\(\frac{1}{3}\)
\(\frac{2}{3}\)
1\(\frac{1}{3}\)
1\(\frac{2}{3}\)
Correct answer is C
\(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)
= \(\frac{\sqrt{2} \times \sqrt{3} + \sqrt{3} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)
= \(\frac{\sqrt{6} + 3}{3}\)
= \(\frac{3 + \sqrt{6}}{3}\)
= Hence, (m + n) = 1 + \(\frac{1}{3}\)
= 1\(\frac{1}{3}\)
Tom will be 25 years old in n years' time. If he is 5 years younger than Bade's present age.
(30 - n)years
(20 - n)years
(25 - n)years
(30 + n)years
Correct answer is A
Let Tom's present agr be x.
Then x = 25 - n
If Tom is 5 years younger than Bade, then Bade's present age is x + 5 = 25 - n + 5
= (30 - n)
In the diagram, VW//YZ, |WX| = 6cm, |XY| = 16cm, |YZ| = 20cm and |ZX| = 12cm. Calculate |VX|
3cm
4cm
6cm
8cm
Correct answer is D
In the diagram, \(\frac{16}{12} = \frac{VX}{6}\) (similar \(\Delta\)s)
VX = \(\frac{16 \times 6}{12}\)
= 8cm
\(\frac{2}{5}\)
\(\frac{7}{60}\)
\(\frac{1}{30}\)
\(\frac{1}{60}\)
Correct answer is C
Hence the probability that only Kebba will hit the target
= P(K)xP(E')xP(O')
= \(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\)
= \(\frac{1}{30}\)
5.6
6.2
6.6
7.0
Correct answer is D
To calculate the mean of grouped data,
- First step to determine the midpoint (x) of each interval or class.
0 - 4 ►2
5 - 9 ► 7
10 - 14 ►12
These midpoints must then be multiplied by the frequencies of the corresponding classes:
2 X 2 = 4
1 X 7 = 7
2 X 12 = 24
Mean = ( 24 + 7 + 4) ÷ ( 2 + 1 + 2 )
: Mean = 7
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