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M = 0.2 kg

u = 4 m/s

t = 10 min = 10 x 60 sec = 600 sec

We know

a = \(\frac{v-u}t\) = \(\frac{0-4}{600}\) = 0.006 m/s²

Then

F = ma

F = 0.2 x 0.006

F = 0.0012 N

S = ut + \(\frac12\)at²

S = 4 x 600 + \(\frac12\) x \(\frac{0.006}{1000}\times600\times600\)

S = 2400 + 180

S = 2580 m

Work done

W = F. 5

W = 0.0012 x 2580

W = 3.096 j

The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. Precision of a measurement is a closeness of two or more measured values to each other.

The true value of a certain length is near 5.678 cm. In one experiment, using a measuring instrument of resolution 0.1 cm, the measured value is found to be 5.5 cm. In another experiment using a measuring instrument of greater resolution, say 0.01 cm, the length is found to be 5.38 cm. We find that the first measurement is more accurate as it is closer to the true value, but it has lesser precision. On the contrary, the second measurement is less accurate, but it is more precise.

The student has 7 choices from the morning courses out of which he can select one course in 7 ways. For the evening course, he has 5 choices out of which he can select one course in 5 ways. 

Hence, he has total number of 7 + 5 = 12 choices.

Answer is (c) 0

Answer is (b) 0°

Answer is (b) 1/6

Answer is (b) 0

Answer is (a) A,B,C are collinear

\(f(x) = \begin{cases} kx + 5 ,& When \,x\leq2.\\ x-1 ,& When\, x >2 \end{cases}\)

Left hand limit of function f(x) at = 2 is f(2⁻) = \(\lim\limits_{x \to2^-}\)f(x) = \(\lim\limits_{x \to2}\)(kx + 5) = 2k + 5.

Right hand limit of function f(x) at = 2 is f(2⁺) = \(\lim\limits_{x \to2^+}f(x)\) = \(\lim\limits_{x \to2}\)(x-1) = 2 - 1 =1.

Since, we know that if function f(x) is continuous at = 2, then left hand limit of function f(x) at x = 2 is equals to the right hand limit of function f(x) at = 2 which is also equals to f(2). 

i.e., f(2 ⁻) = f(2 ⁺) = (2). 

Therefore, 2k + 5 = 1 

⇒ 2k = −4 

⇒ k = −2.

Hence, k = -2 for which the function \(f(x) = \begin{cases} kx + 5 ,& When \,x\leq2.\\ x-1 ,& When\, x >2 \end{cases}\) is continuous at x = 2.

Since,

Cards marked with numbers 13, 14, 15, ….60. 

Therefore,

Total number of cards = 60-12 = 48 

Perfect squares lie in the sequence 13, 14…...,60 are 16, 25, 36, 49. 

Total number of perfect squares lies in 13 – 60 range = 4. 

Hence,

Probability of getting perfect squares 

= \(\frac{Total\,perfect\,squares\,in\,the\,range\,13-60}{Total\,number\,of\,cards}\) 

= \(\frac{4}{48}\) 

= \(\frac{1}{12}\).

Correct answer is: (b) 1/3

P(even composite no) =2/6

=1/3

Let X denote expected amount, which takes the values X : 5 and - 2. With respective probabilities as:

P(x = 5) = P(getting muItiple of 3)

= 2/6 ; {i.e. 3,6}

P(x = 2) = 4/6; {i. e.1,2,4,5}

The probability of distribution is

Expected amount: E(X) = ΣX.P(X)

= 5 x 2/6 -2 x 4/6 = 1 - 67 - 1.33 

= 0.34 i.e. 34 paise gain.

A non - leap year contains 365 days 52 weeks and 1 day more.

(i) We get 53 sundays when the remaining day is Sunday.

Number of days in the week = 7

∴ n(S) = 7

Number of ways getting 53 Sundays. n(E) = 1

∴ P(E) = n(E)/n(S) = 1/7

∴ Probability of getting 53 Sundays = 1/7

(ii) Probability of getting 52 Sundays P(vector E) = 1 – P(E)

= 1 – 1/7 = 6/7.

Correct option   (B) a resident female with annual income Rs 9 lakh

Explanation:

Resident female in between 8 to 10 lakhs haven’t been mentioned. 

Let E₁ = one children family 

 E₂ = two children family and 

 A = picking a child then by Baye’s theorem, required probability is 

P(E₂/A) = (1/2x)/(1/2 . x/2 + 1/2 .x) = 2/3 = 0.667

(iv) The sum of the probabilities of all elementary events of an experiment is  Sum of probabilities of an experiment = 1.

Correct option (D) 2006

Explanation:

2004, (imports - exports)/exports = 10/70 = 1/7

2005,26/76 = 2/7

2006, 20/100 = 1/5

2007, 10/100 = 1/11

Correct option (D) MN = NM

Explanation:

Matrix multiplication is not commutative in general.

