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The decimal number 27.1875 to its binary equivalent is 11011.0011.

The marginal revenue is 66.

Z = 10.50 at x = 3, y = 3

Let the number of units of nuts x & bolts be y it takes 1 hour on machine A for a nut and 3 hours on machine 13 for a nut maximum time available for machine A is 12 hours. 

∴ x + 3y ≤ 12 

3x + y ≤ 12 

Negative units cannot be produced: x, y ≥ 0 our objective is to maximum profit the LPP is Maximize = 2.5 x + y subject to 

x + 3y ≤ 12 

3x + y ≤ 12 

and x ≥ 0, y ≥ 0

sin x + sin 3x + sin 5x = 0 

∴ (sin 5x + sin x) + sin 3x = 0

∴ 2sin \((\frac{5x+x}{2})\) cos \((\frac{5x-x}{2})\) + sin 3x = 0

∴ 2sin 3x cos 2x + sin 3x = 0

∴ sin 3x (2cos 2x + 1) = 0

∴ sin 3x = 0 or 2cos 2x + 1 = 0

∴ sin 3x = 0 or 2cos 2x = \(-\frac{1}{2}\)

∴ sin 3x = 0 or cos 2x = - cos \(\frac{π}{3}\) = cos \((π-\frac{π}{3})\)

∴ sin 3x = 0 or cos 2x = cos \(\frac{2π}{3}\)

Since, sin θ = 0 implies θ = nπ and cos θ = cos α implies θ = 2nπ ± α , n ∈ Z.

∴ 3x = nπ or 2x = 2mπ ± \(\frac{2π}{3}\), 

∴ the required general solution is x = \(\frac{nπ}{3}\) or x = mπ ± \(\frac{π}{3}\) where n, m ∈ Z.

Let  sin^–1 (\(\frac{1}{\sqrt2}\)) = x

∴ sin x = \(\frac{1}{\sqrt2}\)

∴ sin x = -sin \(\frac{π}{4}\)

The principal value branch of sin^–1 x is [\(-\frac{π}{2},\) \(\frac{π}{2}\)] and \(-\frac{π}{2}\) ≤ \(-\frac{π}{4}\) ≤ \(\frac{π}{2}\)

Hence, the required principal value of x is\(-\frac{π}{4}\)

(A) 2x - y = 0

Equation of the curve is y = x² + 4x + 1 

Differentiating w.r.t. x, we get

\(\frac{dy}{dx} = 2x + 4\)

∴ Slope of tangent at (-1, -2) is

\((\frac{dy}{dx})_{(-1,-2)}\) = 2(-1) + 4 = -2 + 4 = 2

Equation of tangent is y - y₁ = \((\frac{dy}{dx})_{(x_1,y_1)}\) (x-x₁)

Here, (x₁, y₁) ≡ (-1, -2)

∴ [ y - (-2)] = 2[x - (-1)] 

∴ y + 2 = 2(x + 1) = 2x + 2

∴ 2x - y = 0

(C) 0.8

n = 10, E(X) = 8      ....(given) 

But, E(X) = np 

∴ 8 = 10 (p)

∴ p = \(\frac{8}{10}=\frac{4}{5} = 0.8\)

Given linear programming is objective function is

Minimize L = 4x + 4y + z

subject to x + y + z \(\leq\) 2

2x + y \(\leq\) 3

x \(\geq\)0, y \(\geq\) 0, z \(\geq\) 0

Converts given minimization problem into maximize problem:

(Put -x in place of x, -y in place of y & -z in place of z in objective function)

Then objective function is 

Maximize - L = -4x - 4y - z

subject to x + y + z \(\leq\) 2

2x + y \(\leq\) 3

x \(\geq\) 0, y \(\geq\) 0, z \(\geq\) 0

Converts inequalities into equations, we get

x + y + z + x₁ + 0x₂ = 2

2x + y + 0z + 0x₁ + x₂ = 3

x, y, z, x₁, x₂ \(\geq\) 0

First simplex table

\(\because\) All Z_j - C_j \(\geq\) 0

And x, y, z are not present in basic variable

\(\therefore\) x = 0, y = 0, z = 0 is the solution of redical maximize linear/programming problem.

\(\therefore\) Max(-L) = -4 x 0 - 4 x 0 - 0 = 0

\(\therefore\) Min L = Max(-L) = 0

Hence, given linear programming is minimized at x = 0, y = 0, z = 0 and minimum value of objective function be 0.

sin x = tan x

∴ sin x = \(\frac{sin\,x}{cos\,x}\)

∴sin x cos x - sin x = 0

∴ sin x (cos x - 1) = 0

∴ sin x = 0 or cos x = 1 

∴ sin x = sin 0 or cos x = cos 0 

Since, sin θ = 0 implies θ = nπ and cos θ = cos α implies θ = 2nπ ± α , n ∈ Z.

