Evaluate \(\frac{3524}{0.05}\) correct to 3 significant figures
705
70,000
70,480
70,500
Correct answer is D
\(\frac{3524}{0.05}\) = 70480
\(\approx\) 70500(3 s.g)
1078
1068
718
178
Correct answer is A
\(\begin{array}{c|c} 8 & 71 \\ 8 & 8 \text{rem} 7\\ 8 & 1 \text{rem} 0\end{array}\)
= 1078
\(\frac{1}{2}\)
\(\frac{2}{3}\)
\(\frac{3}{8}\)
\(\frac{3}{11}\)
Correct answer is C
40 = 20 - x + x + 35 - x
40 = 55 - x
x = 55 - 40
= 15
∴ P(both) \(\frac{15}{40}\)
= \(\frac{3}{8}\)
\(\frac{3}{2}\)
\(\frac{9}{4}\)
\(\frac{5}{2}\)
3
Correct answer is B
\(\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \\ \hline 1 & 2 & 2 & -2 & 4 & 8\\ 2 & 1 & 2 & -1 & 1 & 1\\ 3 & 2 & 6 & 0 & 0 & 0\\ 4 & 1 & 4 & 1 & 1 & 1\\ 2 & 2 & 10 & 2 & 4 & 8\\ \hline & 8 & 24 & & & 18 \end{array}\)
x = \(\frac{\sum fx}{\sum f}\)
= \(\frac{24}{8}\)
= 3
Variance (62) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\)
= \(\frac{18}{8}\)
= \(\frac{9}{4}\)
10\(\frac{1}{2}\)
11\(\frac{1}{2}\)
12
13
Correct answer is C
Median = L + [\(\frac{\frac{N}{2} - f}{fm}\)]h
N = Sum of frequencies
L = lower class boundary of median class
f = sum of all frequencies below L
fm = frequency of modal class and
h = class width of median class
Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5
= 11 + (\(\frac{25 - 21}{20}\))5
= 11 + (\(\frac{(4)}{20}\))
11 + 1 = 12