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Time is Not the data type in JavaScript

In JavaScript, a variable is declared using the keyword var. 

This code snippet is used to display the string, “Welcome to JS” on the screen(monitor).

For storing values you have to declare a variable, for that the keyword var is used. There is no need to specify the data type.

Syntax: 

var <variable name1> [,<variable name2> ,<variable name3> ,etc…] 

Here square bracket indicates optional. 

Eg: var x, y, z; 

x= 11; 

y = ”BVM”; 

z = false; 

Here x is of number type, y is of string and z is of Boolean type.

var keyword is used to declare a variable in JavaScript.

A group of codes with a name is called function

1. Number: Any number(whole or fractional) with or without sign. 

Eg: +1977,-38.0003,-100, 3.14157,etc 

2. String: It is a combination of characters enclosed within double quotes. 

Eg: “BVM”, “[email protected]”, etc 

3. Boolean: We can use either true or false. lt is case sensitive. That means can’t use TRUE OR FALSE

<NPUT Type=”button” onMouseEnter = “checkData()”>

From the list there is only two, number and boolean are the types used in JavaScript.

typeof() function is used to return the data type.

Undefined is a special data type to represent variables that are not defined using var.

(d) %

It is an arithmetic operator, others are logical operator.

(d) C++, It is a programming language others are browsers.

(d) ! 

It is a logical operator, the others are relational operator.

(d) ==

It is a relational operator the others are arithmetic operator.

Importance of median:

(i) It is only average which is used while dealing with qualitative data.

(ii) It is used for typical value in problems concerning wages, distribution of wealth etc.

According to question, average weight of 15 girls

 = 45 kg.

So, \(\bar { X }=\quad \frac { { \Sigma x }_{ i } }{ n } \)

⇒ 45 = \(\frac { { \Sigma x }_{ i } }{ 15 } \)

⇒ Σx_i = 45 × 15 = 675

Total weight of 15 girls Σx_i = 675 kg

Average weight of 30 boys = 52 kg

So, \(\bar { Y }=\quad \frac { { \Sigma y }_{ i } }{ 30 } \)

⇒ 52 = \(\frac { { \Sigma y }_{ i } }{ 30 } \)

⇒ Σy_i = 52 × 30 = 1560

Total weight of boys Σy_i = 1560 kg

Average weight of 45 students

Σx_i + Σy_i = 675 + 1560 = 2235

Average weight of one student = 2235/45 

= 49.67 kg.

Sum of 5 numbers = 18 × 5 = 90

Let removed number is x

Then, sum of four number = 16 × 4 = 64

⇒ Sum of four number + x = 90

⇒ 64 + x = 90

x = 90 – 64 = 26

Thus, fifth number = removed number = 26

Average weight of 25 students of group A = 51 kg

Thus, total of weight of 25 students

W₁ = 51 x 25 = 1275 kg

Average weight of 35 students of group B = 54 kg

Thus, sum of weight of 35 students, W₂ = 54 x 35

= 1890 kg

∴ Average weight of 60 students of class

= \(\frac { { W }_{ 1 }+{ W }_{ 2 } }{ 60 } \)

= \(\frac { 1275+1890 }{ 60 } =\frac { 3165 }{ 60 } \) = 52.75 kg

Thus, average wt. of 60 students = 52.75 kg

Perimeter of a square = (4 × side) units

Side = 8 cm

∴ Perimeter = 4 × 8 cm = 32 cm

Perimeter = 32 cm

Area of a square = (side × side) unit² 

= (8 × 8) cm² = 64 cm²

Area = 64 cm²

i) Perimeter of a scalene triangle = (7 + 8 + 10) m = 25 m

ii) The three sides of the isosceles triangle are 10 cm, 10 cm and 7 cm

∴ Perimeter = (10 + 10 + 7) cm = 27 cm

iii) An equilateral triangle with side 6 cm.

The sides of equilateral triangle are 6 cm, 6 cm and 6 cm

∴ Perimeter = (6 + 6 + 6) cm = 18 cm

Perimeter of the triangle

= (a + b + c) units

= (6 + 8 + 10) feet

= 24 feet

Area of the triangle = 1/2 × b × h sq units

1/2 × 6³ × 8 feet square

= 24 sq. feet

Answer is (A) Arithmetic Mean

Answer is (B) 12

From above table it is clear that frequency 41 is maximum and its corresponding age group of 12. So its mode will be 12.

(i) Here 5 has maximum frequency (4 times) 

so mode of data will be 5.

(ii) Here 6 has maximum frequency (3 times) 

so mode of data will be 6.

(iii) Here 2.5 has maximum frequency (4 times) 

so 2.5 will be mode of data.

Answer is (B) Value having maximum frequency

Value having maximum frequency is called mode.

Answer is (B) 12

From table it is clear that students of age 12 years have maximum frequency 41.

So mode = 12

10, 12, 14, 17, 18, 20, 21 30, 32

N = 9 

Median = \((\frac{N+1}2)^{th}\)

Median = \((\frac{9+1}{2})^{th}\) = 5th item

Therefore, the median is given by the size of the 5th items. Therefore, the Median of the data so given is 18.

Answer is (C) \(\overline { X }\) + \(\frac { n+1 }{ 2 }\) 

(i) 5 has maximum frequency (6), so mode = 5

(ii) 1.3 has maximum frequency (80), so mode = 1.3

Answer is (B) 4

In the distribution 4 has maximum frequency so mode of distribution = 4

Median = size of \((\frac{N+1}{2})^{th}\) item

Median = \(\frac{65+11}2=33^{rd}\) 

While locating the item in the Cumulative Frequency column, the item that exceeded 33rd is 43 that is corresponding to 1800. Thus, the median is 1800.

