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Simora [160]
3 years ago
5

Neptune is an average distance of 4.5 × 10^9 km from the Sun. Estimate the length of the Neptunian year using the fact that the

Earth is 1.50 × 10^8 km from the Sun on average.
Mathematics
1 answer:
zlopas [31]3 years ago
3 0

Answer:

Answer:

164.32 earth year

Step-by-step explanation:

distance of Neptune, Rn = 4.5 x 10^9 km

distance of earth, Re = 1.5 x 10^8 km

time period of earth, Te = 1 year

let the time period of Neptune is Tn.

According to the Kepler's third law

T² ∝ R³

\left ( \frac{T_{n}}{T_{e}} \right )^{2}=\left ( \frac{R_{n}}{R_{e}} \right )^{3}

\left ( \frac{T_{n}}{1} \right )^{2}=\left ( \frac{4.5\times10^{9}}}{1.5\times10^{8}}} \right )^{3}

Tn = 164.32 earth years

Thus, the neptune year is equal to 164.32 earth year.

Step-by-step explanation:

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Answer:

a) P [ X < 24 mm ] = 0,3015          or     P [ X < 24 mm ] =  30,15 %

b) P [ X > 32 mm ]  = 0,1251          or        P [ X > 32 mm ]  = 12,51 %

c) P [  25 < X < 30 ] = 0,4964      or     P [  25 < X < 30 ] = 49,64 %

d) z(s) = 0,84

Step-by-step explanation:

Normal Distribution    N (  μ₀  ;   σ )   is  N ( 26,5 ; 4,8 )

a) P [ X < 24 mm ] = ( X - μ₀ ) / σ

P [ X < 24 mm ] = (24 - 26,5)/ 4,8 = - 0,5208 ≈ - 0,52

P [ X < 24 mm ] = - 0,52  

And from z-table we find area for z score

P [ X < 24 mm ] = 0,3015          or     P [ X < 24 mm ] =  30,15 %

b)P [ X > 32 mm ]  =  1  - P [ X < 32 mm ]

P [ X < 32 mm ]  = ( 32 - 26,5 ) / 4,8

P [ X < 32 mm ]  = 5,5/4,8  =  1,1458 ≈ 1,15

P [ X < 32 mm ]  = 1,15

And from z-table  we get

P [ X < 32 mm ]  = 0,8749

Then:

P [ X > 32 mm ]  =  1  -  0,8749

P [ X > 32 mm ]  = 0,1251          or        P [ X > 32 mm ]  = 12,51 %

c) P [  25 < X < 30 ] = P [ X < 30 ] - P [ X < 25 ]

P [ X < 30 ]  = 30 - 26,5 / 4,8   =  0,73

From  z-table    P [ X < 30 ]  =  0,7673

P [ X < 25 ] = 25 - 26,5 / 4,8  = - 0,3125  ≈  - 0,31

From z-table  P [ X < 25 ] =  0,2709

Then

P [  25 < X < 30 ] =  0,7673 -  0,2709

P [  25 < X < 30 ] = 0,4964      or     P [  25 < X < 30 ] = 49,64 %

d) If  20 %

z- score for 20% is from z-table

z(s) = 0,84

6 0
3 years ago
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