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main points of this exam paper are: Simultaneous Equations, Nearest Minute, Standard Deviation, Time, Normal Distribution, Black Balls, Probability, Faulty, Poisson Distributed, Density Function
Typology: Exams
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(NFQ Level 8)
Answer FIVE questions, selecting three questions from section A and two questions from section B. All questions carry equal marks.
Examiners: Mr. P. Ahern Dr. J. Buckley Dr. A. Kinsella
A^21 and
t
t 2
(i) Evaluate the matrix A^2. (ii) Find the values of t for which det( B ) = 0. (iii) Find the values of t for which the matrix A B^2 is non-invertible. [10 marks]
(b) Find the inverse of the matrix
Hence find the solution of the simultaneous equations
2 3
1 2 3
1 2
− =
x x
x x x
x x [10 marks]
Number of Students 6 17 30 23 14 10 (i) Find the mean time taken and the standard deviation from that mean. (ii) Use these values to find the time within which 95% of values will lie. (Assume a normal distribution) [12 marks] (b) Box 1 contains 2 red balls and 4 black balls; box 2 contains 3 reds and 5 blacks; box 3 contains 6 reds and 2 blacks. A ball is selected and found to be black. Find the probability that it came from box 2. [8 marks]
numbers, is a group. [5 marks] (b) State the properties of an integral domain. Is ( Z (^) 4 ,+ 4 ,× 4 )an integral domain? Note: Z (^) 4 ={ 0 , 1 , 2 , 3 }and + 4 and × 4 represent addition modulo 4 and multiplication modulo 4 respectively. [9 marks] (c) State the condition under which an integral domain is a field. Does ( Z , + ,× ) constitute a field? [6 marks]
π π t^ sin(^10 t ) dt
(^2) and
π π
2 2 t^ cos(^2 t ) dt.^ [4 marks] (b) Show that the functions f ( t )= t and g ( t )= 6 − t are orthogonal on the interval [0,9]. [6 marks] (c) A function f ( t )is defined by
π
π t
f t t 5 , 0
Find its Fourier series representation. [10 marks]
Some formulae:
i
i i
i
i i
f
A c fu
f
fx x
i
i i
i
i i
f
A c fu
f
fx
2 2
2
i i
i i i
i i
i
i i
f f
fu f
fu c
f
f x x s
2 2
i
i i i
i i
i
i i
f
fu f
fu c
f
= (^) n
i i i
i i i PH PE H
1
n i i^ i
1
=
n i i^ i
1
n X p^ q ( ) = X^ n^ X ^
− !
f ( t )= ke −^ kt^2
2
2
f ( z )=^1 e −^ z
where