How many nine-letter arrangements can be made using the letters of the word ISOSCELES? A particle moves in a straight line and its position at time t is given by $x = 4 \sin \left( 2\pi t + \frac{\pi}{3} \right)$ - HSC - SSCE Mathematics Extension 1 - Question 3 - 2003 - Paper 1

The equation given is:

$x = 4 \sin \left( 2\pi t + \frac{\pi}{3} \right)$

This is in the form of:

$x = A \sin(\omega t + \phi)$

where A = 4 (amplitude), ( \omega = 2\pi ) (angular frequency), and ( \phi = \frac{\pi}{3} ) (phase constant). Since the displacement is a sine function of time, the motion is simple harmonic. Therefore, we can conclude that the particle is undergoing simple harmonic motion.

The amplitude of simple harmonic motion is the maximum displacement from the equilibrium position. From the equation:

$x = 4 \sin \left( 2\pi t + \frac{\pi}{3} \right)$

the amplitude A is 4.

Maximum speed in simple harmonic motion occurs when the sine function is at its maximum value of 1. This occurs when:

$2\pi t + \frac{\pi}{3} = \frac{\pi}{2} + 2k\pi \text{ for } k \in \mathbb{Z}$

Solving for t gives:

$2\pi t = \frac{\pi}{2} - \frac{\pi}{3} + 2k\pi = \frac{3\pi - 2\pi}{6} + 2k\pi = \frac{\pi}{6} + 2k\pi$

Thus,

$t = \frac{1}{12} + k$

For the first occurrence (k = 0):

$t = \frac{1}{12} \text{ seconds}$

When rolling two dice, the total number of possible outcomes is 36 (since each die has 6 sides). The combinations that yield a sum of 5 are (1,4), (2,3), (3,2), and (4,1). This gives us 4 successful outcomes. The probability is calculated as follows:

$P(Sum = 5) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{36} = \frac{1}{9}.$

Let X be the random variable representing the number of times the sum of the dice equals 5 in 7 tosses. X follows a binomial distribution with parameters n = 7 and p = \frac{1}{9}.

To find the probability of getting 5 at least twice, we can calculate:

$P(X \geq 2) = 1 - \left( P(X = 0) + P(X = 1) \right).$

Using the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k},$

we calculate:

$P(X = 0) = \binom{7}{0} \left(\frac{1}{9}\right)^0 \left(\frac{8}{9}\right)^7 = \left(\frac{8}{9}\right)^7,$

and

$P(X = 1) = \binom{7}{1} \left(\frac{1}{9}\right)^1 \left(\frac{8}{9}\right)^6 = 7\left(\frac{1}{9}\right) \left(\frac{8}{9}\right)^6.$

Substitute values to find P(X >= 2).

We will prove this by induction.

Base case (n = 1):

For n = 1, the left-hand side is:

$\frac{1}{1 \times 3} = \frac{1}{3}$

The right-hand side is:

$\frac{1}{2 \cdot 1 + 1} = \frac{1}{3}$

Both sides equal, so the base case holds.

Inductive step: Assume it holds for n = k:

$\frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \ldots + \frac{1}{(2k-1)(2k+1)} = \frac{k}{2k + 1}$.

For n = k + 1:

$\frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \ldots + \frac{1}{(2k-1)(2k+1)} + \frac{1}{(2(k+1)-1)(2(k+1)+1)}$

We need to show:

$\frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)} = \frac{k+1}{2(k + 1) + 1}$

Simplifying the left side,

$\frac{(k)(2k + 3) + 1}{(2k + 1)(2k + 3)} = \frac{k(2k + 3) + 1}{(2k + 1)(2k + 3)} = \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)}$

Equating to the right and simplifying will show that both sides match, completing the proof.