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F.3 Measures of Central Tenden
There are 40 students in F3A.In a mathematics test,the mean score of the whole class is 52.4.The mean score for boys is 8 less than that of girls.Let X be the number of boys and Y be the mean score of boys.a) Show that 5Y-X=222b) If P is the sum of both mean scores of the boys and girls, find the minimum... 顯示更多 There are 40 students in F3A.In a mathematics test,the mean score of the whole class is 52.4.The mean score for boys is 8 less than that of girls.Let X be the number of boys and Y be the mean score of boys. a) Show that 5Y-X=222 b) If P is the sum of both mean scores of the boys and girls, find the minimum value of P.
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The number of boys is X, then the number of girls is (40-X) The mean score of boys Y, then the mean score of girls is (Y+8) (a) (X*Y+(40-X)*(Y+8))/40=52.4 XY+40Y+320-XY-8X=2096 40Y-8X=1776 8(5Y-X)=1776 5Y-X=222 (b) P=Y+(Y+8) P=2Y+8 From(a),Y=(222+X)/5 P=2((222+X)/5)+40/5 P=(444+2X+40)/5 P=(484+2X)/5 In F3A, there is at least one boy So, the minimum number of boy is 1, then X=1 P=(484+2(1))/5 P=486/5 P=97.2 So, the minimum value of P is 97.2 2015-01-09 13:34:47 補充： In(b),it is impossible that X=0 If X=0, then Y must be 0. Put X=0 and Y=0 into 5Y-X=222 is impossible. So the value of X must be greater than 0.
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F.3 Measures of Central Tenden
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發問:There are 40 students in F3A.In a mathematics test,the mean score of the whole class is 52.4.The mean score for boys is 8 less than that of girls.Let X be the number of boys and Y be the mean score of boys.a) Show that 5Y-X=222b) If P is the sum of both mean scores of the boys and girls, find the minimum... 顯示更多 There are 40 students in F3A.In a mathematics test,the mean score of the whole class is 52.4.The mean score for boys is 8 less than that of girls.Let X be the number of boys and Y be the mean score of boys. a) Show that 5Y-X=222 b) If P is the sum of both mean scores of the boys and girls, find the minimum value of P.
最佳解答:
The number of boys is X, then the number of girls is (40-X) The mean score of boys Y, then the mean score of girls is (Y+8) (a) (X*Y+(40-X)*(Y+8))/40=52.4 XY+40Y+320-XY-8X=2096 40Y-8X=1776 8(5Y-X)=1776 5Y-X=222 (b) P=Y+(Y+8) P=2Y+8 From(a),Y=(222+X)/5 P=2((222+X)/5)+40/5 P=(444+2X+40)/5 P=(484+2X)/5 In F3A, there is at least one boy So, the minimum number of boy is 1, then X=1 P=(484+2(1))/5 P=486/5 P=97.2 So, the minimum value of P is 97.2 2015-01-09 13:34:47 補充： In(b),it is impossible that X=0 If X=0, then Y must be 0. Put X=0 and Y=0 into 5Y-X=222 is impossible. So the value of X must be greater than 0.
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