Mathematics 2018 Jamb Examination

To find the total number of children, sum the frequencies (f) in the table: \[ \displaystyle \sum f = 3 + 4 + 5 + 6 + 7 + 6 + 5 + 4 = 40 \]

A = (p, q, r, t, u, v)

  B = (r, s, t, u)

  A U B = Elements in both A and B = (p, q, r, s, t, u, v)

  (A U B)1 = elements in the universal set E but not in (A U B)= (w, x, y, z)

  n(A U B) 1 = number of the elements in (A U B)1 = 4

By definition, the locus of points that are at a constant distance from a fixed point is a circle, where the fixed point is the center and the distance is the radius. Therefore, at 4cm from point P, the locus is a circle with a radius of 4cm.

Using the complementary angle identity, \( \displaystyle \sin \theta = \cos(90^\circ - \theta) \). This means the sum of the angles must be \( \displaystyle 90^\circ \): \[ \displaystyle (5x - 28) + (3x - 50) = 90 \] \[ \displaystyle 8x - 78 = 90 \] \[ \displaystyle 8x = 168 \implies x = 21 \]

The first four prime numbers greater than 10 are 11, 13, 17, and 19. \[ \displaystyle \text{Average} = \frac{11 + 13 + 17 + 19}{4} \] \[ \displaystyle \text{Average} = \frac{60}{4} = 15 \]

The mode (or modal value) is the value with the highest frequency. Looking at the table: - The highest frequency (f) is 7. - The age (x) corresponding to this frequency is 5.

Convert both numbers to base 10 first: \[ \displaystyle 243_{\text{six}} = (2 \times 6^2) + (4 \times 6^1) + (3 \times 6^0) = 72 + 24 + 3 = 99 \] \[ \displaystyle 243_{\text{five}} = (2 \times 5^2) + (4 \times 5^1) + (3 \times 5^0) = 50 + 20 + 3 = 73 \] Subtract the base 10 values: \[ \displaystyle 99 - 73 = 26 \]

The integral of \( \displaystyle \frac{1}{x} \) is \( \ln(x) \). Applying the limits: \[ \displaystyle \int_{1}^{2} \frac{5}{x} dx = [5 \ln(x)]_{1}^{2} \] \[ \displaystyle = 5 \ln(2) - 5 \ln(1) \] Since \( \ln(1) = 0 \): \[ \displaystyle = 5 \times 0.6931 = 3.4655 \approx 3.47 \]

For independent events, multiply the probabilities: - \( \displaystyle P(\text{Tail}) = \frac{1}{2} \) - \( \displaystyle P(\text{Even number on a die}) = \frac{3}{6} = \frac{1}{2} \) (Numbers 2, 4, 6) \[ \displaystyle P(\text{Tail and Even}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

First, select the letters: 3 consonants from 7 (\( \displaystyle ^7C_3 \)) and 2 vowels from 4 (\( \displaystyle ^4C_2 \)). Then, arrange the 5 selected letters (\( \displaystyle 5! \)). \[ \displaystyle \text{Ways} = (^7C_3 \times ^4C_2) \times 5! \] \[ \displaystyle = (35 \times 6) \times 120 = 210 \times 120 = 25200 \]