Question

solve the following algebraic equation using trial and error method. 1)1.5z= - 4.5​

Answer 1

Answer:

-3

Step-by-step explanation:

1.5z = -4.5

= 15/10z = -45/10

= z = -45/10 * 10/15

10 is cancelled by 10

so we are left with ,

z = -45/15

15*3=45

So , -45 is cancelled by 15 .

Therefore the answer is -3.

Answer 2

Answer:

TESTING OF HYPOTHESIS

Sampling distributions - Estimation of parameters - Statistical hypothesis - Large sample tests

based on Normal distribution for single mean and difference of means -Tests based on t, Chi-

square and F distributions for mean, variance and proportion - Contingency table (test for

independent) - Goodness of fit.

UNIT - II

DESIGN OF EXPERIMENTS

One way and two way classifications - Completely randomized design – Randomized block

design – Latin square design - 22

factorial design.

UNIT - III

SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS

Solution of algebraic and transcendental equations - Fixed point iteration method – Newton

Raphson method - Solution of linear system of equations - Gauss elimination method – Pivoting

- Gauss Jordan method – Iterative methods of Gauss Jacobi and Gauss Seidel - Eigenvalues of a

matrix by Power method and Jacobi’s method for symmetric matrices.

UNIT - IV

INTERPOLATION, NUMERICAL DIFFERENTIATION AND NUMERICAL

INTEGRATION

Lagrange’s and Newton’s divided difference interpolations – Newton’s forward and backward

difference interpolation – Approximation of derivates using interpolation polynomials –

Numerical single and double integrations using Trapezoidal and Simpson’s 1/3 rules.

UNIT - V

NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS

Single step methods : Taylor’s series method - Euler’s method - Modified Euler’s method -

Fourth order Runge-Kutta method for solving first order equations - Multi step methods :

Milne’s and Adams - Bash forth predictor corrector methods for solving first order equations.