10 ecologically based planning and design techniques

... Instructional techniques Skills training, 204 Stakeholders, 45 Storyboards, 104– creating, 104– defined, 104 illustrated excerpt, 106 page resulting from, 107 templates and, 104 testing before creating, 106 ... Development, 97– 145 content, 127– 45 design elements, 109– 25 guidelines, 99– 125 illustrated, 100 media, 107– prototypes, 108– storyboards, 104– templates, 101– versions, 108– Development tools courseware, ... sophistication and, 176 Technical resources, 54 Technical specs, 162 Templates, 101– as basis for storyboards, 104 benefits, 101 for consistency, 103 creating, 103 illustrated, 102 preexisting, 103 for

Ngày tải lên: 26/01/2023, 15:31

... = 100, u 5 = 8 Find the value of ∆5u 0 Sol We know ∆ = E – 1, therefore, 5 0 u ∆ =(E –1) 5 u0 = (E5 – 5E4 + 10E3 – 10E2 + 5E – 1)u0 =u5 – 5u4 + 10u3 – 10u2 + 5u1 – u0 = 8 – 500 + 2000 – 810 ... argument half way between the arguments of q and r is A+ 1 B 24 , where A is the arithmetic mean of q, r and B is the arithmetic mean of 3q – 2p – s and 3r – 2s – p. Sol On taking h being the interval ...  u x and 2 2 x x u E u   If u x = x 3 and the interval of differencing is unity Find out the expression for both [Ans 6h2 (3x – h), 2 3 3 + + ] 13 If f(x) = e ax , show that f(0) and its

Ngày tải lên: 04/07/2014, 15:20

... the following values: d A Find A for 105. Sol First of all we form the difference table as follow: − 648 766 100 7854 Here, h = 5, a = 100, x = 105 ∴ u = 105 100 1 5 − = Now on applying Newton’s ... Example 5 In an examination, the number of candidates who obtained marks between certain limits are as follows: Marks No of candidates Find no of candidates who obtained fewer than 70 marks. ... known as Gauss Backward difference formula and useful when u lies between 1 and 0 2 4.4.3 Stirling’s Formula This is another central difference formula and useful when | | 1 1 1 u < or− <

Ngày tải lên: 04/07/2014, 15:20

... ) ( ) ) ( ) ( ( ) ( ) ( ) ( ) ( ) ) ( ) 9 10 7 10 8 10 9 + = – 1 2 (x – 8) (x – 9) (x – 10) + 1 2 (x – 7) (x – 9) (x – 10) – 1 2 (x – 7) (x – 8) (x – 10) + 3 2 (x – 7) (x – 8) (x – 9) .(1) Putting ... value of x exist for which f(x) is max.] 7 Given log10 654 = 2.8156, log10 658 = 2.8182, log10 659 = 2.8189 and log10 661 = 2.8202 Find log10 656 by Lagrange’s interpolation formula [Ans 2.8169] ... given at the points (7, 3), (8, 1), (9, 1) and (10, 9) Find the value of y for x = 9.5 using Lagrange’s interpolation formula. Sol We are given ( ) : 7 8 9 10 : 3 1 1 9 x f x Trang 9Here, ( )00 (

Ngày tải lên: 04/07/2014, 15:20

... , x 1 ] = 10 10 yy xx − − = − −− 10 10 10 yy xxxx ⇒ [x 0 , x 1 ] = 1010 1111 yyxx ÷ Similarly, [x 0 , x 1 , x 2 ] = 222 01 2 0 1 2 01 2 0 1 2 111 1 11 yy y x x x xx x x x x÷ and so on. 4. ... days apart round the pds. of summer and winter. Estimate the app. dates and values of max. and min. temperature. 254 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Summer Winter Day Date ... f(3) = 31, f(6) = 223, f(10) = 1011, f(11) = 1343, Sol. The divided difference table is given by () 23 13 28 2 14 331 50510 192 3 64 9 9 1 6 223 133 7 19 788 4 197 8 8 1 10 1011 135 5 27 332 1 332

Ngày tải lên: 04/07/2014, 15:20

... 1.644, 1.859] 268 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7. Apply Hermite’s formula to find a polynomial which meets the following specifications: 110 000 110 iii xyy ′ −− () 35 1 ... (x)= 1 01 xx xb xx ab −− = −− L 1 (x)= 0 10 xx xa xx ba −− = −− ∴ () 0 Lx ′ = 1 ab − and L′ 1 (x) = 1 ba − Hence, () 00 Lx ′ = 1 ab − and L′ 1 (x 1 ) = 1 ba − Therefore from equation ... 266 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Hermite’s interpolation formula is H(x)= () () () () 11 2 2 00 [1 2

