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Median divides the value of a variable into two equal parts

(d) All these

A perfect random number table would be one in which every digit has been entered Randomly

If the range of a series is 10, lowest value equal to 15, then highest value may be 25. 

Range is an absolute measure 

Circular test Is an extension of time reversal test 

5 is the Mean for the following observation ; 3,4,6,7,10 

A variable is any characteristic which can assume Different values. 

b. time series 

A Continuous variable is a variable whose values can theoretically take on an infinite number of values within a given range of values. 

Mean is the most commonly used measure of central tendancy 

Trend Component is used for a short term forecast 

The range of 10, 20, 15, 18, 16, 21, 25 is 15.

The gross national product value is deflated through price index numbers

Fishers ideal index formula satisfies Factor reversal test  and time reversal test  test

 A time series is a set of values arranged in chronological order.

The values which varies with maximum frequency is called mode. 

Quartiles are the values dividing a given set of observation in to four equal parts. 

Fisher‘s formula is called the ideal because of the following reasons: 

(i) It is based on geometric mean which is considered best for constructing index numbers. 

(ii) It fulfills both the time reversal and factor reversal tests. 

(iii) It takes into account both current year as well as base year     prices and quantities. 

(iv) It is free from bias. 

 H.M is the best average to analyse speed. 

A time series Consists of long term changes, short term variation, irregular variation etc. 

Adequacy of data is to be judged in the light of the requirements of the survey and the geographical areas covered by the  Available data. 

In Laspear’s Index number base year quantities are used. 

Quartiles can be determined graphically using ogive.

Splicing means combining 2 or more overlapping series of index number with different years in to one with a common base year. 

Classification means separating items according to similar characteristics and grouping them into various classes.

(a) (S.D/Mean) x 100

In plural sense, statistics means numerical set of data. 

The median of a frequency distribution is found graphically with the help of  Ogive

Skewness means lack of symmetry. 

Skewness means asymmetry of a distribution .

Index numbers are specialized type of averages. 

(b) Not equal

Kurtosis is a measure of peakedness. 

Skewness may be + or –

Draw a circle of radius 4 cm

Draw OA and construct ∠AOB = 120°

Draw ∠OAP = ∠OBP = 90°

PA and PB are required tangents.

\(\frac{n}{2}=\frac{250}{2}\) = 125

⇒ median class is 2 - 3, l = 2, h = 1, cf = 120, f = 62

median = \(l+\frac{\frac{n}{2}-cf}{f}\times i\)

= 2 + \(\frac{5}{62}\)

= \(\frac{129}{62}=2\frac{5}{62}\)m or 2.08 m

50% of students jumped below \(2\frac{5}{62}\)m and 50% above it.

Correct option is (a) x - 1/x = 1

(a) x - 1/x = 1

⇒ x² - 1 = x

⇒ x² - x - 1 = 0 which is a quadratic equation.

(b) x² + 1/x = 1

⇒ x³ + 1 = x

⇒ x³ - x + 1 = 0 which is a cubic equation

(c) x² + 2√x - 1 = 0

⇒ x² - 1 = -2√x

⇒  (x² - 1)² = 4x

⇒ x⁴ - 2x² + 1 = 4x

⇒ x⁴ - 2x² - 4x + 1 = 0 which is a quartic equation

(d) (√x)² - 2² = 0

⇒ x - 4 = 0 which is a linear equation.

Radius of sphere x² + y² + z² = b² is b

and radius  x² + y² + z² = a² is a(b > a)

Enclosed volume = volume of sphere x² + y² + z² = b² - volume of sphere x² + y² + z² = a²

 = \(\frac43\pi b^3-\frac43\pi a^3\) (\(\because\) volume of sphere of radius r = \(\frac43\pi r^3\))

= \(\frac43\pi(b^3-a^3)\) 

 = \(\frac43\pi(b-a)(b^2+a^2+ab)\) cubic units.

