5AM2H_01 Question Working Answer Mark Notes 15 52 2.52 = 18.75 4.33 4 B1 for a right-angled triangle shown or 2.5 or 30 or 60 = 4.33 seen. OR M1 for 52 2.52 5 × cos 30 M1 for or A1 for 4.33 or better OR 5 × sin30 OR M1 for = cos 30 OR M1 for 5 × cos 30 × 2 A1 for 4.33 or better OR M1 for = tan 30 M1 for (x =) A1 for 4.33 or better *16 = 0.68 Steven with explanation 3 M1 for either probability or oe seen A1 for both converted correctly to a comparable form = 0.7 C1 for identification of Steven with explanation based on correct comparison, e.g. > ‘0.7 0.68’ 5AM2H_01 Question Working Answer Mark Notes 17 = = 0.09 3 M1 for = (accept any k 0 or 1) P kr2 P r2 P kr2 ≠ 36 = × 202 M1 (dep) for 36 = × 202 k k = 0.09 A1 for = 0.09 oe P r2 P r2 OR OR M2 for oe, e.g. 202 : = 36 : r2 P A1 for oe
5AM2H_01 Question Working Answer Mark Notes 15 52 2.52 = 18.75 = 4.33 OR 5 × cos 30 OR 5 × sin30 OR × 2 4.33 4 B1 for a right-angled triangle shown or 2.5 or 30 or 60 seen. M1 for 52 2.52 M1 for or A1 for 4.33 or better OR M1 for = cos 30 M1 for 5 × cos 30 A1 for 4.33 or better OR M1 for = tan 30 M1 for (x =) A1 for 4.33 or better *16 = 0.68 = 0.7 Steven with explanation 3 M1 for either probability or oe seen A1 for both converted correctly to a comparable form C1 for identification of Steven with explanation based on correct comparison, e.g. ‘0.7 > 0.68’ 5AM2H_01 Question Working Answer Mark Notes 17 P = kr2 36 = k × 202 P = 0.09 r2 OR P = 0.09 r2 3 M1 for P = kr2 (accept any k ≠ 0 or 1) M1 (dep) for 36 = k × 202 A1 for P = 0.09 r2 oe OR M2 for oe, e.g. 202 : r2 = 36 : P A1 for oe
6849_08b_5am2h_applications_unit_2h_-_november_2011_mark_scheme.docx