\(\frac{2}{9}\)
\(\frac{4}{9}\)
\(\frac{5}{9}\)
\(\frac{7}{9}\)
Correct answer is B
28 - x + x + 25 - x = 45
53 - x = 45
x = 53 - 45
x = 8
chemistry only = 28 - 8
= 20
Probability = \(\frac{20}{45}\)
= \(\frac{4}{9}\)
103\(^o\)
123\(^o\)
133\(^o\)q
143\(^o\)
Correct answer is B
< SQP = 180 - (90 + 33) < on a ----
= 180 - (123)
= 57\(^o\)
Therefore, (m + n) = 123\(^o\)
The diagram shows a circle O. If < ZYW = 33\(^o\) , find < ZWX
33\(^o\)
57\(^o\)
90\(^o\)
100\(^o\)
Correct answer is C
In ZY = 90\(^o\) < subtends In a semi O
ZWY = 180 - (90\(^o\) + 33)
= 57
ZWX = 57 + 33 = 90\(^o\)
In the diagram, XY is a straight line.
60\(^o\)
90\(^o\)
100\(^o\)
120\(^o\)
Correct answer is B
<POX = <POQ; <ROY = QOR
2 <POQ + 2 <ROY = 180
2(<POQ = <ROY) = 180
<POQ + <ROY = 90
The diagram shows a circle centre O. if <STR = 29 and <RST = 45, calculate the value of <STO
12\(^o\)
15\(^o\)
29\(^o\)
34\(^o\)
Correct answer is A
SRT = 180 - (46 + 29) sum of < s in a
= 180 - 75
= 105
SOT = 2 x 46 < at the centre is twice all the circle = 92
RTO = 180 - (96 + 43)
= 41
STO = 41 - 29
= 12\(^o\)
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