Mathematics 2020 Jamb Examination

First, calculate the mean (\(\bar{x}\)) of the data:\[\bar{x} = \frac{7 + 8 + 9 + 10 + 11 + 12 + 13}{7} = \frac{70}{7} = 10\]Now, find the sum of the squares of the deviations from the mean (\(\sum(x - \bar{x})^2\)):\[(7-10)^2 = 9\]\[(8-10)^2 = 4\]\[(9-10)^2 = 1\]\[(10-10)^2 = 0\]\[(11-10)^2 = 1\]\[(12-10)^2 = 4\]\[(13-10)^2 = 9\]Sum of squares = \(9 + 4 + 1 + 0 + 1 + 4 + 9 = 28\).
The variance (\(\sigma^2\)) is the sum of squares divided by the number of items (n):\[\sigma^2 = \frac{28}{7} = 4\]The standard deviation (\(\sigma\)) is the square root of the variance:\[\sigma = \sqrt{4} = 2\]

First, identify the sample space, which is the set of numbers from 20 to 30 inclusive:
{20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}.
The total number of outcomes (n) is 11.
Next, identify the prime numbers within this range. A prime number has only two factors, 1 and itself:
The prime numbers are {23, 29}.
The number of successful outcomes is 2.
Probability is calculated as:\[P(\text{prime}) = \frac{\text{Number of prime numbers}}{\text{Total number of outcomes}} = \frac{2}{11}\]

From the graph, we identify the intercepts. The line crosses the y-axis at (0, 4) and the x-axis at (3, 0).
We can use the intercept form of a linear equation: \(\frac{x}{a} + \frac{y}{b} = 1\), where \(a\) is the x-intercept and \(b\) is the y-intercept:\[\frac{x}{3} + \frac{y}{4} = 1\]To clear the fractions, multiply the entire equation by the LCM of 3 and 4, which is 12:\[12\left(\frac{x}{3}\right) + 12\left(\frac{y}{4}\right) = 12(1)\]\[4x + 3y = 12\]Rearranging to match the options:\[3y = -4x + 12\]

First, multiply the two numbers:\[0.0014 \times 0.011 = 0.0000154\]To express this in standard form (scientific notation), we move the decimal point to the right until there is only one non-zero digit to the left of the decimal point. We move it 5 places to the right:\[0.0000154 = 1.54 \times 10^{-5}\]Alternatively, using indices:\[(1.4 \times 10^{-3}) \times (1.1 \times 10^{-2}) = (1.4 \times 1.1) \times 10^{-3 + (-2)} = 1.54 \times 10^{-5}\]

This problem can be modeled as a right-angled triangle where the height of the tank is the opposite side (h), the distance from the foot is the adjacent side (30m), and the angle of elevation is 60^o.
Using the trigonometric ratio for tangent:\[\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}\]\[\tan 60^o = \frac{h}{30}\]From trigonometric tables or special angles, we know that \(\tan 60^o = \sqrt{3}\):\[\sqrt{3} = \frac{h}{30}\]\[h = 30\sqrt{3}\text{m}\]

First, determine the interest earned (I) by subtracting the principal (P) from the total amount (A):\[I = A - P = 1240 - 1000 = ₦240\]The formula for simple interest is given by:\[I = \frac{P \times R \times T}{100}\]Substituting the known values (I = 240, P = 1000, T = 3) into the formula to find the rate (R):\[240 = \frac{1000 \times R \times 3}{100}\]\[240 = 30R\]Dividing both sides by 30:\[R = \frac{240}{30} = 8\%\]

360\(^o\) - (60\(^o\) + 60\(^o\) + 67 + 50 = 237\(^o\)) 

To solve the inequality, move the terms containing x to one side and constants to the other:\[\frac{1}{3x} - \frac{1}{4x} > -\frac{1}{2}\]Find a common denominator for the left side, which is 12x:\[\frac{4 - 3}{12x} > -\frac{1}{2}\]\[\frac{1}{12x} > -\frac{1}{2}\]Since the question states that x is a positive real number, 12x is always positive. We can safely multiply both sides by 12x without changing the inequality sign:\[1 > -6x\]Divide both sides by -6. Note that dividing an inequality by a negative number reverses the sign:\[-\frac{1}{6} < x \quad \text{or} \quad x > -\frac{1}{6}\]However, the initial condition states x must be a positive real number (x > 0). Since all values of x > 0 are already greater than -1/6, the valid range is simply x > 0.

Start by cross-multiplying the equation:\[\sqrt{2}(x - \sqrt{2}) = 1(x + \sqrt{2})\]Expand the brackets on the left side:\[x\sqrt{2} - 2 = x + \sqrt{2}\]Collect the terms containing x on one side and the constants on the other:\[x\sqrt{2} - x = 2 + \sqrt{2}\]Factorise x out of the left side:\[x(\sqrt{2} - 1) = 2 + \sqrt{2}\]Isolate x by dividing by (\(\sqrt{2} - 1\)):\[x = \frac{2 + \sqrt{2}}{\sqrt{2} - 1}\]Rationalise the denominator by multiplying the numerator and denominator by the conjugate (\(\sqrt{2} + 1\)):\[x = \frac{(2 + \sqrt{2})(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)}\]\[x = \frac{2\sqrt{2} + 2 + 2 + \sqrt{2}}{2 - 1}\]\[x = \frac{3\sqrt{2} + 4}{1} = 3\sqrt{2} + 4\]

Applying the definition of the binary operation a * b = a^b to the given equation a * 2 = 2 - a:\[a^2 = 2 - a\]This results in a quadratic equation. Rearrange it into the standard form \(ax^2 + bx + c = 0\):\[a^2 + a - 2 = 0\]Factorise the quadratic expression by finding two numbers that multiply to -2 and add to 1 (+2 and -1):\[(a + 2)(a - 1) = 0\]Set each factor to zero to find the roots:\[a + 2 = 0 \implies a = -2\]\[a - 1 = 0 \implies a = 1\]The possible values of a are 1 and -2.