Mathematics 2023 Jamb Examination

Let the number of white balls be \(w\). The total number of balls is \(8 + w\).
According to the problem, \(P(W) = \frac{1}{2} P(R)\):\[\frac{w}{8+w} = \frac{1}{2} \left( \frac{8}{8+w} \right)\]\[w = 4\]The total number of balls is \(8 + 4 = 12\). The probability of drawing a red ball then a white ball without replacement is:\[P(R \text{ then } W) = \frac{8}{12} \times \frac{4}{11} = \frac{2}{3} \times \frac{4}{11} = \frac{8}{33}\]

Total frequency \(N = 29 + 40 + 38 = 107\). The median position is \(\dfrac{107}{2} = 53.5\).
The median falls in the 7-8 class (cumulative frequency reaches 69 here).
Using the formula \(\text{Median} = L + \left( \dfrac{\dfrac{N}{2} - C_{f_b}}{f_m} \right)i\):\[\text{Median} = 6.5 + \left( \dfrac{53.5 - 29}{40} \right)2 = 6.5 + \left( \dfrac{24.5}{40} \right)2 = 6.5 + 1.225 = 7.725\]

Let \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\). Performing the multiplication:\[\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 2b & a-b \\ 2d & c-d \end{bmatrix}\]Equating terms: \(2b = 2 \implies b = 1\) and \(a - 1 = -1 \implies a = 0\).
For the second row: \(2d = 1 \implies d = 0.5\) and \(c - 0.5 = 0 \implies c = 0.5\).
Thus, \(A = \begin{bmatrix} 0 & 1 \\ 1/2 & 1/2 \end{bmatrix}\).

To find the number of students who scored "at least 25%", we sum the frequencies of all bars representing 25% or higher.\[\begin{itemize} \item \text{Frequency at 25\% = 8} \item \text{Frequency at 35\% = 3} \item \text{Frequency at 45\% = 5} \end{itemize}\] Sum = \(8 + 3 + 5 = 16\).

By the Remainder Theorem, the remainder when \(f(x)\) is divided by \((x - c)\) is \(f(c)\). Here, \(c = -1\).\[f(-1) = -2(-1)^3 + 6(-1)^2 + 17(-1) - 21\]\[= -2(-1) + 6(1) - 17 - 21\]\[= 2 + 6 - 17 - 21 = -30\]

Commutativity refers to the property where the order of operands does not change the result of the operation. This is expressed as \(a * b = b * a\). Option B describes associativity.

First, move all terms to one side: \(x^2 - x - 6 \le 0\).
Factor the quadratic: \((x - 3)(x + 2) \le 0\).
The critical values are \(x = 3\) and \(x = -2\). Testing the regions, the product is negative or zero between the roots: \(-2 \le x \le 3\).

This is an arithmetic progression with first term \(a = 3\) and common difference \(d = 5\).
The general term formula is \(U_n = a + (n - 1)d\):\[U_n = 3 + (n - 1)5\]\[U_n = 3 + 5n - 5 = 5n - 2\]

The set of all points equidistant from two intersecting lines consists of the two lines that bisect the angles formed by the intersection.

Let the third number be 100.
The first number is \(100 + 35 = 135\).
The second number is \(100 + 80 = 180\).
The ratio is \(135 : 180\).
Dividing both by 45:\[\frac{135}{45} : \frac{180}{45} = 3 : 4\]