Correct Answer: D. 4.7

Explanation

Mean \(\bar{x} = \frac{\sum fx}{\sum f} = \frac{189}{40}\)

= \(4.725 \approxeq 4.7\)

Correct Answer: C. \(\pm 4\)

Explanation

\(2x^{2} + kx + 5 = 0\)

\(\alpha + \beta = \frac{-b}{a} = \frac{-k}{2}\)

\(\alpha \beta = \frac{c}{a} = \frac{5}{2}\)

\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha \beta\)

\(-1 = (\frac{-k}{2})^{2} - 2(\frac{5}{2})\)

\(-1 = \frac{k^{2}}{4} - 5 \implies \frac{k^{2}}{4} = 4\)

\(k^{2} = 16 \therefore k = \pm 4\)

Correct Answer: C. 4.5

Explanation

\(XY = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\) is the distance between a point \(X(x_{1}, y_{1})\) and \(Y(x_{2}, y_{2})\).

\(XY = \sqrt{(3 - 5)^{2} + (5 - 1)^{2}} = \sqrt{20}\)

= \(2\sqrt{5} = 4. 467 \approxeq 4.5\)

Correct Answer: B. 12.6 years

Explanation

\(Mean = \frac{\text{sum of ages}}{\text{no of pupils}}\)

Let the sum of the ages of the 15 pupils be x and the sum of the 16 pupils be y.

\(\frac{x}{15} = 14.2 \implies x = 14.2 \times 15 = 213\)

\(\frac{y}{16} = 14.1 \implies y = 14.1 \times 16 = 225.6\)

\(\text{Age of new pupil} = 225.6 - 213 = 12.6\)