2143.57333 whole number
1 answer:
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Answer:
Step-by-step explanation:
Given that there is a t distribution with 7 degrees of freedom.
P(-1.29 < t < 1.29)=1-0.2380
=0.7620
b) Now there is a different t distribution with 18 df.
When
c=-1.733
Answer:
The minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.
Step-by-step explanation:
Consider the provided information.
The length of the half of the major axis is 2,876,769,540 kilometers, and the eccentricity is 0.0444.
The eccentricity (e) of an ellipse is the ratio of the distance from the center to the foci (c) and the distance from the center to the vertices (a).
Substitute a = 2,876,769,540 and e = 0.0444 in above formula and solve for c.
Minimum distance of Uranus from the sun is:
Hence, the minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.