Find the mean deviation of the set of numbers 4, 5, 9
zero
2
5
6
Correct answer is B
x = \(\frac{\sum x}{N}\)
= \(\frac{18}{3}\)
= 6
\(\begin{array}{c|c} x & x - x & x - x \\ \hline 4 & -2 & 2\\ 5 & 1 & 1\\ 9 & 3 & 3\\ \hline & & 6\end{array}\)
M.D = \(\frac{|x - x|}{N}\)
= \(\frac{6}{3}\)
= 2
1
2
3
4
Correct answer is B
| x | 1 | 2 | 3 | 4 | 5 | Total |
| f | y + 2 | y - 1 | 2y - 3 | y + 4 | 3y - 4 | 8y - 2 |
| fx | y + 2 | 2y - 2 | 6y - 9 | 4y + 16 | 15y - 20 | 28y - 13 |
Mean = \(\frac{\sum fx}{\sum f}\)
\(\therefore \frac{28y - 13}{8y - 2} = \frac{43}{14}\)
\(\implies 14(28y - 13) = 43(8y - 2)\)
\(392y - 182 = 344y - 86\)
\(392y - 344y = -86 + 182 \implies 48y = 96\)
\(y = 2\)
Evaluate \(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
3\(\frac{2}{3}\)
4
4\(\frac{1}{3}\)
4\(\frac{2}{3}\)
Correct answer is D
\(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)
= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3} + 2x^2 + x]\)
= [\(\frac{4}{3}\) + 2 + 1] - [\(\frac{-4}{3}\)+2 -1]
= \(\frac{13}{3}\) + \(\frac{1}{3}\)
= \(\frac{14}{3}\)
= \(4 \frac{2}{3}\)
Integrate \(\frac{1 - x}{x^3}\) with respect to x
\(\frac{x - x^2}{x^4}\) + k
\(\frac{4}{x^4} - \frac{3 + k}{x^3}\)
\(\frac{1}{x} - \frac{1}{2x^2}\) + k
\(\frac{1}{3x^2} - \frac{1}{2x}\) + k
Correct answer is C
\(\int \frac{1 - x}{x^3}\)
= \(\int^{1}_{x^3} - \int^{x}_{x^3}\)
= x-3 dx - x-2dx
= \(\frac{1}{2x^2} + \frac{1}{x}\)
Find the point (x, y) on the Euclidean plane where the curve y = 2x2 - 2x + 3 has 2 as gradient
(1, 3)
(2, 7)
(0, 3)
(3, 15)
Correct answer is A
Equation of curve;
y = 2x2 - 2x + 3
gradient of curve;
\(\frac{dy}{dx}\) = differential coefficient
\(\frac{dy}{dx}\) = 4x - 2, for gradient to be 2
∴ \(\frac{dy}{dx}\) = 2
4x - 2 = 2
4x = 4
∴ x = 1
When x = 1, y = 2(1)2 - 2(1) + 3
= 2 - 2 + 3
= 5 - 2
= 3
coordinate of the point where the curve; y = 2x2 - 2x + 3 has gradient equal to 2 is (1, 3)