Ngày tải lên: 04/07/2014, 15:20

... COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Approximate f by a polynomial P(x) = a + bx + cx 2 such that x1 may ≤ () () 3 fx Px 5 10 − −≤× Sol. The given function f(x)= 246810 0 1 1 ... series. 278 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 6. Prove that () () () 2 nn1 n 1xTx U x xU x + −=− . Sol. If x = cos θ , we get T n (cos θ ) = cos n θ and U n (cos θ ... = 1 – 24 6 8 10 6 30 168 1080 7920 xx x x x +− + − + (1) given that ∈ = 5 × 10 –3 ⇒ ∈ = 0.005 Now, truncating the series (1) at x 8 , we have P(x) = 1 – 24 6 8 6 30 168 1080 xx x x +−

Ngày tải lên: 04/07/2014, 15:20

... − −− − + −+ + −− 288 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES S’ (x i )= () () 1 1 2 6 ii ii i i Sx Sx PP h h − − + + + (12) For equation (9) and (10) c i = 3a i–1 = h i 2 + 2b ... 290 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ⇒ () () () 123 1 2 3 462 MMM fx fxfx ++= − +   but M 0 = 0 M 3 then it becomes 12 1 2 412418 MM andM M+= = = M 1 = 2 and M 2 = ... COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 1. S(x i ) = f(x i ); i = 0, 1, 2, n 2. On each subinterval [x i–1 ,x i ], 1 ≤ i ≤ n, S(x) is a polynomial in n of degree at most n . 3. S(x) and

Ngày tải lên: 04/07/2014, 15:20

... as: 1( ) 10 x= − −y z 1 (51 2 ) 10 y= − x z− 1 ( ) 10 z= − −x y If we start by assuming y0 = =0 z0 then, we obtain 1 ( ) 1 44 0 0 4.4 10 Now we susbtitute x=4.4 and z0 =0 for y1 and we obtain ... identical systems x1−x2=1 x1−x2=1 1 1.00001 x − x2 =0 and x1−0.99999 x2 =0 Respective solutions are: (100001, 100000) and (–99999, –100000) obviously the two solutions differ very widely Therefore ... iterations in Gauss-Seidel method Trang 10Example 8 Solve the following system of equations using Gauss-Seidel method: 10x + y + 2z = 44 2x +10y + z = 51 x + 2y + 10z = 61 Sol Given system of equations

Ngày tải lên: 04/07/2014, 15:20

... bx yae = Taking logarithm on both the sides, we get 10 10 10 log log log yabxe=+ i.e., YABx=+ (1) where 10 10 log , log YyAa== and 10 log Bb e= . The normal equations for () 1 are, ... = 10,x = ∑ 30,y = ∑ 120,xy = ∑ 2 30, x = ∑ 2 434, xy= ∑ 3 100, x = ∑ 4 354. x = ∑ Putting all these values in () 2 , () 3 and () 4 , we get 30 5 10 30abc=+ + (5) 120 10 30 100ab ... Taking 0 uxx=− and 0 , vyy=− therefore 1933ux=− and 357vy=− Then the equation 2 yabxcx=+ + is transformed to 2 vABuCu=+ + (1) 394 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Normal

Ngày tải lên: 04/07/2014, 15:20

... − σ , where x and y are the means of x-series and y-series respectively; r is the coefficient of correlation between x and y; σx and σy are the standard deviations of x-series and the y-series ... coefficient b yx and b xy and hence find the correlation coefficient between x and y: (1, 2), (2, 4), (3, 8), (4, 7), (5, 10), (6, 5), (7, 14), (8, 16), (9, 2), (10, 20). Sol Here n = 10 We may prepare ... 3Then, y i = +a bx i and 2i i i i x y =ax +bx for each i=1, 2, ,n therefore Equations ( )2 and ( )3 are normal equations for this line Solving ( )2 and ( )3 for a and b and putting these values

Ngày tải lên: 04/07/2014, 15:20

... 30. (ii) The age of husband, when the age of wife is 19. Trang 5Sol We have 256 25.6 10 i x x x 17.2 10 i y y n Let the assumed mean of x- series and y- series be 26 and 17 respectively Then, ... two regression coefficients are 0.8 and 0.2, what would be the value of coefficient of 7 x and y are two random variables with the same standard deviation and correlation coefficient r Show that ... x-series be x and that of y-series be y Then, each of the given lines passes through ( , )x y On solving (1) and (2), we get x = 5 and y = 1 3. Therefore mean of x-series is 5 and mean of y-series

Ngày tải lên: 04/07/2014, 15:20

... 40–50 40 90–100 05 Sol. Since the class intervals are equal throughout no adjustment in frequencies are required. 462 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 70 60 50 40 30 20 10 0 8 12 ... showing the sales and net profits of private industrial companies. Year Sales Net Profits 1995–1996 14% 49% 1996–1997 10% –25% 1997–1998 13% –1% Sol. 60% 50% 40% 30% 20% 10% 0% –10% –20% –30% 1995-1996 ... observations and then squares are drawn with sides proportional to these square roots, on an appropriate scale, which must be satisfied. 458 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES

Ngày tải lên: 04/07/2014, 15:20

... x∞ ) + f ′′( x∞ )Ek − ≈ Ek f ′( x∞ ) + f ′′( x∞ )( Ek + Ek − ) . (10) .(11) .(12) 82 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES f ′′( x∞ )(Ek + Ek −1 ) is negligible compared to f ′( ... x1 = and f(x0) = 20, f(x1) = – 1, then by Secant method the next approximation is given by   x1 − x0 x = x1 –   f(x1)  f ( x1 ) − f ( x0 )  84 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ... î + å n ) f ( î + å n ) − f ( î + å n −1 ) .(2) On expanding f (î + å n ) and f (î + å n−1) in Taylor’s series about the point î in (2), and using f (ξ) = 0, we obtain εn + 1 (εn − εn −1 ) εn...

