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3 years ago

Centered 7 meters above the ground, a Ferris wheel of radius 6 meters is rotating with angular speed 24 degrees per second.

Mathematics

1 answer:

borishaifa [10]3 years ago

4 0

24 degrees per second is 360 degrees in 15 seconds (pretty fast for a Ferris wheel!).

a) The distance above the ground can be modeled using a cosine function.
.. h(t) = 7 -6cos(2πt/15)

b) h(t) = 10
.. 10 = 7 -6cos(2πt/15)
.. (10 -7)/-6 = cos(2πt/15)
.. 2πt/15 = π ±π/3 +2kπ
. t = 7.5*((1 ±1/3) +2k) . . . . . . . . seconds; k = any integer

You are 10 meters above the ground at t=5 seconds and every 15 seconds thereafter, as well as at t=10 seconds and every 15 seconds thereafter.

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Consider the following game, played with three standard six-sided dice. If the player ends with all three dice showing the same

ddd [48]

Answer:

$P(A) = \frac{1}{36}$

$P(B) = \frac{15}{36}$

$P(U) = \frac{11}{36}$

Step-by-step explanation:

From the question we are told that

The number of dice is $n = 3$

Generally for all three dice to show the same number the second and the third dice must have the same outcome as the outcome of the first dice.

This means that the number of outcome the first die can have is 6 (6 sides), the number of outcome which the second and the third dice can have is 1 (they must match the first )

So the probability that all three dice show the same number on the first roll is mathematically represented as

$P(A) = \frac{6}{6} * \frac{1}{6} * \frac{1}{6}$

=> $P(A) = \frac{1}{36}$

Generally for two of the three dice show the same number after the first roll then the number of outcome for one of the dice would be is 6 , the number of outcome for another one must be 1(i.e it must match the first ) , the number of outcome for the remaining one must be 5 (i.e it can show any of the remaining 5 sides which the first and second dice are not showing )

Now the number of ways of selecting this 2 dice that show the same number from the 3 dice is mathematically represented as

$N = ^3C_2$

Here C stands for combination

$N = ^3C_2 = 6$

So the probability that exactly two of the three dice show the same number after the first roll is mathematically represented as

$P(B) = N \frac{1}{6} * \frac{5}{6}$

=> $P(B) = \frac{15}{36}$

Generally from the question we are told that if two dice match, the player re-rolls the die that does not match.

Now the probability that the die that did not match the first time will match the second time is

$P(E ) = \frac{1}{6}$

Generally if that one die does not show the same number in the second round , the probability that it will match in the third round is

$P(R) = \frac{5}{6} * \frac{1}{6}$

Generally the probability that he wins (i.e when all three are showing the same number ) and exactly two is showing the same number is mathematically represented as

$P(K) = P(E)* P(B) + P(R)* P(B)$

=> $P(K) = \frac{1}{6} * \frac{15}{36} + \frac{5}{6} * \frac{1}{6} * \frac{15}{36}$

=> $P(K)= \frac{165}{1296}$

So the probability that the player wins, conditioned on exactly two of the three dice showing the same number after the first roll is mathematically represented as

$P(U) = \frac{\frac{165}{1296}}{\frac{15}{36} }$

=> $P(U) = \frac{11}{36}$

6 0

3 years ago

Select the correct answer.

trasher [3.6K]

Answer:

this question is very unorganized please make it more clear, thanks :)

Step-by-step explanation:

4 0

2 years ago

What times is equal to 75

Ne4ueva [31]

Answer: 5 x 15= 75

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.

Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

$t=\sqrt{\frac{2h}{a}}$ ,where h=height of release

a=acceleration

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

Where I=moment of inertia

a for hoop

$a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}$

$a=\frac{g\sin \theta }{2}$

a for Uniform solid cylinder

$a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}$

$a=\frac{2g\sin \theta }{3}$

a for spherical shell

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}$

$a=\frac{3g\sin \theta }{5}$

a for Uniform Solid

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}$

$a=\frac{5g\sin \theta }{7}$

time taken will be inversely proportional to the square root of acceleration

$t_1=k\sqrt{2}=1.414k$

$t_2=k\sqrt{\frac{3}{2}}=1.224k$

$t_3=k\sqrt{\frac{5}{3}}=1.2909k$

$t_4=k\sqrt{\frac{7}{5}}=1.183k$

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0

3 years ago

ABC has the vertices A(1, 4), B(3, 4), and C(1, 1). Find the coordinates of each point of concurrency. 13. circumcenter of ∆ABC

oee [108]

Triangle ABC is a right angle triangle. The point of circumcentre and the point of orthocentre is the same point, as shown in the diagram below. The coordinate of the point is (2.5, 2.5)

6 0

3 years ago