12.2
12.27
12.9
13.4
Correct answer is B
\(\begin{array}{c|c} \text{Class intervals} & F & \text{Class boundary}\\ \hline 5 - 7 & 17 & 4.5 - 9.5\\ 10 - 14 & 32 & 9.5 - 14.5\\ 15 - 19 & 25 & 14.5 - 19.5\\ 20 - 24 & 24 & 19.5 - 24.5 \end{array}\)
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
19
18
13
12
Correct answer is A
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
find the equation of the curve which passes through by 6x - 5
6x2 - 5x + 5
6x2 + 5x + 5
3x2 - 5x - 5
3x2 - 5x + 3
Correct answer is D
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
Evaluate ∫\(^{\pi}_{2}\)(sec2 x - tan2x)dx
\(\frac{\pi}{2}\)
\(\pi\) - 2
\(\frac{\pi}{3}\)
\(\pi\) + 2
Correct answer is B
∫\(^{\pi}_{2}\)(sec2 x - tan2x)dx
∫\(^{\pi}_{2}\) dx = [X]\(^{\pi}_{2}\)
= \(\pi\) - 2 + c
when c is an arbitrary constant of integration
Differentiate \(\frac{x}{cosx}\) with respect to x
1 + x sec x tan x
1 + sec2 x
cos x + x tan x
x sec x tan x + secx
Correct answer is D
let y = \(\frac{x}{cosx}\) = x sec x
y = u(x) v (x0
\(\frac{dy}{dx}\) = U\(\frac{dy}{dx}\) + V\(\frac{du}{dx}\)
dy x [secx tanx] + secx
x = x secx tanx + secx