1995 - JAMB Mathematics Past Questions and Answers - page 4
∫dy = ∫(4x - 3)dx, y = 2x2 - 3x + C
when y = 5, x = 2, C = 3
y = 2x2 - 3x + 3
Users' Answers & Comments(\pi) = (\int^{3}_{1}(3x^2 - 2x + 1))
dx = [x3 - 3x + x + C]
= 22
Users' Answers & CommentsThe frequency distribution above shows the ages of students in a secondary school. In a pie chart constructed to represent the data, the angles corresponding to the 15 years old is
Find the standard deviation of the data using the table above
(\begin{array}{c|c} \text{class intervals} & Fre(F) & \text{class-marks(x)} & Fx & (x - x)& (x - x)^2 & F(x - x)^2 \ \hline 1 - 3 & 5 & 10 & 10 & -3 & 9 & 90\ 4 - 6 & 8 & 40 & 40 & 0 & 0 & 0 \ 7 - 9 & 5 & 40 & 40 & 3 & 9 & 360 \ \hline & 18 & & 90 & & & 450 \end{array})
x = (\frac{\sum fx}{\sum f})
= (\frac{90}{18})
= 5
S.D = (\frac{\sum f(x - x)^2}{\sum f})
= (\frac{450}{18})
= (\sqrt{25})
= 5
Users' Answers & CommentsFind the mode of the distribution above to find the mode of the distribution.
Mode = a + (b - a)(fm - Fb)
2Fm - Fa - Fb
= 3.0 + (\frac{(3.4 - 3)(15 - 4)}{2(15) - 4 - 10})
= 3 + (\frac{(6.4)(11)}{30 - 14})
= 3 + (\frac{4.4}{16})
= 3 + 0.275
= 3.275
= 3.3cm
Users' Answers & CommentsThe median of the distribution above is
Median = (\frac{a + b}{fm}) ((\frac{1}{2} \sum f - CF_b))
= 2.95 + (\frac{0.5}{15})(2.-7)
= 2.95 + (\frac{0.5}{15}) x 13
= 2.95 + 0. 43
= 3.38
= 3.4
Users' Answers & Comments