How many bit strings of length 12 contain exactly three "1s"
2 answers:
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Theoretically, the number of the vertex comes at the top must be equal for infinitely many numbers of trials, but it's a tedious job to roll the dice for a large number of times. Rolling the tetrahedron only 10 times will not predict the correct result.
Han suspects that this tetrahedron is weighted in some way to make 1 appear more often than the other numbers.
So, check his suspect experimentally with the help of the concept of center of mass. A fair tetrahedron must have the center of mass at the center of the body.
The given tetrahedron has 4 equilateral triangular faces, so at first, measure all the sides of triangular faces and make sure that all are having the same length.
Then, conduct an experiment to check the location of the center of mass.
Hang the tetrahedron by attaching a thin string at the vertex 1 as shown in the figure and observe the base surface ( triangle 234) whether it is parallel to the ground or not.
Now, repeat the same by hanging it with vertices 2, 3, and 4 and observe the base surface.
For the center of mass to be located at the center of the tetrahedron, the base must be parallel to the ground (flat ground).i.e the hanging string must be perpendicular to the base for all the four cases.
If this is so, then the tetrahedron is a fair tetrahedron otherwise not.
The addison see to the horizon at 2 root 2mi.
We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.
We have to find the how much farther can addison see to the horizon
<h3>Which equation we get from the given condition?</h3>Where, we have
d- the distance they can see in thousands
h- their eye-level height in feet
For Kaylib
For Addison h=85(1/3)
Subtracting both distances we get
Therefore, the addison see to the horizon at 2 root 2mi.
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