4. 5. 6. a3 2 b3 5 a3 2 a2b 1 a2b 2 ab2 1 ab2 2 b3 5 a2( a 2 b) 1 ab ( a 2 b) 1 b2( a 2 b) 5 ( a 2 b)( a2 1 ab 1 b2) (2x 1 y)(4x2 2 2xy 1 y2) (x 2 2y)(x2 1 2xy 1 4y2) (5 2 3d)(25 1 15d 1 9d2) (4x 1 3y)(16x2 2 12xy 1 9y2) a4 2 b4 5 a4 2 a3b 1 a3b 2 a2b2 1 a2b2 2
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.
Assignment One Answer 1. i) Range = 90 – 15 = 75 Mean = (15 + 24 + 30 + 45 + 66 + 90) / 6 = 45 Variance = ( (15 – 45)2 + (24 – 45)2 + (30 – 45)2 + (45 – 45)2 + (66 – 45)2 + (90 – 45)2 ) / 6 = 672 Standard deviation = √(672) = 25.923 ii) Range = 120 – 2 = 118 Mean = (2 + 29 + 49 + 67 + 90 + 109 + 120) / 7 = 66.571 Variance = ( (2 – 66.571)2 + (29 – 66.571)2 + (49 – 66.571)2 + (67 – 66.571)2 + (90 – 66.571)2 + (109 – 66.571)2 ) + (120 – 66.571) ) / 7 = 1584.816 Standard deviation = √(1584.816) = 39.810 2. i) A∪B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} (ii) A∩B = ∅ (iii) A∩C
2 NO + 1O2 ( 2 NO2 4. ____ Al(NO3)3 + ____ Ba(OH)2 ( ____ Al(OH)3 + ____ Ba(NO3)2 5. 1 H2 + 1Br2 ( 2 HBr 6. ____ HCl + ____MnO2 ( ____MnCl2 + ____ H2O + ____Cl2 7. ____ F2 + ____H2O ( ____ O2 + ____HF 8.
PRL 12. Cortisol 13. Melatonin 14. MSH 15. hGH 16. Calcitonin 17.
Adding the two cases above, we arrive at the answer: un = un−1 + un−2 . (c): Use either (a) or (b) to determine the number of bit strings of length 7 that do not contain two consecutive zeros. SOLUTION: We note directly that u1 = 2 and u2 = 3. Then u3 = 2 + 3 = 5, u4 = 3 + 5 = 8, u5 = 5 + 8 = 13, u6 = 8 + 13 = 21, and u7 = 13 + 21 = 34. Problem 3.
–necrosis 1.12. –rrhage 1.13. –ostomy 1.14. –otomy 1.15. –plasty 1.16.