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Answer is (b) 2

Life expectancy of new born baby is called ‘longevity’.

A secular trend is a variable that evidences a consistent pattern within a given period of time. It is a statistical tendency that can be easily identified and it is not subject to seasonal or cyclical effects.

Answer is (d) (x - y)(y - z)(z - x)

Type I Error is taking a wrong decision to reject the null hypothesis when it is actually true. 

Type II Error is taking a wrong decision to accept the null hypothesis when it is actually not true.

The coordinates of the point are (0,0,0)

(z₁², + z₂²,) ~ χ²₍₂₎ d.f. ie; is a chi-square distribution with n = 2 degress of freedom mean = n = 2.

Answer is (c) cot^-1 1/x

The method of least squares gives the most satisfactory measurement of the secular trend in a time series when the distribution of the deviations is approximately normal. The least-squares estimates are unbiased estimates of the parameters.

It is a chi-square distribution with 2 d.f. Mean = n = 2.

• The assumptions made in interpolation and extrapolations are: 

• There are no sudden jumps in the values of dependent variable(Y) from one period to another(X). 

• The rate of change of figures (Y) from one period to another(X) is uniform. 

• There will be no consecutive missing values in the series.

In the typical application of the Bernoulli distribution, a value of 1 indicates a "success" and a value of 0 indicates a "failure", where "success" refers that the event or outcome of interest. The parameter p in the Bernoulli distribution is given by the probability of a "success".

p(x) = p^x q^{1 – x} ; x = 0,1, : p(x) = 0.4^x 0.6^{1 – x} ; x = 0,1

Here = 1 – p = 1 – 0.4 = 0.6

Answer is (d) π/3

Answer is (b) (x - 3)/2

Answer is (d) One-one

Answer is (a) one-one

Answer is (a) One-One

n is the parameter, Mean = 0, Range = (-∞, ∞)

‘While estimating the unknown parameter, if a specific value is proposed as an estimate, which is called Point estimation’. ‘While estimating the unknown parameter instead of a specific value, an interval is proposed, which is likely to contain the parameter is called Interval estimation’.

Set of all admissible values of parameter is called parameter space.

Given k = 10 ΣX̄ = 256.8 , Σ R = 7.85 , n = 5

Here standards are not given, 5 im

control limits for X̄ = charts are ,

C.L = X̄ = 25.68

L.C.L = X̄ – A₂R̄= 25.68 

= 0.58 x 0.785 = 25.22

U.C.L = X̄ + A₂R̄ 

= 25.68 – 0.58 x 0.785 

= 26.14

correct option: (a) 0

H₀: μ = 50,.

Estimation of unknown parameter by proposing a specific value as an estimate is called point estimation. While estimating the unknown parameter by proposing an interval, which is likely to contain the parameter is called interval estimation.

It is the method of controlling the quality of the products using statistical technique.

1. It is used in interval estimation, to write down the confidence intervals. 

2. It is used in testing of hypothesis to test whether the difference between the sample statistic and the population parameter is significant or not.

By N.W.C.R first allocation is made in the cell (1,1) as: Xu 

= Min(a, b) = m, n (500,400) 

= 40, and replace 500 by (500-400)= 100
Next allocation is made at (1,2)

Controlling the quality of the product during the manufacturing process itself is the Process control. Controlling the quality of the finished products/ manufactured products is called product control.

(b) 1/4tan^-1 x/4+k

Correct option (B)

1. Matrix-Minim method and 

2. North-west corner rule.

Option: (c) Symmetric 

correct option: (a) π/4

option: (b) e^sinx

Let least even number is 2n - 4.

Therefore 5 consecutive even numbers are 2n - 4, 2n - 2, 2n, 2n + 2 and 2n + 4.

\(\therefore\) Their sum = (2n - 4) + (2n - 2) + 2n + (2n + 2) + (2n + 4)

 = 10 n

\(\therefore\) 10n = 80 (\(\because\) Given that sum of 5 consecutive even numbers is 80)

⇒ n = 80/10 = 8

\(\therefore\) least even number = 2n - 4 = 2 x 8 - 4 = 16 - 4 = 12

Let \(\sqrt{-7+21i}\) = x + iy

⇒ (x + iy)² = –7 + 21i  (By squaring both sides)

⇒ x² + i²y² + 2ixy = –7 + 21i (\(\because\) (a + b)² = a² + b² + 2ab)

⇒ x² – y² + 2xyi = –7 + 21i  (\(\because\) i² = –1)

⇒ x² – y² = –7 .........(1)

And 2xy = 21 ..........(2)     

(By comparing real and imaginary part of complex numbers)

\(\because\) (x² + y²)² = (x² – y²)² + 4x²y²    (\(\because\) (a + b)² = (a – b)² + 4ab)

= (–7)² + (21)²   (From equation (1) and (2))

= 49 + 441

= 490

⇒ x² + y² = \(\sqrt{490}\) \(=7\sqrt{10}\) .......(3)

By adding equations (1) and (3), we get

2x² = –7 + \(7\sqrt{10}\) 

⇒ x² = \(\frac{7}{2}\)(–1 + \(\sqrt{10}\))

⇒ x \(=\pm\sqrt{\frac{7}{2}(-1+\sqrt{10})}\)  (\(\because \) \(\sqrt{10}\) > 1 ⇒ x is a real number)

By putting the value of x² in equation (3), we get

y² = \(7\sqrt{10}-\mathrm x^2\)

\(=7\sqrt{10}-\frac{7}{2}(-1+\sqrt{10})\)

\(=\frac{14\sqrt{10}+7-7\sqrt{10}}{2}\)

\(=\frac{7(\sqrt{10}+1)}{2}\)

⇒ y \(=\pm\sqrt{\frac{7}{2}(\sqrt{10}+1)}\)

\(\because \) xy = \(\frac{21}{2}> 0\)   (From equation (2))

\(\therefore\) x & y have same sign.

