Convert 2710 to another number in base three
10013
10103
11003
10003
Correct answer is D
\(\begin{array}{c|c}
3 & \text{27 rem 0} \\
\hline
3 & \text{ 9 rem 0} \\
\hline
3 & \text{ 3 rem 0} \\
\hline
3 & \text{ 1 rem 1}\\
\hline
& 0
\end{array}\)
Hence the correct answer is 10003
\(\frac{2}{15}\)
\(\frac{1}{10}\)
\(\frac{1}{3}\)
\(\frac{2}{5}\)
Correct answer is C
Sample space S = {10, 11, 12, ... 30}
Let E denote the event of choosing a number divisible by 3
Then E = {12, 15, 18, 21, 24, 27, 30} and n(E) = 7
Prob (E) = \(\frac{n(E)}{n(S)}\)
Prob (E) = \(\frac{7}{21}\)
Prob (E) = \(\frac{1}{3}\)
\(\frac{1}{5}\)
\(\frac{1}{2}\)
\(\frac{2}{5}\)
\(\frac{3}{4}\)
Correct answer is C
Let E demote the event of obtaining at least a 4
Then n(E) = 16 + 10 + 14 = 40
Hence, prob (E) = \(\frac{n(E)}{n(S)}\)
\( = \frac{40}{100}\)
\( = \frac{2}{5}\)
In how many ways can a team of 3 girls be selected from 7 girls?
\(\frac{7!}{3!}\)
\(\frac{7!}{4!}\)
\(\frac{7!}{3!4!}\)
\(\frac{7!}{2!5!}\)
Correct answer is C
A team of 2 girls can be selected from 7 girls in \(^7C_3\)
\( = \frac{7!}{(7 - 3)! 3!}\)
\( = \frac{7!}{4! 3!} ways\)
Find the standard deviation of 5, 4, 3, 2, 1
\(\sqrt{2}\)
\(\sqrt{3}\)
\(\sqrt{6}\)
\(\sqrt{10}\)
Correct answer is A
Mean x = \(\frac{\sum x}{n}\)
\( = \frac{5 + 4 + 3 + 2 + 1}{5}\)
\( = \frac{15}{5}\)
= 3
\(\begin{array}{c|c}
x & d = x - 3 & d^2 \\
\hline
5 & 2 & 4 \\
4 & 1 & 1 \\
3 & 0 & 0 \\
2 & -1 & 1 \\
1 & -2 & 4 \\
\hline
& & \sum d^2 + 10
\end{array}\)
Hence, standard deviation;
\( = \sqrt{\frac{\sum d^2}{n}} = \sqrt{\frac{10}{5}}\)
\( = \sqrt{2}\)