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Pepsi [2]
3 years ago
10

A tetrahedron is a triangular pyramid with 4 equilateral faces. Han has a box of tetrahedrons with each of the 4 vertices marked

with one of the numbers 1, 2, 3, and 4. He rolls one tetrahedron and makes note of the number on the vertex that points up. After rolling this tetrahedron 10 times, he notices that the number 1 appears five times. Han suspects that this tetrahedron is weighted in some way to make 1 appear more often than the other numbers. Using another tetrahedron from the box that he has confirmed is fair (all of the numbers tend to be rolled with equal frequency), explain a process Han could use to gather evidence to show whether this tetrahedron is also fair.

Mathematics
1 answer:
Daniel [21]3 years ago
8 0

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.

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Novay_Z [31]

Answer:

The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

Given:

Initial speed of debris is, s=120\ ft/s

The height 'h' of the debris above the ground is given as:

h(t)=-16t^2+120t

As per question, h(t)=56\ ft. Therefore,

56=-16t^2+120t

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0

Using quadratic formula to solve for 't', we get:

t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s

Therefore, the debris will reach a height of 56 ft twice.

When time t=0.5\ s during the upward journey, the debris is at height of 56 ft.

Again after reaching maximum height, the debris falls back and at t=7\ s, the height is 56 ft.

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