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DaniilM [7]
3 years ago
9

My computer has a mass of 0.031080997078386 slug the Earth's surface.

Engineering
1 answer:
poizon [28]3 years ago
4 0

Answer:

The answer is "0.187 lbm and 1 lbf".

Explanation:

The mass = 0.031080997078386\  slug

Calculating mass on Mars:

\to m=m_g\frac{g}{g_e}

        =0.031080997078386 \times \frac{32.2}{5.35}\\\\=0.187 \ lbm

\to W=mg_e

        =0.187 \times 5.35\\\\=1 \ lbf

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Answer:

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

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

36

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

36

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

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

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​

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

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

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

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5

​

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

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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

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+9×

18

1

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12

1

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+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

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​

=

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​

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2

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2

=54.833−(7)

2

=54.833−49

=5.833

∴ Standard deviation =

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=

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5 0
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Inessa05 [86]

Answer:

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Explanation:

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