Find the least value of n for which \(^{3n}C_{2} > 0, n \in R\)
\(\frac{1}{3}\)
\(\frac{1}{6}\)
\(\frac{2}{3}\)
1
Correct answer is C
\(^{3n}C_{2} > 0 \implies \frac{3n!}{(3n - 2)! 2!} > 0\)
\(\frac{3n(3n - 1)(3n - 2)!}{(3n - 2)! 2} > 0\)
\(\frac{3n(3n - 1)}{2} > 0\)
\(3n(3n - 1) > 0 \implies n > 0; n > \frac{1}{3}\)
The least number in the option that satisfies \(n > 0; n > \frac{1}{3} = \frac{2}{3}\)
(-6, -15)
(-6, 5)
(20, 5)
(20, 15)
Correct answer is C
No explanation has been provided for this answer.
0.10
0.30
0.60
0.70
Correct answer is B
Let the number of students that like both Science and History = x
Number of Science only = 35 - x
Number of History only = 30 - x
35 - x + 30 - x + x = 50
65 - x = 50
x = 15
P(Science and History) = \(\frac{15}{50} = 0.30\)
mn
m + n
m + n - 2mn
1 - mn
Correct answer is C
P(only one alive) = P(husband and not wife) + P(wife and not husband)
= m (1 - n) + n ( 1 - m)
= m - mn + n - mn
= m + n - 2mn
\(\frac{1}{3}\)
\(\frac{4}{9}\)
\(\frac{5}{9}\)
\(1\)
Correct answer is A
Differentiate distance twice to get the acceleration and then equate to get p.
\(s = 7 + pt^{3} + t^{2}\)
\(\frac{\mathrm d s}{\mathrm d t} = v(t) = 3pt^{2} + 2t\)
\(\frac{\mathrm d v}{\mathrm d t} = a(t) = 6pt + 2\)
\(a(3) = 6p(3) + 2 = 8 \implies 18p = 8 - 2 = 6\)
\(p = \frac{6}{18} = \frac{1}{3}\)