9

1

+100×

12

1

+121×

18

1

+144×

36

1

=

9

1

+

2

1

+

3

4

+

9

25

+5+

6

49

+

9

80

+9+

3

25

+

18

121

+4

=

18

987

=

6

329

=54.833

Then, Var(X)=E(X

2

)−[E(X)]

2

=54.833−(7)

2

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

=

5.833

=2.415

4 0

2 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

A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind

Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

$N_F=n_F\times \frac{N_S}{n_S}$ here,

$n_F$ = the number of unseeded faults = 6

$N_S$ = number of seeded faults = 30

$n_s$ = number of seeded faults found = 15

So NF will be calculated as,

$N_F=6\times \frac{30}{15}=12$

And the estimate of faults remaining is $N_F-n_F$ = 12 - 6 = 6

8 0

3 years ago

4. A banking system provides users with several services:

VLD [36.1K]

A diagram showing a use case diagrams for these requirements is given in the image attached.

<h3>What is system Case diagram?</h3>

A use case diagram is known to be a kind of graphical illustration of a users in terms of their various possible association or interactions within any given system.

A use case diagram in banking can be used to prepare, depict and also to know all the functional requirements of the banking system.

Therefore, Give the use case specification for the banking system services and paying a bill online is given in the image attached.

Learn more about Case diagram from

brainly.com/question/12975184

#SPJ1

4 0

2 years ago