Find the mean deviation of a set of numbers: 14, 15, 16, 17, 18, and 19.
2.5
1.7
1.5
3.5
Correct answer is C
x̄ = \(\frac{ 14 + 15 + 16 +17 + 18 + 19}{6} = \frac{99}{6}\) = 16.5
M . D = \(\frac{ (14 - 16.5) + (15 - 16.5) + (16 - 16.5) + (17 - 16.5) + (18 - 16. 5) + (19 - 16.5) }{ 6}\)
M.D = \(\frac{(-2.5) + (-1.5) + (-0.5) + (0.5) + (1.5) + (2.5)}{6}\)
Taking the absolute value of the deviations
M.D = \(\frac{2.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5}{6}\)
M.D = \(\frac{9}{6}\) = 1.5
288
400
300
225
Correct answer is D
\(M ∝ n^2\sqrt{q}\)
\(M = Kn^2\sqrt{q}\)
K = \(\frac{M}{n^2\sqrt{q}}\)
K = \(\frac{24}{2^2\sqrt4}\)
k = \(\frac{24}{8} = 3\)
Now, let's find M when n = 5 and q = 9
M = \(Kn^2\sqrt{q}\)
M = \( 3\times5^2\sqrt9\)
\(M = 3\times25\times3\)
Therefore, M = 225.
The diagonals of a rhombus are 16 cm and 12 cm find the length of the side.
20cm
8cm
14cm
10cm
Correct answer is D
In a rhombus, the diagonals are perpendicular bisectors of each other, and they bisect the angles of the rhombus. This means that a rhombus is essentially made up of four congruent right-angled triangles.
We can use the Pythagorean theorem to find the length of one side of the rhombus (s)
\(s^2 = 8^2 + 6^2\)
\(s^2 = 64 + 36\)
\(s^2 = 100\)
s = \(\sqrt{100}\)
s =10 cm
So, the length of each side of the rhombus is 10 cm.
If 2x - 3y = -11 and 3x + 2y = 3, evaluate \( (y - x)^2\)
16
25
9
4
Correct answer is A
2x - 3y = -11 --- (i)
3x + 2y = 3 --- (ii)
Multiply equation (i) by 3 and equation (ii) by 2
6x - 9y = -33 --- (iii)
6x + 4y = 6 --- (iv)
Subtract equation (iii) from (iv)
13y = 39
y = \(\frac{39}{13}\) = 3
substitute (3) for y in equation (ii)
3x + 2(3) = 3
3x + 6 = 3
3x = 3 - 6
3x = -3
x = \(\frac{-3}{3}\) = - 1
Now,
\((y - x)^2 = (3 - (-1))^2\)
= \((3 + 1)^2\)
= \(4^2\)
= 16
\(\frac{2}{5}\)
\(\frac{5}{2}\)
\(\frac{2}{7}\)
\(\frac{7}{2}\)
Correct answer is A
Gradient(slope) m = \(\frac{ y_2 - y_1}{ x_2 - x_1}\)
the points are \((\frac{1}{2}, \frac{- 1}{3}) and ( 3 , \frac{2}{3})\)
m = \(\frac{\frac{2}{3} - (\frac{-1}{3})}{3 - \frac{1}{2}}\)
= \(\frac{\frac{2}{3} + \frac{1}{3}}{3 - \frac{1}{2}}\)
= \(1 \div\frac{5}{2}\) = \(1\times\frac{2}{5}\)
Therefore, m = \(\frac{2}{5}\)
Find the value of m in the diagram above.
\(40^0\)
\(50^0\)
\(130^0\)
\(140^0\)
Correct answer is C
∠EHI = ∠DEH = 40° (alternate angles are equal)
∠BEH = 90° (given)
∠BED = 90° - 40° = 50°
∠BEF = 180° - 50° = 130° (sum of angles on a straight line is 180o)
∠BEF = ∠ABE = 130° (alternate angles are equal)
∴ m = 130°
The length of the diagonal of a square is 12 cm. Calculate the area of the square.
\(36 cm^2\)
\(48 cm^2\)
\(72 cm^2\)
\(18 cm^2\)
Correct answer is C
Let each side be l, then area = \(l^2\)
Using Pythagoras theorem
\(l^2 + l^2 = 12^2\)
\(2l^2 = 144\)
divide both sides by 2
\(l^2 = 72\)
Therefore, the Area of the square is \(72 cm^2\)
Find the quadratic equation whose roots are \(\frac{2}{3} and \frac{- 3}{4}\)
\(12y^2 - y - 6 = 0\)
\(12y^2 - y + 6 = 0\)
\(12y^2 + y - 6 = 0\)
\(y^2 + y - 6 = 0\)
Correct answer is C
Let p = \(\frac{2}{3}\) and q = \(\frac{- 3}{4}\)
using (y - p)(y - q) = 0
= ( y - \(\frac{2}{3})\)( y - (\(\frac{- 3}{4})) = 0\)
= (\( y - \frac{2}{3})( y + \frac{3}{4})\) = 0
\( y^2 + \frac{3}{4}y - \frac{2}{3}y - \frac{6}{12} = 0 \)
\( y^2 + \frac{1}{12}y - \frac{1}{2}\) = 0
= multiply through by the l. c. m of 3 and 4 = 12
∴ the quadratic equation is \(12y^2 + y - 6 = 0\)
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