HSC Commerce (English Medium) १२ वीं कक्षा - Maharashtra State Board Question Bank Solutions for Mathematics and Statistics

Identify the regression equations of X on Y and Y on X from the following equations :
2x + 3y = 6 and 5x + 7y – 12 = 0 

Using the rules of negation, write the negatlon of the following: 

(a) p ∧ (q → r)

(b)  ~P ∨ ~q

Write the Truth Value of the Negation of the Following Statement :

The Sun sets in the East. 

Find the feasible solution for the following system of linear inequations:
0 ≤ x ≤ 3, 0 ≤ y ≤ 3, x + y ≤ 5, 2x + y ≥ 4

Write the truth value of the negation of the following statement : 

cos² θ + sin² θ = 1, for all θ ∈ R 

If Σ_x1 = 56 Σ_y1 = 56, Σ`x_1^2` = 478,
Σ`y_1^2` = 476, Σ_x1y1 = 469 and n = 7, Find
(a) the regression equation of y on x.
(b) y, if x = 12.

Find graphical solution for following system of linear inequations :
3x + 2y ≤ 180; x+ 2y ≤ 120, x ≥ 0, y ≥ 0
Hence find co-ordinates of corner points of the common region.

Compute the product moment coefficient of correlation for the following data: 
n = 100, `bar x` = 62, `bary` = 53, `sigma_x` = 10, `sigma_y` = 12

`Sigma (x_i - bar x) (y_i - bary) = 8000`

Information on v:ehicles [in thousands) passing through seven different highways during a day (X) and number of accidents reported (Y) is given as follows :   

`Sigmax_i` = 105, `Sigmay_i` = 409, n = 7, `Sigmax_i^2` = 1681, `Sigmay_i^2` = 39350 `Sigmax_iy_i` = 8075

  Obtain the linear regression of Y on X.

Rewrite the following statement without using if ...... then.

If a man is a judge then he is honest.

Rewrite the following statement without using if ...... then.

It 2 is a rational number then `sqrt2` is irrational number.

Rewrite the following statement without using if ...... then.

It f(2) = 0 then f(x) is divisible by (x – 2).

Without using truth table prove that:

(p ∨ q) ∧ (p ∨ ∼ q) ≡ p

Without using truth table prove that:

(p ∧ q) ∨ (∼ p ∧ q) ∨ (p ∧ ∼ q) ≡ p ∨ q

Without using truth table prove that:

∼ [(p ∨ ∼ q) → (p ∧ ∼ q)] ≡ (p ∨ ∼ q) ∧ (∼ p ∨ q)

Using rules in logic, prove the following:

p ↔ q ≡ ∼(p ∧ ∼q) ∧ ∼(q ∧ ∼p)

Using rules in logic, prove the following:

∼p ∧ q ≡ (p ∨ q) ∧ ∼p

Using rules in logic, prove the following:

∼ (p ∨ q) ∨ (∼p ∧ q) ≡ ∼p

Using the rules in logic, write the negation of the following:

(p ∨ q) ∧ (q ∨ ∼r)

Using the rules in logic, write the negation of the following:

p ∧ (q ∨ r)