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Quiz 7 Key

# Quiz 7 Key - Quiz 7 CE 361 December 13th 2007 25 pts...

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Unformatted text preview: Quiz 7 CE 361 December 13th, 2007 ' 25 pts Available Name: BE SURE T05§CLUDE ALL UNITS IN YOUR CALULATIONS AND SHOWALL WORK?! Multiple Choice Questions 52 points each) 1. According to the US. Water Advisory Committee on Water Data, which is the recommended technique for ﬂood frequency analysis? a. Gumbel Distribution b. Log—Normal Distribution c. Log-Pearson Type III Distribution d. Normal Distribution 2. The concept of virtual water helps us realize what? a. How much water is needed to produce different goods and services b. The best method to allocate drinking water in times of scarcity c. The flow of water within a watershed d. Technique to determine which countries have the most available freshwater 3. Which of the following sectors historically consumes the most water? a. Industrial b. Domestic 0. Fire Department d. Agriculture 4. Skew coefficient is to the size of the sample; thus it is obtain accurate estimate from small samples a. Very sensitive, Difﬁcult b. Proportional; Necessary c. The reciprocal; Preferred (1. Independent; Possible 5. Flood damages are usually reported as to property. a. Pisks b. Proximity c. Value of goods near (1. Direct damage to 6. The mean, standard deviation, and skew coefficient of the observed log—transformed discharges (in ftIi/s) for a river are 3.0, 0.5, and 0.3 respectively. ComputeVZS—year peak discharge. You do not need to calculate a weighted skew.(5pts) ‘” waﬁhn 19...... Note) \$20= 5.0 + disused) = 3°13 10306 5 %15\\ ('95 fa.“—-—-_—— ‘_______— Table 1.0.4.1 11".. Values for Pearson Type III Dis‘lrihulion {cwuirrrmd} MW Recurrence Interval in Years ___—_________,_._____._._.————-—-———~—-— LDIOI 1.03211 1.1111 1.2500 2 5 IO 35 50 100 2.110 E-xmudanue waubilily lew 80:111. .99 .95 .90 .811 .50 .20 .10 .114 .1112 .111 ' .0115 2.2 41.9115 41.1182 #11“ w{1.7152 43.3311 11.574 1.234 2.340 2.9m 3.705 4.4414 2.1 419-115 4.1.914 41361.1 41.165 41.319 0.5.92 1 .2911 2.230 2.9422. 3.6513 1372 1.0 41.991] 41.949 4.1.895 41.77? «11.307 0 .609 '1 .3102 2.219 22.912 3 .6115 4 .298 1.9 ml .031? 411.934 “11.920 41.788 41.294 11.627 1.31 1.1) 2.107 2.381 3.5213 4.223 1.3 “1.1187 4.1120 41.945 41.799 41.282 0.643 1.3113 2.193 3.8415 3.499 4.147 1.7 —1 .141} wI .1151“: ~09?le 41.8118 41.2133 (1.6611 1.324 2.179 2.1315 3.444 4.1169 1.6 H.191 ~l .0911 41.994 41.81“! 41.254 11.675 1.329 3.136 3.71111 21.3811 3.9911! 1.5 —| .256 —1.1.11 —1 .1113 —D.825 41.240 0.6911 1.333 2.1416 22.7613 3.3110 3.910 1.4 —1.3I1-1- «1.1651 --1 .041 «41.832 “0.225 0.705 1.33? 2.118 2.706 3.271 3.8325 1.3 4.3331 «1.261) 4.06:1 41.838 41.210 11.719 1.331) 2.1118 2.665 3.21 1 3.745 1.3 —1 .6141) «1243 4.1.1116 «0.844 41.195 (1.731 1.3411 2.1187 2.626 3.149 3.661 1.1 ~15 It." —I .2811 "1.111? 41.1148 421.131) 11.745 1.341 2.066 2.585 3.08? 3.575 1.0 w 1 .5113 —1 .3 .17 ~1 .128 411.852 41.164 0.?551 1.340 2.1113 2.542 3.022 3.489 .9 ml .{yt-AJ —- I .353 "1.14? 41.3511 41.148 0.7m 1.339 2.018 2.493 2.95? 3.41111 .11 -1 .133 «1.3811 “1.166 «(1.8561 41.13”); 0.11811 1.3313 1.993 2.-=153 2.891 3.312 3 "1.811131 4.423- 4.183 “0.85? 41.116 (1.791} 1.333 | .9157 2am 2.824 3.123 .1: ~—1 .331) —1 .458 4.2013! 41.2157 41.09:} 0.301} 1.323 1.939 2.359 2.755 3.132 .5 ml .955 ~I .1191 "1.216 —0.35{i w{1.08:1 13.808 1.323 1.910 2.311 2.686 3.041 .4 4.029 ml .524 — I .23! «11.1555 41.066 11.816- 1.31? 1.330 2.1121]. 2.615 2.949 .3 4. 1114 «1.555 w4:13:15 ~0.853- --11.1)5(1 11.824 1.309 1.3.49 2.211 2.544 2.856 .2 —2. 178 «1.536 -—1.258 «0.85U 41.033 0.8313 1.301 1.1118 2.159 2.4"?! 2.763 .1 «2.252. ~— I .(1 I I5: -l .3110 41.841“: «11.01"! (1.836 1.292. 1.785 2. [[17 2.400 2.611) .0 6.326 —~1.()=15 MLZSE 41.842 (1 1.1.842 1.282 1.751 2.1154 2.326 2.576 7. What is the difference between green and blue water? Give an example of each.' (3pm) Tm. (mom-\nomﬂ Mitt-5 Ewan... RDL'MVM ON} Mmiﬁcm'k' / £09m¥f3 Ls O(\ \O\U\‘L— Wﬁtu’ . r\ -mpt~e.%<:\\‘ m, wmw g S We is Why/660:} back to M ( CA\'MoﬁP\/-z(c *hrovgh "the Ohmitwdmol SeJ 36% Of\o\ JCFmgext‘c/Q'xm ‘ ‘ Ekum‘ekp'. (3&“x Ww‘r-Uﬁ 60mm*rmsQ\ta* “am PMM PAW vouW". WOV\‘LC Myth-«d ox WWOQ\(- 8. Risk—based design approaches integrate the procedures of What two analysis techniques in design? (Zpts) an mr+utm'\':) A (“\$33 ‘3 *— QQA‘OJOA‘ 3pm) Anm\3‘\$‘3 9. Describe how Frequency analysis is an information problem? Also detail what assumptions must be made. (5pts) , Euwﬁg k-Q— 0mg. \rxotcx m Sud-Homing \Om NLMA 09 pilotﬁé Plougg‘ (230.0? \ng Flow 5 'hmﬂeﬁ p max/mm. d1g¥t§bm3flgq {:of C)\ (J‘CDC\\3 dJLaSVQKMuxLA, NW 5m ossme -\'m5r N3 Ohm 5 paved Ol- mtofcl. 1h moada‘ak» do‘s‘w‘ Eilrg ’ Doe, OLLM dvf ~ “:3 hm“ m '\r\Sv\Pthtm¥\} mat) in) datum. W “SK 0? ex‘rmm m’r ...
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