25x² - 10x + 1 = 0

25x² - 5x - 5x + 1

5x(5x - 1) -1(5x - 1)

(5x - 1)(5x - 1)

(5x - 1)²

Ascending order of outcomes is 15, 16, 16, 17, 18, 18

\(\because\) Total possible outcomes is n = 6(even)

\(\therefore\) Median = \(\frac{n/2^{th}+(n/2+1)^{th}observations}2\) 

  = \(\frac{3^{rd}observation+4^{th}observation}2\) = \(\frac{16+17}2\) = 16.5

Diameter of hemisphere = 56 cm

Its radius = d/2 = 56/2

Radius = 28 cm

T. S. A of hemisphere = 3πr²

= 3 × 22/7 × 28 × 28

= 7392 cm²

C. S. A of hemisphere = 2πr²

= 2 × 22/7 × 28 × 28

= 4928 cm²

Volume of hemisphere = 2/3 πr²

= 2/3 × 22/7 × 28 × 28 × 28

= 45994.666 cm³

\(x\) + \(\frac{1}{x}\) = \(\frac{17}{4}\)

⇒ \(​​\frac{x^2 + 1}{x}\) = \(\frac{17}{4}\)

⇒ \(x^2 - \frac{17x}{4} + 1\) = \(0\)

⇒ \( 4x^2 - 17x + 4 = 0\)

⇒ \(4x^2 - 16x - x + 4 = 0\)

⇒ \(4x(x - 4) - 1(x - 4) = 0\)

⇒ \((4x - 1) (x - 4) = 0\)

∴ \( x = \frac{1}{4}, 4\)

\(\lim\limits_{x \to \infty}(-ln \, x + 2x\frac 12)\) (∞ - ∞ case)

Take y = 1/x then limit converts to y → 0

= \(\lim\limits_{y \to 0}(-ln (\frac 1y)+ 2(\frac 1y)^{\frac 12}\)

= \(\lim\limits_{y \to 0}(-ln\, 1 - ln\,y) + \frac {2}{y\frac 12}) (∵ ln \frac AB = ln\,A-ln\,B)\)

= \(\lim\limits_{y \to 0}(-ln\, y+ \frac {2}{y\frac12})\) (∞ - ∞ case) (∵ ln 1 = 0)

= \(\lim\limits_{y \to 0}(\frac {y\frac12\,ln\,y+2}{y\frac 12})\)

Now,  \(\lim\limits_{y \to 0} y\frac 12 \, ln\,y \) (0 x ∞)

= \(\lim\limits_{y \to 0} \frac {ln\,y}{y\frac{-1}{2}}\) (∞/∞ type)

= \(\lim\limits_{y \to 0} \frac {\frac 1y}{\frac {-1}{2}y\frac{-3}{2}}\) (By using D.L.H Rule)

= \(\lim\limits_{y \to 0} \frac {-2\,y\frac 32}{y}\)

= \(\lim\limits_{y \to 0} -2\,y\frac 12\)

= -2 x 0 = 0

∴ \(\lim\limits_{y \to 0} \frac {y\frac 12\, ln\,y+2}{y\frac 12} = \frac {0+2}{0} = \frac 20 = \infty\) (not defined)

The given statement is false

A number having exactly two factors, that is the number 1 and the number itself is called a prime number. Prime Numbers are those Natural numbers that have not more than 2 factors 1 and the number itself or Prime numbers are greater than 1 that can not be formed by multiplying by two smaller

Natural numbers: The Natural Number 2 is the smallest prime number. 1 is divided itself only hence 1 is not a prime number.

f(x) = \(\begin{cases} \frac{x^2-4}{x-2}, & \quad x\neq2\\ 2, & \quad x = 2 \end{cases}\) 

\(\lim\limits_{x\to2}f(x)=\lim\limits_{x\to2}\frac{x^2-4}{x-2}\)

\(=\lim\limits_{x\to2}\frac{(x-2)(x+2)}{x-2}\) 

\(=\lim\limits_{x+2}\,x+2\)

= 2 + 2 = 4

But f(2) = 2

Therefore \(\lim\limits_{x\to 2}f(x)\neq f(2)\) 

Thus, f(x) has a removable discontinuity.

We have to change the value of function at x = 2 which is 4.

\(\lim\limits_{{(x,y)}\to(0,0)} f(x,y) \) = \(\lim\limits_{{(x,y)}\to(0,0)}\frac {sin^{-1}(x+2y)}{tan^ {-1}(2x+4y)}\)

= \(\lim\limits_{x\to0}\frac {sin^{-1}(x+2mx)}{tan^ {-1}(2x+4mx)}\)(By taking y = mx)

= \(\lim\limits_{x\to0}\frac {sin^{-1}(x(1+2m))}{tan ^{-1}(x(2+4m)}\)( 0/0 type)

= \(\lim\limits_{x\to0}\frac {(1+2m)(1+(x(2+4m))^2)}{(2+4m)\sqrt{1-x^2(1+2m)^2}}\)( By using D.L.H Rule)

= \(\frac {(1+2m)(1+0)}{2(1+2m)(1-0)^{\frac 12}}\) ( By taking limit)

= 1/2

= f (0, 0) 

∴ Given function is continuous at (0, 0)