2011 - WAEC Mathematics Past Questions and Answers - page 5
< NOP = 180 - 125 = 55o(< s on a straight line)
But < NOP = < ONM (alternate < s)
< ONM = 55o
< M + < N + < MQN = 180o (sum of interior < s of a (\bigtriangleup)) i.e
65o + 55o + < MQN = 180
< MQN = 180 - 12o = 60
< MQR + < MQN = 180 (< s on straight line)
< MQR + 60 = 180
< MQR + 60 = 180
< MQR = 180 - 60
= 120o
The graph represents the relation y = x\(^2\) - 3x - 3. What is the equation of the line of symmetry of the graph?
< TUW = 110o = 180o (< s on a straight line)
< TUW = 180o - 110o = 70o
In (\bigtriangleup) XTU, < XUT + < TXU = 180o
i.e. < YTS + 70o = 180
< XTU = 180 - 110o = 70o
Also < YTS + < XTU = 180 (< s on a straight line)
i.e. < YTS + < XTU - 180(< s on straight line)
i.e. < YTS + 70o = 180
< YTS = 180 - 70 = 110o
in (\bigtriangleup) SYT + < YST + < YTS = 180o(Sum of interior < s)
SYT + 40 + 110 = 180
< SYT = 180 - 150 = 30
< SYT = < XYW (vertically opposite < s)
Also < SYX = < TYW (vertically opposite < s)
but < SYT + < XYW + < SYX + < TYW = 360
i.e. 30 + 30 + < SYX + TYW = 360
but < SYX = < TYW
60 + 2(< TYW) = 360
2(< TYW) = 360o - 60
2(< TYW) = 300o
TYW = (\frac{300}{2}) = 150o
< SYT
student that like swimming = x + 2
where 2 is the number of students who like reading, dancing and swimming. To find x from the venn diagram of swimming
6 +3 + 2 + x = 16
11 + x = 16
x = 16 - 11 = 5
no. of students that like dancing and swimming
x + 2 = 5 + 2 = 7
< QPR = < STU = 65o (Corresponding angles)
245 + < QPR = x = 360o (< s at a point)
i.e. 245 + 65 + x = 360
x = 360 - (245 + 65)
x = 360 - 310
x = 50o
< QPR = < STU = 65o (Corresponding angles)
245 + < QPR = x = 360o (< s at a point)
i.e. 245 + 65 + x = 360
x = 360 - (245 + 65)
x = 50o
Using the cumulative frequency curve, estimate the median of the data represented on the graph.
Using the cumulative frequency curve. What is the 80th percentile?