734 812 + 50 + 1 062 328 = A 1 977 180 C 1 797 180 B 1 797 190 ( D 1 779 190 9. 2 450 000 − 36 000 − 300 500 = A 1 789 500 C 2 113 500 ( B 1 798 500 D 2 383 950 10. 158 780 ÷ 25 = 6 351 remainder . What is the number need to be written in the above? A 1 C 10 B 5 ( D 15 11.
Data Table #1 Trial Mass of Stopper Hanging Mass Radius T1 T2 T3 Revolutions 1 0.014kg 0.05 kg 0.595m 12.93s 13.10s 14.6s 20 2 0.014kg 0.1 kg 0.595m 11.53s 11.78s 10.8s 20 3 0.014kg 0.15 kg 0.595m 8.59s 9.37s 9.35s 20 4 0.014kg 0.2 kg 0.595m 8.0s 8.03s 7.72s 20 5 0.014kg 0.1 kg 0.765m 12.97s 13.81 ------- 20 6 0.014kg 0.1 kg 0.50m 10.78s 9.72s ------- 20 7 0.014kg 0.1 kg 0.60m 10.0s 10.25s ------- 20 *For the last 3 trials, we only had time to take 2 time intervals. IV. Analysis and Interpretation 1. Organizing Data- Calculate the weight of the hanging mass for each trial. This weight is the force that maintains circular motion, Fc.
Given: sides, 5,6,7 in miles Island A = 78.46° Island B = 135.58° Island C = 57.12° Given the information… From Island B I would travel a Northwest Bearing to Island C. 8. Heron’s formula: sq.root of {s(s-a)(s-b)(s-c)} Sqrt((a^2+b^2+c^2)^2 - 2 (a^4+b^4+c^4))/4 = sqrt ((25+25+16)^2-2(625+625+256))/4 = sqrt ((66)^2-3012)/4 = sqrt (4356-3012)/4 = 36.66060556/4 = 9.16515139 square feet 9 feet
| 29 | 13 | 1.1 | 1.1 | 24.7 | | 30 | 27 | 2.2 | 2.4 | 27.1 | | 31 | 13 | 1.1 | 1.1 | 28.2 | | 32 | 19 | 1.6 | 1.7 | 29.9 | | 33 | 17 | 1.4 | 1.5 | 31.4 | | 34 | 23 | 1.9 | 2.0 | 33.4 | | 35 | 22 | 1.8 | 1.9 | 35.4 | | 36 | 14 | 1.2 | 1.2 | 36.6 | | 37 | 20 | 1.7 | 1.8 | 38.4 | | 38 | 20 | 1.7 | 1.8 | 40.1 | | 39 | 25 | 2.1 | 2.2 | 42.3 | | 40 | 36 | 3.0 | 3.2 | 45.5 | | 41 | 15 | 1.2 | 1.3 | 46.8 | | 42 | 16 | 1.3 | 1.4 | 48.2 | | 43 | 26 | 2.1 | 2.3 | 50.5 | | 44 | 27 | 2.2 | 2.4 | 52.9 | | 45 | 24 | 2.0 | 2.1 | 55.0 | | 46 | 23 | 1.9 | 2.0 | 57.1 | | 47 | 30 | 2.5 | 2.6 | 59.7 | | 48 | 19 | 1.6 | 1.7 | 61.4 | | 49 | 20 | 1.7 | 1.8 | 63.1 | | 50 | 39 | 3.2 | 3.4 | 66.6 | | 51 | 9 | .7 | .8 | 67.4 | | 52 | 13 | 1.1 | 1.1 | 68.5 | | 53 | 19 | 1.6 | 1.7 | 70.2 | | 54 | 13 | 1.1 | 1.1 | 71.3 | | 55 | 19 | 1.6 | 1.7 | 73.0 | | 56 | 23 | 1.9 | 2.0 | 75.0 | | 57 | 14 | 1.2 | 1.2 | 76.3 | | 58 | 21 | 1.7 | 1.9 | 78.1 | | 59 | 11 | .9 | 1.0 | 79.1 | | 60 | 23 | 1.9 | 2.0 | 81.1 | | 61 | 21 | 1.7 | 1.9 | 83.0 | | 62 | 10 | .8 | .9 | 83.9 | | 63 | 14 | 1.2 | 1.2 | 85.1 | | 64 | 4 | .3 | .4 | 85.4 | | 65 | 12 | 1.0 | 1.1 | 86.5
Descriptive Statistics: Income ($1000), Size, Credit Balance ($) Variable Mean StDev Minimum Median Maximum Range Mode Income ($1000) 43.74 14.64 21.00 43.00 67.00 46.00 55 Size 3.420 1.739 1.000 3.000 7.000 6.000 2 Credit Balance($) 3976 932 1864 4090 5678 3814 3890 Variable Mode Skewness Kurtosis Income ($1000) 4 0.05 -1.29 Size 15 0.53 -0.72 Credit Balance($) 2 -0.14 -0.74 The statistics indicates that the mean income is 43.74; the median income is 43.00 and the mode income is 55. The standard deviation is given approximately as 1.74. Maximum income is 67,000 and the minimum income is 21,000 and a Standard Deviation of 14.64. The second individual variable is household size the mean household size of the sample is 3.42, the median household size of the sample is 3 and the mode household size of the sample is 2 and the standard deviation is 1.739. The maximum household size is 7 and the minimum household size is
You would equate this as 82 x 152 = 172. (8 x 8) + (15 x 15) = (17 x 17) 64 + 225 = 289 2) 7, 24, and 25. The equation would be 72 x 242 = 252 (7 x 7) + (24 x 24) = (25 x 25) 49 + 576 = 625 3) 28, 45, and 53. The equation would be 282 x 452 = 532 (28 x 28) + (45x 45) = (53 x 53) 784 + 2025 = 2809 4) 48, 64, and 91. The equation would be 482 x 642 = 802 (48 x48) + (64 x 64) = (80 x 80) 2304 + 4096 = 6400 5) 9, 40, and 41.
