JAMB Mathematics Past Questions & Answers - Page 301

mean (x) = \(\frac{1 + x + 1 + 2x + 1}{3}\)

= \(\frac{3x + 3}{3}\)

= 1 + x

\(\begin{array}{c|c} X & (X -X) & (X -X)^2\\ \hline 1 & -x & x^2 \\ x + 1 & 0 & 0\\2x + 1 & x & x^2\\ \hline & & 2x^2\end{array}\)

S.D = \(\sqrt{\frac{\sum(x - 7)^2}{\sum f}}\)

= \(\sqrt{(6)}^2\)

= \(\frac{2x^2}{3}\)

= 2x2

= 18

x2 = 9

∴ x = \(\pm\) \(\sqrt{9}\)

= \(\pm\)3

mean (x) = \(\frac{\sum x}{N}\)

= k + k + 1 + k + 3

= \(\frac{3k + 3}{3}\)

= k + 1

\(\begin{array}{c|c} X & (X -X) & (X -X)^2\\ \hline k & -1 & 1 \\ k + 1 & 0 & 0\\ k + 2 & 1 \\ \hline & & 2\end{array}\)

Variance (52) = \(\frac{\sum (x - x)^2}{N}\)

= \(\frac{2}{3}\)

\(\begin{array}{c|c} \text{Class intervals} & F & \text{Class boundary}\\ \hline 5 - 7 & 17 & 4.5 - 9.5\\ 10 - 14 & 32 & 9.5 - 14.5\\ 15 - 19 & 25 & 14.5 - 19.5\\ 20 - 24 & 24 & 19.5 - 24.5 \end{array}\)

mode = 9.5 + \(\frac{D_1}{D_2 + D_1}\) x C

= 9.5 + \(\frac{5(32 - 17)}{2(32) - 17 - 25}\)

= 9.5 + \(\frac{75}{27}\)

= 12.27

\(\approx\) 12.3

mean(x) = \(\frac{\sum x}{N}\)

= \(\frac{48}{8}\)

= 5.875

re-arranging the numbers;

2, 3, 5, 6, 2, 7, 8, 9

median = \(\frac{6 + 7}{2}\)

= \(\frac{1}{2}\)

= 6.5

m + 2n = 5.875 + (6.5)²

= 13 + 5.875

= 18.875

= \(\approx\) = 19

m = \(\frac{dy}{dv}\) = 6x - 5

∫dy = ∫(6x - 5)dx

y = 3x² - 5x + C

when x = 2, y = 5

∴ 5 = 3(2)² - 5(2) +C

C = 3

∴ y = 3x² - 5x + 3