\(\frac{3}{25}\)
\(\frac{1}{5}\)
\(\frac{6}{25}\)
\(\frac{2}{5}\)
Correct answer is D
No. of teams that scored at most 3 goals = 3 + 1 + 6 = 10
Hence, probability that a team selected at random scored at most 3 goals
= \(\frac{10}{25}\) = \(\frac{2}{5}\)
\(\frac{9}{25}\)
\(\frac{1}{5}\)
\(\frac{6}{25}\)
\(\frac{2}{5}\)
Correct answer is A
Prob. (team scored 4 goals) = Prob. (team scored 7 goals) = \(\frac{3}{25}\)
Hence, probability that a team selected at random scored either 4 or 7 goals;
= \(\frac{6}{25} + \frac{3}{25}\)
= \(\frac{9}{25}\)
The mean of 1, 3, 5, 7 and x is 4. Find the value of x
2
4
6
8
Correct answer is B
Mean = \(\frac{\sum x}{n}\)
4 = \(\frac{1 + 3 + 5 + 7 + x}{5}\)
4 x 5 = 16 + x
20 - 16 = x
4 = x
x = 4
1
2
3
9
Correct answer is A
Using the two - point from
\(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\)
\(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\)
\(\frac{y - 2}{-4} = \frac{x - 4}{-12}\)
\(\frac{-12(y -2)}{-4}\) = x - 4
3(y -2) = x -4
3y - 6 = x - 4
3y = x - 4 + 6
3y = x + 2...
By comparing the equations;
3y = px + , p = 1
If M and N are the points (-3, 8) and (5, -7) respectively, find |MN|
8 units
11 units
15 units
17 units
Correct answer is D
|MN| = \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\)
= \(\sqrt{(-3 -5)^2 + (8 - 7)^2}\)
= \(\sqrt{(-8)^2 + (8 + 7)^2}\)
= \(\sqrt{64 + (15)^2}\)
= \(\sqrt{64 + 225}\)
= \(\sqrt{289}\)
= 17 units
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