If \(2\sin^{2} \theta = 1 + \cos \theta, 0° \leq \theta \leq 90°\), find the value of \(\theta\).
90°
60°
45°
30°
Correct answer is B
\(2\sin^{2} \theta = 1 + \cos \theta\)
\(2 ( 1 - \cos^{2} \theta) = 1 + \cos \theta\)
\(2 - 2\cos^{2} \theta = 1 + \cos \theta\)
\(0 = 1 - 2 + \cos \theta + 2\cos^{2} \theta\)
\(2\cos^{2} \theta + \cos \theta - 1 = 0\)
Factorizing, we have
\((\cos \theta + 1)(2\cos \theta - 1) = 0\)
Note: In the range, \(0° \leq \theta \leq 90°\), all trig functions are positive, so we consider
\(2\cos \theta = 1 \implies \cos \theta = \frac{1}{2}\)
\(\theta = 60°\).
Simplify: \(^{n}C_{r} ÷ ^{n}C_{r-1}\)
\(\frac{n(n-r)}{r}\)
\(\frac{n}{r(n-r)}\)
\(\frac{1}{r(n-r)}\)
\(\frac{n+1-r}{r}\)
Correct answer is D
\(^{n}C_{r} = \frac{n!}{(n-r)! r!}\)
\(^{n}C_{r - 1} = \frac{n!}{(n - (r - 1))! (r - 1)!}\)
\(^{n}C_{r} ÷ ^{n}C_{r - 1} = \frac{n!}{(n - r)! r!} ÷ \frac{n!}{(n-(r-1))!(r-1)!}\)
= \(\frac{n!}{(n-r)! r!} \times \frac{(n-(r-1)! (r-1)!}{n!}\)
= \(\frac{(n + 1 - r)! (r - 1)!}{(n - r)! r!}\)
= \(\frac{(n+1-r)(n-r)! (r-1)!}{(n-r)! r (r - 1)!}\)
= \(\frac{n + 1 - r}{r}\)
24
23
21
19
Correct answer is C
Let the sum of the 8 numbers = y.
\(\frac{y}{8} = 20 \implies y = 20 \times 8 = 160\)
\(160 - 17 = 143 (\text{the sum of the other 7 numbers})\)
\(mean = \frac{143 + 25}{8} = \frac{168}{8}\)
= 21
Four fair coins are tossed once. Calculate the probability of having equal heads and tails.
\(\frac{1}{4}\)
\(\frac{3}{8}\)
\(\frac{1}{2}\)
\(\frac{15}{16}\)
Correct answer is B
Let \(p(head) = p = \frac{1}{2}\) and \(p(tail) = q = \frac{1}{2}\)
\((p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2}q^{2} + 4pq^{3} + q^{4}\)
The probability of equal heads and tails = \(6p^{2}q^{2} = 6(\frac{1}{2}^{2})(\frac{1}{2}^{2})\)
= \(\frac{6}{16} = \frac{3}{8}\).
Given that \(y = 4 - 9x\) and \(\Delta x = 0.1\), calculate \(\Delta y\).
9.0
0.9
-0.3
-0.9
Correct answer is D
\(y = 4 - 9x\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{\Delta y}{\Delta x} = -9\)
\(\frac{\Delta y}{0.1} = -9 \implies \Delta y = -9 \times 0.1 = -0.9\)