Maths

Part 1 Question 1 (Numerical relationships) Rainf either in Ross-On-Wye (mm) stratum| 1997| 1998| 1999| 2000| 2001| 2002| 2003| 2004| 2005| 2006| 2007| Rainfall(mm)| 17| 16| 6| 36| 79| 56| 66| 44| 61| 34| 161| (i) To calcul imbibe the normal amount of pelting sort the poesy in rising slope order as follows: 6, 16, 17, 34, 36, 44, 56, 61, 66, 79, 161 The median is the core entertain, in this gaucherie the median = 44 To calcul ingest the modest add all the value to playher and divide by the total amount of sums as follows: 576÷11 ?52.3 The taut is higher than the median as the stiff represents the rainfall of all the long time and as the rainfall in 2007 was especially high this impart influence the mean norm. This ordain non affect the average at the median, which l unmatchedsome(prenominal) looks at the middle value and does not take the extremes at either end into consideration.

compute of portions of harvest and vegetables e haven by a sample of people bod of portions fruit and vegetables, n| 0| 1| 2| 3| 4| 5| 6| 7| Number of people, f| 27| 16| 16| 31| 24| 20| 9| 7| (ii) (a) To calculate the circumstances of people who ate the recommended 5 or more portions of fruit and vegetables during the day as follows: 20+9+7=36 36÷cl×100=24 therefrom 24% of people ate the recommended 5 or more portions of fruit and vegetables per day. (b) To calculate the mean number of fruit and vegetables as follows: ?nf?f = 0×27+1×16+2×16+3×31+4×24+5×20+6×9+7×727+16+16+31+24+20+9+7 Or 440150 ?2.93 (2d .p.) Therefore the mean average is 2.93 (2d.! p.) (iii) If there atomic number 18 780 hedgehogs on an island and the world increases by 3% each year, the population after one year will be: 780×1.03 ? 803.4 (1d.p.) Therefore the number of hedgehogs will be 803. After 10 years the population will have increased by: 780×1.03^10 ?1048.3 (1d.p.) Therefore the number of hedgehogs will be 1048 after 10 years. (iv) If...If you want to get a full essay, order it on our website: BestEssayCheap.com

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