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\(^o\)
1
2
3
4
Correct answer is C
(n \(\oplus\) 4) \(\oplus\) 3 = 0 (mod 5)
(3 \(\oplus\) 4) \(\oplus\) 3
12 \(\oplus\) 3 = 15 (mod 5)
(5 x 3 + 0) = 0 (mod 5)
7.83\(m^2\)
32.29\(m^2\)
50.29\(m^2\)
82.50\(m^2\)
Correct answer is D
Area of path = Area of (pond+path) - Area of pond
The area of the pond with the path: The radius = (4 + 2.5)m = 6.5m
Area = \(\pi \times r^{2}\) = \(\frac{22}{7} \times 6.5^{2} \approxeq 132.79m^{2}\)
Area of the pond = \(\frac{22}{7} \times 4^{2} \approxeq 50.29m^{2}\)
Area of the path = (132.79 - 50.29)m^{2} = 82.50m^{2}\)
Calculate the variance of 2, 3, 3, 4, 5, 5, 5, 7, 7 and 9
2.2
3.4
4.0
4.2
Correct answer is D
x = \(\frac{2 + 3 + 3 + 4 + 5 + 5+ 5+ 7 + 7 + 9}{10}\)
=\(\frac{50}{10}\)
= 5
Variance = \(\frac{\sum{(x - x})^2}{N}\)
\(\frac{9 + 4+ 4+ 1 + 4 + 4 + 16}{10}\)
= \(\frac{42}{10}\)
= 4.2
N2,500.00
N2,600.00
N4,500.00
N4,600.00
Correct answer is D
P = \(\frac{100l}{RT} = \frac{100 \times 500}{5 \times 2.5} = \frac{50,000}{12.5}\)
= 4,000
Annual interest is \(\frac{500}{5}\) = 100
for 6 year = 4,000 + 100 + 100 + 100 + 100 + 100 + 100
= N 4,600
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