Mathematics 2013 Jamb Examination

To find the probability that a picked fruit is neither an apple nor an orange, we must first determine the total number of fruits in the basket and the number of fruits that satisfy the given condition.
Step 1: Calculate the total number of fruits in the basket:\[\text{Total} = 9 \text{ (apples)} + 8 \text{ (bananas)} + 7 \text{ (oranges)} = 24\]Step 2: Identify the fruits that are neither apples nor oranges. In this basket, if a fruit is not an apple and not an orange, it must be a banana. The number of bananas is 8.
Step 3: Calculate the probability using the formula:\[P(\text{neither apple nor orange}) = \frac{\text{Number of bananas}}{\text{Total number of fruits}}\]\[P = \frac{8}{24}\]Step 4: Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 8:\[P = \frac{1}{3}\]Thus, there is a 1 in 3 chance of picking a fruit that is neither an apple nor an orange.

To determine the correct quadratic equation for the graph, we identify the key features: the x-intercepts and the y-intercept. From the graph, the curve crosses the x-axis at \(x = -1\) and \(x = 2\), and the y-axis at \(y = -2\).
If the x-intercepts are \(x_1 = -1\) and \(x_2 = 2\), the quadratic equation can be written in factored form as: \[ y = a(x - x_1)(x - x_2) \] \[ y = a(x - (-1))(x - 2) \] \[ y = a(x + 1)(x - 2) \] To find the value of \(a\), we use the y-intercept \((0, -2)\). Substituting \(x = 0\) and \(y = -2\) into the equation: \[ -2 = a(0 + 1)(0 - 2) \] \[ -2 = a(1)(-2) \] \[ -2 = -2a \implies a = 1 \] Now, substitute \(a = 1\) back into the factored form and expand the expression: \[ y = 1(x + 1)(x - 2) \] \[ y = x^2 - 2x + x - 2 \] \[ y = x^2 - x - 2 \] This matches the standard form of the quadratic function shown in the graph.

In the diagram above, < CDE = < CED = 15^o

(base < s of isos. △)

< ECD = 180^o - (15 + 15)^o

= 180^o - 30^o = 150^o

But x + 110^o = 150^o

(Sum of opp. interior < s of a△ = opp. exterior < )

x = 150^o - 110^o = 40^o

To find the total time allocated, we sum the minutes for each subject shown on the bar chart:
* Mathematics: 180
* English: 120
* Biology: 80
* Chemistry: 100
* Physics: 120
* Geography: 120

Summing these values: \[ 180 + 120 + 80 + 100 + 120 + 120 = 720 \text{ minutes} \]

Sum of the angles in a triangle = 180°

180° - x + 60 + 70 = 180
130° - x = 180° - 180° = 0°
x = 130°

5x° + (16x - 24)° + 5x° + (4x + 12)° + (6x + 12)° = 360°

The sum of the interior angles of any quadrilateral is always 360°. Therefore:\[(3y + 10) + (2y + 30) + (y + 20) + 4y = 360\]Combine like terms:\[10y + 60 = 360\]\[10y = 300\]\[y = 30^\circ\]

First, convert all measurements to centimeters: Length = 720 cm, Width = 420 cm.
Area of floor = \(720 \times 420 = 302,400 \text{ cm}^2\).
Area of one tile = \(30 \times 30 = 900 \text{ cm}^2\).
Number of tiles = \(\dfrac{302,400}{900} = 336\).

In a circle, a chord and two radii form an isosceles triangle. Since the angle is 90°, it is a right-angled isosceles triangle. Using Pythagoras theorem where the chord is the hypotenuse:\[c^2 = r^2 + r^2 = 8^2 + 8^2 = 64 + 64 = 128\]\[c = \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2} \text{ cm}\]

Radius \(\displaystyle r = \frac{\text{diameter}}{2} = 2\sqrt{3}\).
The minor sector angle is 120°, so the major sector angle is \(360^\circ - 120^\circ = 240^\circ\).\[\text{Area} = \frac{240}{360} \times \pi r^2 = \frac{2}{3} \times \pi \times (2\sqrt{3})^2\]\[\text{Area} = \frac{2}{3} \times \pi \times 12 = 8\pi \text{ cm}^2\]