K12_1821161 Jeopardy Jan 9, 2014 1:03:23 PM Page 4. Blackboard Collaborate ?? K12_1821161 $100 Question from 1 Jan 9, 2014 1:03:23 PM Page 5. Blackboard Collaborate ?? K12_1821161 $100 Answer from 1 Jan 9, 2014 1:03:23 PM Page 6.
65% 12% (a) (f) 10. (a) £100 30 m 40% (b) (g) (b) 150 kg £8 (c) (h) 30p 52p Ben: £10 (d) (i) (c) 10p £60 (e) (j) 1.5 kg 150 kg 90% 80% (b) (f) 70% 60% (c) (g) 55% 75% (d) 95% Adam: £15; Adam: £18; Ben: £12 17.4 Decimals, Fractions and Percentages 1. (a) (f) 2. (a) (f) 0.42 0.1 14% 90% (b) (g) (b) (g) 0.37 0.22 72% 18% (c) (h) (c) (h) 0.2 0.03 55% 4% 4 (d) (i) (d) (i) 0.05 0.15 40% 70% (e) 0.08 (e) 3% © The Gatsby Charitable Foundation MEP: Demonstration Project Teacher Support Y7B, P17 17.4 1 2 1 4 3 10 16 25 4 5 49 50 7 10 14 25 Answers (a) (f) (b) (g) (b) (e) (h) (k) (c) (h) (c) (f) (i) (l) (d) (i) (e) 3 20 3. 4.
(a) 11010001 (b) 000101010 6. (a) Negative (b) Negative 7. (a) 550ns (b) 600ns (c) 2.7us (d) 10V 8. 2ms 9. 250Hz 10.
13. The mean is 3.16667, the median is 3.25, the sum of squared deviations is 0.5333333334, the variance is 0.10667, and the squared deviation is 0.32660. 16. The Governors’ mean is 43 and the standard deviation is 6.83. The CEOs’ mean is 44 and the standard deviation is 12.65.
8-92, TSIS Ch. 2-3, and SFW pp. 233-48. Due: RR #3: ??? 6 9/28 The Rhetorical Situation and ILHL Read: ILHL pp.
72 PSATP = P part(O2) + P part(CO2) + P part(N2) P part(N2) = PSATP - P part(O2) - P part(CO2) P part(N2) = 101.3 kPa – 79 kPa – 0.75 kPa = 21.55 kPa Answer: P part(N2) = 21.55 kPa 73 PV=nRT n = PV/RT P = 2500 kPa V = 12 L R = 8.314 J/mol*K T = 22 + 273.15 = 295.15 K n = (2500000*12*10-3m3)/ (8.314 J/mol*K*295.15 K) = 12.22 mol. Answer: n = 12.22 mol. 74 C3H8 + 4O2 => 3CO2 + 2H2O n (O2) = 3.5 mol * 4 = 14 mol PV = nRT V(O2) = nRT/P V(O2) = (14 mol * 8.314 J/mol*K *(28 + 273.15) K) / 1.013 * 105 Pa V(O2) = 0.3311 m3 = 331.1 L Answer: V(O2) = 331.1 L 75 PV=nRT n = PV/RT n = (1.013*105 Pa * 65 * 10-3 m3) / (8.314 J/mol*K * 298 K) = 2.65 mol Answer: n = 2.65 mol 76 PV=nRT n = PV/RT n = (1.013*105 Pa * 7.5 * 10-3 m3) / (8.314 J/mol*K * 298 K)
Multiply each term by the common ratio to find the next three terms. 25 × 0.5 = 12.5 12.5 × 0.5 = 6.25 6.25 × 0.5 = 3.125 The next three terms of the sequence are 12.5, 6.25, and 3.125. 7. 4, −1, ,… SOLUTION: Calculate the common ratio. The common ratio is × × × = = = .
SOLUTION: We observe directly that t1 = t2 = 0 and t3 = 1. Now according to the relation in (a), t4 = 1 + 0 + 0 + 21 = 3. Likewise, t5 = 3 + 1 + 0 + 22 = 8, t6 = 8 + 3 + 1 + 23 = 20, and t7 = 20 + 8 + 3 + 24 = 47. So the answer is 47. Problem 2.
(3d) 97, 127 B.C.A.C. 76, 207 W.A.C. 76, 175 D.L.R. (4th) 1, 25 C.R. (5th) 215, 69 B.C.L.R.
We can conclude that the data are Poisson distributed. Chi-Square test of independence Problem 12.12 Use the following contingency table to determine whether variable 1 is independent of variable 2. Let α = .01 | Variable 2 | Variable1 | 24 | 13 | 47 | 58 | | 93 | 59 | 187 | 244 | Step 1 Ho: the two classifications are independent Ha: the two classifications are dependent Step 2 d.f = (r – 1) (c – 1) Step 3 α = 0.01 x 2 0.01, 3df = 11.3449 Step 4 Reject Ho if x 2 > 11.3449 | Variable 2 | Total | Variable1 | 24 (22.92) | 13 (14.10) | 47 (45.83) | 58 (59.15) | 142 | | 93 (94.08) | 59 (57.90) | 187 (188.17) | 244 (242.85) | 583