Ngày tải lên: 04/07/2014, 15:20

... (remember, not individual and the retirement plan • Expand the discussion based on the discussion created by the above statement Do not talk about LTCI Establish the connection based on your strength ... love you and want to make sure you are safe The plan I want to put together recognizes that one or more may be available It can allow them to provide the care you need and want longer and better ... and assisted care Even if you decide you want to go to a nursing home when you first need long term care you will have to transfer most of your assets including qualified funds and low cost based...

Ngày tải lên: 07/03/2014, 14:20

... 4835/24, Ansari Road, Daryaganj, New Delhi - 1100 02 Visit us at www.newagepublishers.com Preface The present book ‘Computer Based Numerical and Statistical Techniques is primarily written according ... concepts, definitions and large number of examples in the best possible way have been discussed in detail and lucid manner so that the students should feel no difficulty to understand the subject A ... subject A unique feature of this book is to provide with an algorithm and computer program in Clanguage to understand the steps and methodology used in writing the program Thorough care has been...

Ngày tải lên: 04/07/2014, 15:20

... it is to COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES be cut-off to a manageable size such as 0.29, 0.286, 0.2857, etc The process of cutting off super-flouts digits and retaining as many ... 7326853000 becomes 7327 × 106 1.3 ERRORS Error = True value – Approximate value A computer has a finite word length and so only a fixed number of digits are stored and used during computation ... machine dependent and is called machine epsilon After the computation is over, the result in the machine form (with base b) is again converted to decimal form understandable to the users and some more...

Ngày tải lên: 04/07/2014, 15:20

... equation X2 – 2X + log10 = are given by, X = δX Therefore, = ± − log10 2 = ± − log 10 δ(log 2) < 0.5 × 10 4 − log δ(log 2) < × 0.5 × 10 4 (1 − log 2)1/2 < 0.83604 × 10 −4 or ≈ 8.3604 × 10 5 Example 21 ... ≤ x + x + + x n 10 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 0.0005 0.0005 + = 0.0002501 3.724 4.312 Therefore, Relative Error, Er = (Because Error = 1 × 10 − n = × 10 −3 = 0.0005) ... 5V δu = ∂u δV = (12V − 5)δV ∂V 14 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES  12V −  ∂u δV × 100 100 =   2V − 5V   u (12 − 5) × 0.05 × 100 = − × = −11.667% (2 − 5) Hence maximum...

Ngày tải lên: 04/07/2014, 15:20

... − b − ac 2a a = = 0 .100 0 × 101 b = 400 = 0.4000 × 103 c = 1= 0 .100 0 × 101 b2 – 4ac = 0.1600 × 106 – 0.4000 × 101 = 0.1600 × 106 (To four-digit accuracy) b − 4ac = 0.4000 × 10 On substituting these ... as x = 2c b + b − 4ac 20 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES x = 0.2000 × 101 0.4000 × 103 + 0.4000 × 103 x = or 0.2000 × 101 = 0.0025 0.8000 × 10 This is the exact root of the ... δa = 0.1 a× 100 δb = ⇒ δb = 0.2 b× 100 (1) 19 ERRORS AND FLOATING POINT ∂A −a ∂A = = and 2 ∂b b b − a ∂a b −a Substituting these values in equation (1), we have δA < 0.00135 + 0.0 0100 < 0.00235...

Ngày tải lên: 04/07/2014, 15:20

... x2 – 100 0x + 25 = Sol Given that, x2 – 100 0x + 25 = x= ⇒ Now 100 0 ± 106 − 102 106 = 0.000e7 and 102 = 0 .100 0e3 Therefore 106 – 102 = 0 .100 0e7 ⇒ Hence roots are:  0 .100 0e + 0 .100 0e   0 .100 0e ... 0.36143447 × 10 gives 0.0001 1101 × 107 On shifting the fractional part three places to the left we have 0.1 1101 × 104 which is obviously a floating-point number Also 0.0001 1101 × 107 is a floating-point ... 0 .100 0e7 ⇒ Hence roots are:  0 .100 0e + 0 .100 0e   0 .100 0e − 0 .100 0e    and   2     106 − 10 = 0 .100 0e which are 0 .100 0e4 and 0.0000e4 respectively One of the roots becomes zero due to...

Ngày tải lên: 04/07/2014, 15:20