∴ x = nπ or x = 2mπ ± 0

∴ The required general solution is x = nπ or x = 2mπ, where n, m ∈ Z.

Given, \(\displaystyle\sum_{i=1}^{n}\) p_i = 1

∴ k + 2k + 3k + 4k + 5k = 1

∴ 15k = 1

∴ K = \(\frac{1}{15}\)

∴ P(x ≤ 4) = P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)

= \(\frac{1}{15}\) + \(\frac{2}{15}\) + \(\frac{3}{15}\) + \(\frac{4}{15}\)

= \(\frac{10}{15}\)

= \(\frac{2}{3}\)

(r ∧ q) ↔ ~ p

≡ (T ∧ F) ↔ ~ T

≡ (T ∧ F) ↔ F

≡ F ↔ F

≡ T

Hence, the truth value is ‘T’

Since, there are six patients

n = 6 

p = P(success) = 0.5 and q = 1 - p = 0.5 

X ~ B(n = 6, p = 0.5) 

The p.m.f. of x is given as P(X = x) = ⁿC_x(p)^x (q)^n-x

P(X = x) = P(x) = ⁶C_x (0.5)^x (0.5)^6-x, 

x = 0, 1, 2, 3, 4, 5, 6

a. P(none will recover) 

P(X = 0) = ⁶C₀ (0.5)⁰ (0.5)^6-0 

= (1) (1) (0.5)⁶

= 0.015625

b. P(half of them will recover)

P(X = 3) = ⁶C₃ (0.5)³ (0.5)^6-3 

= \(\frac{6!}{3!.3!}\) (0.5)³ (0.5)³ 

= \(\frac{6\times5\times4\times3!}{3\times2\times1\times3!}\)(0.125) (0.125) 

= 20 x 0.015625 

= 0.3125

Answer is: (B) \(\frac{1}{\sqrt{10}}\)

s = \(\frac{a+b+c}{2}=\frac{18+24+30}{2}=36\)

sin \(\frac{A}{2} = \sqrt \frac{(s-b)(s-c)}{bc}=\sqrt\frac{(36-24)(36-30)}{24\times30}=\sqrt\frac{12\times6}{24\times30}=\frac{1}{\sqrt{10}}\)

Given:

Given values are 2, 2, 3, 4, 5, 5, 7, 8

Formula used:

Mean = (Sum of values)/(Number of values)

Calculations:

Mean = (2 + 2 + 3 + 4 + 5 + 5 + 7 + 8)/8

⇒ 36/8

⇒ 4.5

∴ The mean of 2, 2, 3, 4, 5, 5, 7, 8 is 4.5

(C)

sin^-1 (1-x) -2 sin^-1 x = \(\frac{π}{2}\)

Let x = sin θ

∴ sin^-1 (1 - sin θ) - 2 sin^-1 (sin θ) = \(\frac{π}{2}\)

∴ sin^-1 (1 - sin θ) - 2θ = \(\frac{π}{2}\)

∴ sin^-1 (1 - sin θ) = \(\frac{π}{2}\) + 20

∴ 1 - sin θ = sin (\(\frac{π}{2}\) + 20)

∴ 1 - sin θ = cos 2θ    ...[∵ sin (\(\frac{π}{2}\) + θ) = cons θ]

∴ 1 - sin θ = 1 -2sin² θ

∴ 2sin²θ - sin θ = 0

∴ sin θ (2 sin θ - 1) = 0

∴ sin θ = 0 or sin θ = \(\frac{1}{2}\)

∴ x = 0 or x = \(\frac{1}{2}\)

For x = \(\frac{1}{2}\), sin^-1 (1 - x) -2 sin^-1 x = sin^-1 (1 - \(\frac{1}{2}\)) - 2sin^-1 (\(\frac{1}{2}\)) 

sin^-1(\(\frac{1}{2}\)) - 2 sin^-1(\(\frac{1}{2}\)) = - sin^-1(\(\frac{1}{2}\)) = - \(\frac{π}{6} \) ≠  \(\frac{π}{2}\)

∴ x ≠ \(\frac{1}{2}\)

∴ x = 0

Given:

Sample size = 60

\(\sum x^2=18000, \ \sum x=960\)

Formula used:

\({\rm{Variance\;}} = {\rm{\;}}\frac{1}{{\rm{n}}}\sum {{\rm{x}}^2}\; - \;{\left( {\frac{{\sum {\rm{x}}}}{{\rm{n}}}} \right)^2}\)

Where, 

n = sample rate

Calculation:

Variance

⇒ (1/60) × 18000 - (960/60)²

⇒ 300 - (16)²

⇒ 300 - 256

⇒ 44

∴ The variance is 44.