Correct option (B) 1/3

Concept:

• The ways of arranging n different things = n!
• The ways of arranging n things, having r same things and rest all are different = \(\rm n!\over r!\)
• The no. of ways of arranging the n arranged thing and m arranged things together = n! × m!
• The number of ways for selecting r from a group of n (n > r) = nCr 
• To arrange n things in an order of a number of objects taken r things = nPr  

Calculation:

For no consonants to be together the word can be formed as 

'1' A '2' I '3' O '4', where 1, 2, 3, 4 are the places where 2 consonants P and N could be placed

∴ Vowels arranged in = 3! ways

Ways of arranging consonants = ⁴P₂

The total number of ways formed N = 3! × ⁴P₂ 

N = 6 × 12 = 72

Explanation:

• Positional Power—Result of the manager's position within organization. 
• Legitimate power—Manager's position within the organization and the authority that lies with that position.
• Reward Power—People in power are after able to give out rewards. Raises, promotions, desirable assignments, training opportunities and simple compliments—these are all examples of rewards controlled by people "in power". 
• Coercive Power—This source of power is also problematic and can be abused. Threats and punishment are common coercive tools.
• Informational Power—Having control over information that others need or want puts you in a Powerful position.
•  Personal Power—Is a source of influence and authority a person has over his or her followers. 
• Expert Power—Power lay virtue of knowledge.
• Referent Power—Power bestowed by virtue of love and respect.
• Expert and referent power are by willful acceptance of others and therefore do not require either to be pre-justified or to be post-justified others are by virtue of perception of the powerful and may not be acceptable to others. 

Let coordinates of a point P changes from (x, y) to (X, Y) in new coordinate axes whose origin has the coordinates h = 3, k = -1. Therefore, we can write the transformation formulae as x = X + 3 and y = y – 1.

Substituting, these values in the given equation of the straight line, we get 2(X + 3) – 3 (Y – 1) + 5 = 0 or 2X – 3Y + 14 = 0. Therefore, the equation of the straight line in new system is 2x – 3y + 14 = 0.

Put x = X + 1, y = Y + 1 in the given equation.

We get

2(X + 1)² + 4(X + 1)(X + 1)(Y + 1) + 3 (Y + 1)² = 0

⇒ 2X² + 4xy + 3Y² + 8X + 10Y + 9 = 0

Let (X, Y) be the new coordinates of the point (x, y).

=> x = X + 3, y = Y + 4

The transformed equation is 2(X + 3)² + 4(X + 3)(Y + 4) + 5(Y + 4)² = 0

=> 2(X² + 6X + 9) + 4(XY + 4X + 3Y +12) + 5(Y² + 8y + 16) = 0

=> 2X² + 12X + 18 + 4XY + 16X + 12Y + 48 + 5Y² + 40Y + 80 = 0

=> 2X² + 4XY + 5Y² + 28X + 52Y + 146 = 0

Any one of the 5 persons can leave the cabin in 7 ways independent of other.
Hence the required number of ways = 7 × 7 × 7 × 7 × 7 = 75.

Each of the first three questions can be answered in 4 ways and each of the next three questions can be answered in 5 different ways.
Hence, the required number of different sequences of answers = 4 × 4 × 4 × 5 × 5 × 5 = 8000.

Answer is:

SMTP

Answer is:

SHOW TABLES

Disadvantages of lock stitch machine are:

1. It leaves unfinished seam allowances 

2. Undesirable in fabrics that rave easily 

3. The operation need to be stopped frequently to rewind the bobbin.

A process by serging machine, it sews the fabric together, cuts off the fabric to make a smooth edge and wraps the thread around the edge in one operation.

Early intervention and education for children. An inclusive curriculum for early childhood programs must support the right of all children, regardless of their abilities to participate enthusiastically in natural settings within their communities. Natural settings include, but are not limited to: home, preschool, nursery school. 

1. Putting the emotional needs of the students in the forefront 

2. The most fruitful way of building an inclusive learning environment comes from establishing meaningful connections with our students. 

3. Students must be engaged with information in a responsive learning environment that provides rich ways for students to progress from a very early stage 

4. Teachers need to use multiple approaches of learning

Concept:

Combinations: The number of ways in which r distinct objects can be selected simultaneously from a group of n distinct objects, is:

nCr = \(\rm \frac {n!}{r!(n-r)!}\).

Permutations: The number of ways in which r objects can be arranged in n places (without repetition) is:

ⁿP_r = \(\rm \frac{n!}{(n - r)!}\).

• nPr = nCr × r!.

• n! = 1 × 2 × 3 × ... × n.

• 0! = 1.

Calculation:

There are 26 letters in the English alphabet. If we separate the group (a, some 5 letters, b), we will be left with 19 more letters.

These 20 objects (1 group + 19 letters) can be arranged among themselves in 20! ways.

Since either a or b can be at the beginning or the end of the group of 7 letters (a, some 5 letters, b), the number of possible arrangements of the group  will be 2 × (¹P₁ × ⁵P₅ × 1P1) = 2 × 5!.

Also, each group of 5 letters can be selected from the remaining 24 letters (except a and b) in ²⁴C₅ ways.

Required total number of ways = (2 × 5! × 24C5) × 20!

= 2 × 24P5 × 20!.