75o
150o
160o
68o
Correct answer is B
Interior angle = 2(75°)
= 150°
In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40°. Calculate ∠QSR
30o
40o
45o
50o
Correct answer is D
< Q = < R (OQ = OR = radii)
< QOR = 180° - 2(40°) = 100°
< QSR = < RPQ = \(\frac{1}{2}\) < QOR
= \(\frac{100}{2} = 50°\)
I and II only
III only
II only
II and III only
Correct answer is B
No explanation has been provided for this answer.
\(\frac{1}{10}\)
\(\frac{3}{10}\)
\(\frac{9}{20}\)
\(\frac{2}{3}\)
Correct answer is A
P(Kodjo passing) = \(\frac{3}{4}\); P(Adoga passing) = \(\frac{3}{5}\)
P(Kodjo failing) = \(\frac{1}{4}\); P(Adoga failing) = \(\frac{2}{5}\)
P(both fail) = \(\frac{1}{4} \times \frac{2}{5}\)
= \(\frac{1}{10}\)
Calculate the standard deviation of the following marks; 2, 3, 6, 2, 5, 0, 4, 2
1.5
1.7
1.8
1.9
Correct answer is C
| x | 2 | 3 | 6 | 2 | 5 | 0 | 4 | 2 | 24 |
| x - \(\bar{x}\) | -1 | 0 | 3 | -1 | 2 | -3 | 1 | -1 | |
| \((x - \bar{x})^2\) | 1 | 0 | 9 | 1 | 4 | 9 | 1 | 1 | 26 |
Mean = \(\frac{24}{8} = 3\)
Standard deviation = \(\sqrt{\frac{26}{8}}\)
= \(\sqrt{3.25}\)
= 1.802 \(\approeq\) 1.8
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