# Sp94Ex2sol - {I Sp 94 EGN 3420 Exam 2 Name “bah 1" 5...

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Unformatted text preview: {I- Sp 94 EGN 3420 Exam 2 Name “bah 1" 5 B) C] Problem 1 {51] pts} Test scores of several students In two classes are given below. Geometry {G} Chmﬁstrs (C) so so ss '12 93 mo 7'4 as as 93 ‘ so or Find the least squares line for predicting Chemistry scores {C} from Geometry.r scores {G}. gr.“ a, war a. -.-. Hod-i Hang deluge... 13%,?31 sun. s14: -;_‘BoﬁEJo1-.ml‘“¢ Find the coefﬁcient of determination r:B and the standard error of the estimatesyg r1: ss-r-sse' = pass,a‘l-a3.a'i"‘f' 551' mass? = C_ﬁh‘§i¢ Ans. C=—3.aoaﬂ+1.i1‘lls G Ans. r3=o.sese, 5m: 29H“! Use the resulting least squares regression line to ﬁnd the expected score in Chemistry for a student who made ill] on a Geometry test. Ans. {2: Base ghaesaensmeteoﬁ=eex-3e Work Area e GE 6.0. C.- (Cr-Ci}: (are? so 130% Ls‘oﬂ 52m 1. [Lee T's-‘15 T2 ~1wa Heme 13.07.! Luann 1,3. ‘1'. mo gauge asst: mLL‘t 1. Sine maﬁa 82 sun-“a sees wane i-La‘eﬁ‘f “LLB“- .93 11‘1'4: 'qule: "fatal-“f 0.1159 1H5. .35" '- ET use”: HGLG £95,517 0... iﬂoﬁ Ugh 11 Lit-H 311315 34,132 1.3.3‘11‘1 lbb3.3'{ Problem 2 {25 pts} The following data was obtained from an unknown function fix}. 1"; = {[31} 1 mun-Ila” Ki -1 n 1 1 1 2 5 A} Find the Lagrange intelpolatmg polynomial fax} = Efﬁe} Lo} You do not have to simplifyr the answer. HHS; f3 [35) = B} Estimate {(1.1}. Ans. 111.1} = ﬂ ,9, Lane}: tartan-rxe‘ILY-da‘r a 1 3Lx+tlLK*L'} = .Lg‘a-E‘ﬁ +1.13 ”v‘ﬁlwchlﬂrgj [ht—n} E-n-qol—L‘l {a {I L‘UL‘} *- {1—iaLx-Hue-Fﬂ 3 = fi+tl£ﬁ*n(£‘l) a; 1 L}: -'L}LL_-;Lt-L3 ‘—u—.— {ll-1‘93 Lil'tﬂu‘u‘xgl loH‘J‘LO‘VHD‘L" Z L L : _ 1. a} {a Humane-x3) : ”H‘Hmibn u {x} 7&1 M : ._ _ -L LtL—ﬁellﬁrhlih'lsl [Htllxw‘r “-13 1 L115]: [-. a —X\{x*xr. 3 xxnx . ‘5 1 {wanna-n __ _ = —' (Lava) '- {11 4%} [1‘3"‘xi 1' [F.3- xL) {1+1} ("L-“:1 {1'1} l:- -; _ 3U“: '- “faLuLm Hill-n”) + YLLLL‘H 4.- 19.3 L1H -_-.'I ILLS-311': 3* 1- “*- - 3'1‘ 1" E + Mutt: 1% 1H”! in. 1 1M alga: 1‘: E3 SEEM-1.x ': 1.1.5“! .r-eo Hohlem 3 {25 pts] lGriwen the data points [11,11], {1,0}, [2,15] and {3.55;}, the resulting third order Newton Divided Difference polynomial is {3(3) = 3xtx-1] + xtx-le—Z} Find 5'4. Ans. 3'" = 1r Lt 93(9) = suite—a +3(3—~5£3+L‘: =1} IE +- To 1‘11, «<1 no ...
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