Answer is (B) \(\frac { { x }_{ i }-a }{ h }\)

Here no. of members 6 has maximum frequency 10 so mode = 6.

Median = size of \((\frac{N+1}{2})^{th}\) item

Q₁ = size of \((\frac{N+1}4)^{th}\) item

Or, Q₁ = size of \((\frac{135+1}{4})^{th}\) item

= Q₁ = 34^th item

Upper Quartile

Q₁ = size of 3 \((\frac{N+1}{4})^{th}\) item

or, Q₁ = size of 3 \((\frac{135+1}{4})^{th}\) item

= Q₁ = 102^nd item

This corresponds to 115 in the cumulative frequency. Hence, upper quartile is 30.

Answer is (C)Median

Since 17 occurs the highest number of times in the series i.e. 4 times. Hence, Mode = 17

Answer is (A) 0

7×34 is multiple of 9 

=> 7 + x + 3+ 4 is a multiple of 9 

=> 14 + x = 0, 9, 18, 27, 

=> x = -1, 4, 13, 

Since, x is a digit 

x = 4

(c) More than 50%

Correct answer is (b) zero

Correct option is: (C) 90

Let the cost price be Rs.100 

Gain = 10%

SP = \(\frac{100\,+\,Gain\%}{100}\times CP\)

= \(\frac{100\,+\,10}{100}\times100\)

= Rs.110 

Now, according to question make the selling price double 

= 110 × 2 

= Rs.220 

Now, Gain will be 

= 220 – 100 

= Rs.120

\(Gain\%=\frac{Gain\,\times100}{CP}\)

= \(\frac{120\,\times\,100}{100}\)

= 120%

Correct option is: (B) 300%

Kinetic energy , K = \(\frac{P^2}{2m}\)

When momentum p is doubled , the kinetic energy becomes

\(K^1 = \frac{(2p)}{2m}\)

\(K^1 = 4 (\frac{p^2}{2m})\)

\(K^1 = 4K\)

increases in kinetic energy = \(\frac{K^1 - K}{K}\)

\(= \frac{4K - K}{K} \times 100\)

\(\left(\frac{3K}{K}\right) \times 100\)

300%

Correct option is: (B) 300%

The given system of equations is: 

x − 3y – 3 = 0 

3x − 9y − 2 = 0 

The above equations are of the form 

a₁ x + b₁ y − c₁ = 0 

a₂ x + b₂ y − c₂ = 0 

Here, a₁ = 1, b₁ = −3, c₁ = −3 

a₂ = 3, b₂ = −9, c₂ = −2 

So according to the question, we get 

\(\frac{a_1}{a_2}\) = \(\frac{1}{3}\) 

\(\frac{b_1}{b_2}\) = \(\frac{−3}{−9}\) = \(\frac{1}{3}\) and, 

\(\frac{c_1}{c_2}\) = \(\frac{−3}{−2}\) = \(\frac{3}{2}\) 

⇒ \(\frac{a_1}{a_2}\) =\(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\) 

Hence, we can conclude that the given system of equation has no solution.

f(x) = \(\sqrt{x-1}\); f(x) = 4

(i) f(0) = 4

(ii) f(3) = \(\sqrt{3-1}\) = \(\sqrt{2}\)

(iii) f(a + 1) = \(\sqrt{a+1-1}\) = \(\sqrt{a}\)

(3) f(xy) ≤ f(x).f(y)

√(1 + x²y²) ≤ √(1 + x²) √(1 + y²)

⇒ f(xy) ≤ f(x).f(y)

Given as f(x) = x (x – 1) on [1, 2]

= x² – x

The every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. Therefore it is continuous in [1, 2] and differentiable in (1, 2). Therefore both the necessary conditions of Lagrange’s mean value theorem is satisfied.

So, there exist a point c ∈ (1, 2) such that:

f'(c) = (f(2) - f(1))/(2 - 1)

f'(c) = (f(2) - f(1))/1

f (x) = x² – x

Differentiate with respect to x

f’(x) = 2x – 1

For the f’(c), put the value of x=c in f’(x):

f’(c)= 2c – 1

For the f(2), put the value of x = 2 in f(x)

f (2) = (2)² – 2

= 4 – 2

= 2

For the f(1), put the value of x = 1 in f(x):

f (1) = (1)² – 1

= 1 – 1

= 0

∴ f’(c) = f(2) – f(1)

⇒ 2c – 1 = 2 – 0

⇒ 2c = 2 + 1

⇒ 2c = 3

c = (3/2) ∈ (1,2)

Thus, lagrange's mean value theorem is verified.

Given as f(x) = 2x – x² on [0, 1]

The every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. Therefore it is continuous in [0, 1] and differentiable in (0, 1). Therefore both the necessary conditions of Lagrange’s mean value theorem is satisfied.

So, there exist a point c ∈ (0, 1) such that:

f'(c) = (f(1) - f(0))/(1 - 0)

⇒ f’(c) = f(1) – f(0)

f (x) = 2x – x²

Differentiate with respect to x:

f’(x) = 2 – 2x

For the f’(c), put the value of x = c in f’(x):

f’(c)= 2 – 2c

For the f(1), put the value of x = 1 in f(x):

f (1)= 2(1) – (1)²

= 2 – 1

= 1

For the f(0), put the value of x = 0 in f(x):

f (0) = 2(0) – (0)²

= 0 – 0

= 0

f’(c) = f(1) – f(0)

⇒ 2 – 2c = 1 – 0

⇒ – 2c = 1 – 2

⇒ – 2c = – 1

c = -1/-2 = (1/2) ∈ (0,1)

Thus, lagrange's mean value theorem is verified.