(A) 

Equations of the co-ordinate axes are x = 0 and y = 0. 

Equations of the lines passing through the point (2, 3) and parallel to the co-ordinate axes are x = 2 and y = 3 

i.e., x - 2 = 0 and y - 3 = 0.

∴ the joint equation of these lines is 

(x - 2) (y - 3) = 0 

∴ xy – 3x – 2y + 6 = 0

 \(AB = \begin{bmatrix} {1} & {2} & 3 \\[0.3em] {1} & -2 & {-3} \end{bmatrix}\)\(\begin{bmatrix} 1 & -1 \\[0.3em] 1 & 2 \\[0.3em] 1 & -2 \end{bmatrix}\)= \(\begin{bmatrix} 6 & -3 \\[0.3em] -4 & 1 \\[0.3em] \end{bmatrix}\)

Consider, |AB| = \(\begin{bmatrix} 6 & -3 \\[0.3em] -4 & 1 \\[0.3em] \end{bmatrix}\) =(6x1) - (-4 x -3) = 6 - 12 = - 6 ≠ 0

∴ AB is a non-singular matrix

∴ (AB)^-1 = \(\frac{1}{|AB|}\) (Adj AB) = \(\frac{1}{-6}\) \(\begin{bmatrix} 1 & 3 \\[0.3em] 4 & 6 \\[0.3em] \end{bmatrix}\)

Let \(\overline {a}\) = \(3\hat i + 2\hat j +\hat k\) , \(\overline {n}\) \(4\hat i + 3\hat j +2\hat k\)

Vector equation of the plane passing through the point A (\(\overline {a}\)) and perpendicular to the

vector \(\overline {n}\) is \(\overline {r} .\overline{n}=\overline {a}.\overline{n}\)

∴ \(\overline {r}\) .  (\(4\hat i + 3\hat j +2\hat k\)) = (\(3\hat i + 2\hat j +\hat k\)) . (\(4\hat i + 3\hat j +2\hat k\))

= 3(4) - 2(3) + 1(2)

= 12 - 6 + 2

∴ \(\overline {r}\) .  (\(4\hat i + 3\hat j +2\hat k\)) = 8,

which is the vector equation of the plane

R= {(2, 2), (2, 4), (3, 3), (4,4)}.

No of ways of selecting one vowel out of six vowels (3A’S, 21’s, 10’s) = ⁶C₁ = 6.

P(1 vowel) = ⁶C₁/¹³C₁ = 6/13

 mean = 50

\(\therefore\) \(\cfrac{3480+30a+70b}{120}\) = 50

30a + 70b = 6000 – 3480

30a + 70b = 2520 (1)

17 + 32 + 19 + a + b = 120 

68 + a + b = 120,

a + b = 52 (2) 

(2) x 30 => 30a + 30b = 1560

30a + 70b = 2520,

30a + 30b = 1560,

40b = 960,

b = 960/40 = 24,

a + 24 = 52

a = 52 – 24 = 28,

\(\therefore\) a = 28, b = 24

Total number of babies = 70 

Half =35

So median is the weight of 35 baby. 

The weight of 35 111 child be in between 29 and 40 placed child, its weight will be 3 kg.

\(\therefore\) Median of the weight = 3 kg.

Mean = \(\cfrac{6.10+6.20+6.18+6.20+6.25+6.21+6.15+6.10}{8}\)

= \(\cfrac{49.39}{8}\)= 6.17

If distances are written in ascending order 6.10, 6.15, 6.18, 6.20, 6.21, 6.25 Median = \(\cfrac{6.18+6.20}{2}\) = 6.19 m

Mean and median are gives the average distance covered by a person. Hence they will not have much difference..

Half of the number of houses = 20

We have to find the electricity usage of the 20th house. According to this, we can divide

Total number of workers = 30 

Half= 15

So median is the wage of 15th worker. 

The daily wages between the place 11 and 18 = 700 rupees

Median of daily wages = 700 rupees

Write the number in order.

