2014 - WAEC Mathematics Past Questions and Answers - page 4
p = { 1, 2, 3}
Q = {2, 3, 5}
prob(prime number) = prob(1 and 2) or
= prob(2 and 3) or
= prob(3 and 2)
There are three possibilities of the sum being prime
Total possibility = 9
probability(sum being prime) = (\frac{3}{9})
= (\frac{1}{3})
(3 \sqrt{0.0005957})
(\begin{array}{c|c} \log No(0.0005957) & \frac{1}{3} \log \frac{1}{3} \ \hline (0.0005957)^{\frac{1}{3}} & 4.7750 \times \frac{1}{3} \ & 6 + 2.7750 \hline & 3 \hline & 2.9250\end{array})
prob(p) = (\frac{1}{5})
prob(Q) = (\frac{1}{4})
Prob(neither p) = 1 - (\frac{1}{5})
(\frac{5 - 1}{5} = \frac{4}{5})
prob(neither Q) = 1 - (\frac{1}{4})
(\frac{4 - 1}{4} = \frac{3}{4})
prob(neither of them) = (\frac{4}{5} \times \frac{3}{4} = \frac{12}{20})
= (\frac{3}{5})
number of sides = (\frac{360^o}{\theta} = \frac{360^o}{306o})
n = 12o
Sum of interior angle = (n - 2) 180o
(12 - 2) 180v = 10 x 180o
= 1800o
T1, T2, T3
2, -9, -20 .... -141
l = a + (n - 1)d
first term, a = 2
common difference d = T3 - T2
= T2 - T1
= -20 - (-9) = -9 -2
= -20 + 9
= -9 -2
= -20 + 9
= -11
-11 = -11
d = -1
last term l = -141
-141 = 2 + ((\cap) - 1) (-11)
-141 = 2 + (-11 (\cap) + 11)
= 2 - 11(\cap) + 11
-141 = 13 - 11(\cap)
-141 - 13 = -11(\cap)
-154 = -11(\cap)
(\cap) = (\frac{-154}{-11})
(\cap) = 14
volume of cylinder = (\pi r^2h)
volume of cylinder p = (\pi r^2 \times 10)
= 10(\pi r^2)
volume of cylinder Q = (\pi (2r)^2 h)
= 4(\pi r^2)h
4(\pi r^2 = 10 \pi r^2 h)
h = (\frac{10}{4} = 4.5cm)
score 0 = 4 students
score 1 = 2 students
score 2 = 5 students
score 3 = 6 students
score 4 = 3 students
= 20 students
at most 2 marks = 5 + 2 + 4 students = 11 students
probability(at most 2 marks) = (\frac{11}{20})
y + z = 180o
z = 180o - y
x + k = 180o
k = 180o - x
the sum of angles in the (\bigtriangleup) are: n + z + k = 180
but n + 180o - y + 180o - x = 180o
n + 180o - y - x = 0
but we want to find n
n + 180o = x + y
but x + y = 220o
n + 180o = 220o
n = 220o - 180o
n = 40o