Collects their dy dx terms in an equation and

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Collects theirdydxterms in anequation and factorises3.1aM1Completes convincing argumentto obtain required result byfactorisingthen simplifyingyAG2.1R19(b)(ii)Begins argument by setting0dydx=to form an equation forxandyPIby2.1M1For stationary points0dydx=()322222220221212xyyyxx yxx yx y+==+=⇒ −=Since22x y< 0 there can be nostationary points.Obtainsory =√−2𝑥𝑥orx=𝑦𝑦221.1bA1Substitutesorx=𝑦𝑦22into equation for curve1.1aM1Completes convincing argumentto deduce the required result2.2aR19(b)(iii)Substitutesy= 1 into equation ofcurve to obtain correct quadraticACF3.1aM1()()211203073071330yxxxxdydxyx= ⇒+====−Deducesx= 3PIby substituting theirxin theirdy/dx2.2aR1Substitutes theirxandy= 1 intheir dy/dx1.1aM1Obtains correct equation oftangentACFISW1.1bA1Total150x=3222,24xyyxxy+= −+22yx= −22yx= −320xyy+=10
MARK SCHEME – A-LEVEL MATHEMATICS – 7357/3 – JUNE 2019QMarking InstructionsAOMarkTypical Solution10Ticks correct box1.2B1Strong negativeTotal1QMarking InstructionsAOMarkTypical Solution11Circles correct answer1.2B1QuotaTotal1QMarking InstructionsAOMarkTypical Solution12(a)Calculates correct value of mean(accept 161)1.1bB1x=160.6sd = 6.8160.6 – 2 × 6.8 = 147146 < 147Hence Ann is an outlierCalculates correct value ofstandard deviation(accept 7.2 or better)1.1bB1Usestheir𝑥𝑥𝑎𝑎𝑛𝑛𝑑𝑑their𝑠𝑠.𝑑𝑑in𝑥𝑥2 ×𝑠𝑠.𝑑𝑑(accept 146.2)1.1bM1Compares 146 with theircalculation and correctlyconcludes that Ann’s height is anoutlierFTtheir𝑥𝑥and their s.d2.1R1F12(b)States correctly that the meanwouldincreasewith a validreasonorincreasesto 162.2Accept the mean would increaseas the lower/lowest value hasbeen removed or other validreason2.2bB1The mean would increase becauseAnn’s height is less than the meanStandard deviation would decreasebecause Ann’s height is an outlierStates correctly that the standarddeviation woulddecreasewith avalid reasonordecreasesto 5.03Accept the standard deviationwould decrease because the datais less spread out or other validreason2.2bB1Total611
MARK SCHEME – A-LEVEL MATHEMATICS – 7357/3 – JUNE 2019QMarking InstructionsAOMarkTypical Solution13(a)(i)Obtains correct mean1.1bB1613(a)(ii)Obtains correct variance1.1bB14.813(b)(i)Uses the Binomial formula withn= 30,p= 0.2 or𝑃𝑃(𝑋𝑋 ≤10)− 𝑃𝑃(𝑋𝑋 ≤9)PIby correct answer1.1aM1𝑃𝑃(𝑋𝑋= 10) =30100.2100.820= 0.0355Obtains correct probabilityAWFW[0.035, 0.036]1.1bA113(b)(ii)Calculates either𝑃𝑃(𝑋𝑋 ≤4) = 0.255or𝑃𝑃(𝑋𝑋 ≤5)= 0.4275using the Binomialdistribution3.1bM1𝑃𝑃(𝑋𝑋 ≤4) = 0.255𝑃𝑃(𝑋𝑋 ≥5) = 1− 𝑃𝑃(𝑋𝑋 ≤4)= 10.255= 0.745States𝑃𝑃(𝑋𝑋 ≥5)= 1𝑃𝑃(𝑋𝑋 ≤4)orsubtracts their stated value of𝑃𝑃(𝑋𝑋 ≤4)from 11.1bM1Obtains correct probabilityAWFW[0.74, 0.75]1.1bA113(c)(i)Raises their 0.745 to power of 53.1bM10.7455= 0.229Obtains their correct probabilityFTtheir0.745AWRTtheir0.2291.1bA1F13(c)(ii)Gives a valid reason thatprobability/likelihood/chances maychange/increase/decrease as a resultof external factor change over 5 dayperiod or Patrick improves3.5bE1Probability may change asPatrick improvesTotal1012

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MARK SCHEME, Kompakt

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