2100, 2100, 2200, 2300, 2500, 2800, 3300, 3300, 3500 (1)

Median = 2500

Mean = Total amount of rain / No. of districts

 = 700/14 = 50

In ascending order: 13.5, 20.5, 30.5, 33.5, 45.5, 50.3, 53.4, 56.3, 56.4, 56.9, 56.9, 66.7, 70.6, 89

Median = \(\cfrac{53.4+56.3}{2}\) = 54.85

The mean is less than median because the number contains are far small and large numbers than mean.

 Total number of workers = 30 

Half = 15

So median is the wage of 15th worker. The daily wage between the place 10 and 18 = 600 rupees

Median of daily wage = 600 rupees

a. Mean = 5 

b. Median = 3.5 

Suitable average median = 3.5

Let a, a+d, a+3d, a+4d are the numbers of an arithmetic sequence, then median =

\(\cfrac{a+a+4d}{2}\) = \(\cfrac{2a+4d}{2}\)= a+2d

Median will be the term which is at center = a + 2d

\(\therefore\) A set of numbers in arithmetic sequence, the mean and median are equal.

In the table monthly income up to 18^th place be 6000 rupees. That is 18^th place family also includes in the middle of total number of families.

\(\therefore\) Median of the income = Rs. 6000

a. Mean = \(\cfrac{sum}{number} = \cfrac{248000}{10} = 24800\)

b. 9 families have monthly income less than the mean income. So in this situation this is not a suitable average.

A null hypothesis is a hypothesis that says there is no statistical significance between the two variables in the hypothesis. In the example, Susie's null hypothesis would be something like this: There is no statistically significant relationship between the type of water I feed the flowers and the growth of the flowers.

1. aren’t I? 

2. haven’t you? 

3. wasn’t it? 

4. Isn’t she? 

5. didn’t she? 

6. arent you?

7. mustn’t I? 

8. wasn’t he? 

9. did I?

10. have you? 

11. don’t you?

12. aren’t we?

13. will It?

14. did you?

15. didn’t she?

16. weren’t they?

1. do you? 

2. wasn’t it?

3. shan’t I? 

4. won’t they?

5. doesn’t he?

6. isn’t it? 

7. doesn’t he? 

8. aren’t)? 

9. weren’t we?

10. am I?

Correct option is A) bargain

1. I do not recollect of ever seeing my mother by the light of the day. 

2. He was a cruel man hardened by a long life of slave holding. 

3. Before I go, is there anything I can do for you?

OR

Is there anything I can do for you before I go?

4. I thought you picked fruit for a living.

5. Nicola’s smile was steady and engaging.

6. I felt I could not bear to Intrude upon this happy family party.

7. Nicola was glaring at his younger brother in vexation.

8. What struck one most was their unremitting willingness to work. 

9. Have I not told you to keep away from her? 

10. The Goddess stood before me.

1. isn’t it?

2. aren’t I?

3. was he?

4. can you?

5. can it?

6. doesn’t it?

7. doesn’t it?

8. Isn’t it?

9. wasn’t he?

10. didn’t he?

1. I ate a full meal consisting of chapatis and meat curry.

2. Babar Ah teaches his students under the open sky.

3. Babar All gives lessons just the way he has heard from his teachers.

4. Ba bar Ahi started his school at the mere age of nine.

5. The teaching staff of nine is made up of high school volunteers.

6. Babar Airs’ tale is a testament to the difference that one person can make in his or her world.

7. Isn’t it high time that we be the change that we want to see In this world?

8. He sprang off and ran up the bill like a buck.

9. My mother was named Harriet Bailey.

10. I have no accurate knowledge of my age.

1. Learning to write an address in English was considered great education there.

2. There stood a fair complex, oned man, six foot tall, with a red turban and white trousers.

3. You have done a great deed.

4. The owner gave a loud guffaw startling everyone around.

5. Babar Ah is the first member in his family to get a proper education. 

6. Nasiruddiri Sheikh believes that education Is a man’s true religion. 

7. Babar Ah took the initiative of starting his own school.

8. Peace Is costly but is worth the expense.

9. Krishna bid goodbye to his job as a rickshaw driver.

10. I have some vague notions about human beings.

Correct option is C) exactly even

1. She has promised to wait till she sees me back there.

2. I am the Goddess of the lake and that river Veda is my plaything. 

3. My mother was of a darker complexion than either my grandmother or grandfather.

4. I do not remember to have ever met a slave who could tell of his birthday. 

5. They greeted us with friendly faces.

6. You must be saving to emigrate to America. 

7. There is every hope that she will walk and sing again.

8. I looked at the people around me. 

9. I called my son in and shut the door of my hut.

10. What have I said to offend you so much?

Correct option is E) were/would help

Correct option is C) we would learn three English songs

Correct option is A) prevent it at the start