Pre-Lab Questions: 1. Absorbance = - Log10T = 2-Log10 (%T) a) 2 – Log (89.95) = .0460 f) 2 – Log (28.18) = .5501 b) 2 – Log (80.91) = .0920 g) 2 – Log (20.51) = .6880 c) 2 – Log (59.02) = .2290 h) 2 – Log (17.38) = .7600 d) 2 – Log (47.75) = .3210 i) 2 – Log (14.45) = .8401 e) 2 – Log (34.83) = .4580 2. Molar Concentration 1.213 g Cu (1 mol Cu/ 63.546 g Cu) (1 mol [Cu (H2O6)2+]/ 1 mol Cu) (1/ .100 L) = .191 M [Cu (H2O6)2+] a) M2 = (M1V1)/V2 M2 = (.191 M [Cu (H2O6)2+])( 1.0 mL)/ (50.0 mL) = .0038 M b) (.191 M [Cu (H2O6)2+]) (2.0 mL)/ (50.0 mL) = .0076 M c) (.191 M [Cu (H2O6)2+]) (5.0 mL)/ (50.0 mL) = .0191 M d) (.191 M [Cu (H2O6)2+]) (7.0 mL)/ (50.0 mL) = .0267 M e) (.191 M [Cu (H2O6)2+]) (10.0 mL)/ (50.0 mL) = .0382 M f) (.191 M [Cu (H2O6)2+]) (12.0 mL)/ (50.0 mL) = .0458 M g) (.191 M [Cu (H2O6)2+]) (15.0 mL)/
2) Case Shipments = -80,315.68 + 996.56*49 + 3877.95*113 + .096*960,000 + .076*154,455 - .048*345,028.8 - .02*368,337.5 = 483,432 Standard Error = 34,733.12 given by Excel Confidence Interval of 95% -> T Value of 97.5% for 42 degrees of freedom is 2.018 2.018 * 34,733.12 * SQRT (1 + (1/48)) = 70,818 95% Confidence Interval = Estimate +/- Standard Error*T-Value*SQRT(1+(1/n)) The 95% Confidence Interval is from 412,614 and 554,250 cases. 3) $1 invested into the Dealer Allowance would increase current sales by .076 cases and decrease sales in two months by .02 cases, resulting in a net increase of .056 cases. Assuming that Harmon Foods is able to sell their cereal for $2/pack, the value of a case is $48 ($2 * 24 packs/case). Therefore, $1 invested in Dealer Allowance equates to $2.69 in added sales revenue (assuming all other
In your conclusion, answer the (why?) in procedure steps 7 & 8 Results: Length of string (cm) Time for 15 swings (s) Time for 1 swing (s) Accepted value of T (s) % error calculated (%) T2 (your value from column 3) 100 30.9 2.06 2.00 3.0 4.00 90 28.1 1.87 1.90 1.6 3.61 80 26.3 1.75 1.80 2.7 3.24 70 25.3 1.69 1.68 0.6 2.82 60 24.0 1.60 1.55 3.2 2.40 50 22.1 1.47 1.42 3.5 2.01 40 19.2 1.28 1.27 0.8 1.61 30 16.5 1.10 1.10 0.0 1.21 20 13.5 0.90 0.89 1.1 0.79 10 9.6 0.64 0.64 0.0
5²+12²=13² → 25 + 144 = 169 a=7, b=(49-1)/2=24 and c=(49+1)/2. 7²+24²=25² → 49 + 576 = 625 a=9, b=(81-1)/2=40 and c=(81+1)/2=41. 9²+40²=41² → 81+1600=1681 a=11, b=(121-1)/2=60 and c=(121+1)/2=61. 11²+60²=61² → 121+3600=3721 a=13, b=(169-1)/2=84 and c=(169+1)/2=85. 13²+84²=85² → 169+7056=7225 Since there are an infinite number of