How thick a layer would Earth form as it wraps around the neutron star’s surface is: 6.67 10⁻³ m.
<h3>Density of the Neutron star</h3>
Density
ρ = m / V
Where:
ρ= density
m = mass of the planet 5.98 10²⁴ km
V =volume of the spherical layer
Volume of a sphere
Volume = 4/3 π r³
Mass = 1.5 = 1.5 1,991 10³⁰
Mass= 2.99 10³⁰ kg
Density:
ρ = 2.99 10³⁰ / [4/3 π (10 10³)³]
ρ is = 7.13 10 17 kg / m³
V = 5.98 10²⁴ / 7.13 10¹⁷
V = 8,387 10⁶ m³
Thickness of the layer
V = 4π r² e
e = V / 4π r
e = 8,387 10⁶ / [4π (10 10³)²]
e = 6.67 10⁻³ m
Inconclusion how thick a layer would Earth form as it wraps around the neutron star’s surface is: 6.67 10⁻³ m.
Learn more about Density of the Neutron star here:brainly.com/question/15700804
Answer:
b. The number of digits in a randomly selected row until a 3 is found.
Explanation:
A random variable often used in statistics and probability, is a variable that has its possible values as numerical outcomes of a random experiment or phenomenon. It is usually denoted by a capital letter, such as X.
In statistics and probability, random variables are either continuous or discrete.
1. A continuous random variable is a variable that has its possible values as an infinite value, meaning it cannot be counted.
2. A discrete random variable is a variable that has its possible values as a finite value, meaning it can be counted.
Also, any random variable that meets certain conditions defined in a research study.
Hence, an example of a geometric random variables is the number of digits in a randomly selected row until a 3 is found.
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