\(\therefore\) If x \(=\sqrt{\frac{7}{2}(-1+\sqrt{10})}\) then y \(=\sqrt{\frac{7}{2}(\sqrt{10}+1)}\)

Or

If x \(=-\sqrt{\frac{7}{2}(-1+\sqrt{10})}\) then y \(=-\sqrt{\frac{7}{2}(1+\sqrt{10})}\)

Hence, \(\sqrt{-7+21i}\) \(=\sqrt{\frac{7}{2}(\sqrt{10}-1}\) \(+i\sqrt{\frac{7}{2}(\sqrt{10}+1)}\)

\(=\sqrt{\frac{7}{2}}(\sqrt{\sqrt{10}-1}+i\sqrt{\sqrt{10}+1})\)

or  \(\sqrt{-7+21i}\) \(=-\sqrt{\frac{7}{2}(\sqrt{10}-1)}-i\sqrt{\frac{7}{2}\sqrt{10}+1}\)

\(=-\sqrt{\frac{7}{2}(\sqrt{10}-1)}+i\sqrt{\sqrt{10}+1}\)

Hence, the square root of –7 + 21i is \(\sqrt{\frac{7}{2}}\left(\sqrt{\sqrt{10}-1}+i\sqrt{\sqrt{10}+1}\right)\) or \(-\sqrt{\frac{7}{2}}\left(\sqrt{\sqrt{10}-1}+i\sqrt{\sqrt{10}+1}\right)\).

Angles of quadrilateral are,

(4x)°, 5(x+2)°, (7x – 20)° and 6(x+3)°

∴ 4x + 5(x + 2) + (7x - 20) + 6(x + 3) = 360°

4x + 5x + 10 + 7x - 20 + 6x + 18 = 360°

22x + 8 = 360°

22x = 360° - 8°

22x = 352°

x = 16°

Hence angles are,

(4x)° = (4 × 16)° = 64°

5(x + 2)° = 5(16 + 2)° = 90°

(7x - 20)° = (7 × 16 - 20)° = 92°

6(x + 3)° = 6(16 + 3) = 114°

Yes, it is a polynomial.
x(x-1)+x=x²-x+x=x²
As the above expression has a positive integer exponent for x, we can say that it is a polynomial.

Favourable outcome (Getting a tail) =1
Total number of outcomes [H,T]=2
Probability = 1/2

Concept:

A natural number is a number that occurs commonly and obviously in nature. As such, it is a non-negative number. The set of natural numbers can be denoted by

N = {1, 2, 3, 4,....}

The set of natural numbers is bounded below and not bounded above in R.

We can prove not bounded above in R using contradiction.

Proof:

Assume by way of contradiction that N is a bounded above. Then, since N is not empty, it follows from the completeness axiom that sup(N) exists. Thus there must be m ∈ N such that

sup(N) - 1 < m (sup(N) means supremum of N or least upper bound)

⇒ sup(N) < m + 1

As m ∈ N , also m + 1 ∈ N, Which is a contradiction.

∴ N is not bounded above

Concept:

Logarithm Rule

log mⁿ = n log m

Calculations:

Let x, y, z are three consecutive positive integers.

⇒ y = x + 1 and z = y + 1

⇒ z = x + 2

Consider, log (1 + xz)

= log [1 + x(x+2)]

= log [1 + x² + 2x]

= log (1 + x)²

= 2 log (1 + x)

= 2 log y

Hence, If x, y, z are three consecutive positive integers, then log (1 + xz) is 2 log y

f(x) is continuous only at one integral value of x which is x=1

f(x)=[x²]−[x]²
 
at x →1⁻f(x)=0
at x →1⁺f(x) =1−1=0
so continuous at 1

We know that, for a relation to be transitive,

(x,y) ∈ R and (y, z) ∈ R then (x,z) ∈ R.

Here, (1,2) ∈ R and (2,1) ∈ R but (1,1) ∈ R.

R is transitive.

\(\int_0^2 \frac{5x+4}{x^2+4}dx\)

= \(\frac{5}{2}\)\(\int_0^2 \frac{2x}{x^2+4}dx\) + 4 \(\int_0^2 \frac{1}{x^2+4}dx\) 

= \(\frac{5}{2}\)\([log(x^2+4)]_0^2\) + \(\frac{4}{2}[tan^{-1}\frac{x}{2}]_0^2\) 

= \(\frac{5}{2}\)(log 8 - log 4) + 2 (tan^-11 - tan^-10)

= \(\frac{5}{2}\)log \(\frac{8}{4}\) + 2(\(\frac{\pi}{4}-0\))

= \(\frac{5}{2}\) log 2 + \(\frac{\pi}{2}\)

Mass number= 17+ 18 =35. Q. The atom of an element X contains 17 protons, 17 electrons and 18 neutrons whereas the atom of an element Y contains 11 protons, 11 electrons and 12 neutrons.

no of electrons = 17

No of nuetrons = 18

no of protons = no of electrons

So no of protons = 17

Atomic number = no of electrons = no of protons

So atomic number = 17

Mass number = atomic no + no of nuetrons

So

Mass numb

er= 17+ 18 =35

Element of nucleus :

Protons:17

Neutron:18

So,  

=>Mass No. =17+18 = 35

And ,

=>No.  of electrons =17.  (as no. of electrons =no. of protons) 

And, 

=>Atomic No.  = 17

=>Name of element = Cl(chlorine)  

Hope it helps u.... ✌