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To evaluate the relation $R$ on the set $A = {0, 1, 2, 3, 4, 5}$, we first need to understand the conditions for an element $(x, y)$ to be in $R$. Specifically, $(x, y) ∈ R$ if and only if $max{x,y} ∈ {3,4}$. Considering this, let's list the pairs: For $max{x, y} = 3$, the possible pairs are: $(0, 3), (3, 0), (1, 3), (3, 1), (2, 3), (3, 2), (3, 3)$ For $max{x, y} = 4$, the possible pairs are: $(0, 4), (4, 0), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4)$ Combining these, the set $R$ consists of the following elements: $R = {(0, 3), (3, 0), (1, 3), (3, 1), (2, 3), (3, 2), (3, 3), (0, 4), (4, 0), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4)}$ This gives us a total of 16 elements in $R$, not 18 as initially claimed in statement $S1$. Next, we analyze the properties of the relation $R$: Reflexivity: A relation is reflexive if $(x, x) ∈ R$ for all $x ∈ A$. For example, $(0, 0), (1, 1), (2, 2)$ are not in $R$, so $R$ is not reflexive. Symmetry: A relation is symmetric if whenever $(a, b) ∈ R$, then $(b, a) ∈ R$ as well. For all pairs $(x, y)$ listed, both $(x, y)$ and $(y, x)$ are present. Thus, $R$ is symmetric. Transitivity: A relation is transitive if whenever $(a, b) ∈ R$ and $(b, c) ∈ R$, then $(a, c) ∈ R$. An example where transitivity fails is $(0, 3)$ and $(3, 1)$ are in $R$ but $(0, 1)$ is not in $R$. Therefore, $R$ is not transitive. In conclusion, statement $S2$ is correct as $R$ is symmetric but neither reflexive nor transitive.

Probability distribution $x = 0$, $p = \frac{^7C_2}{^{10}C_2} = \frac{42}{90}$ $x = 1$, $p = \frac{^7C_1 \times ^3C_1}{^{10}C_2} = \frac{42}{90}$ $x = 2$, $p = \frac{^3C_2}{^{10}C_2} = \frac{6}{90}$ Now, $\mu = \sum x_i p_i = \frac{42}{90} + \frac{12}{90} = \frac{54}{90}$ $\sigma^2 = \sum p_i x_i^2 - \mu^2 = \frac{42}{90} + \frac{24}{90} - (\frac{54}{90})^2$ $\Rightarrow \frac{66}{90} - (\frac{54}{90})^2$ $\sigma^2 \Rightarrow \frac{28}{75}$

$i^{k_1} + i^{k_2} \ne 0 \Rightarrow i^{k_1} \rightarrow 4$ option for $i, -1, -i, 1$ Total cases $ \Rightarrow 4 \times 4 = 16$ Unfovourble cases $ \Rightarrow i^{k_1} + i^{k_2} = 0$ $ \{\begin{array}{c}1, -1 \\ -1, 1 \\ i, -i \\ -i, i\end{array}\}$ 4 Cases $\Rightarrow$ Probability $ = \frac{16-4}{16} = \frac{3}{4}$

A, E,G R D N Probabllity $(P) = \frac{\text{favourable case}}{\text{Total case}}$ (when A & E are in order) Total case = $6!$ Favourable case = ${6}C_2 . 4! = (15)4! = (30)4!$ $P = \frac{(15)4!}{(30)4!} = \frac{1}{2}$ Probability when not in order = $1 - \frac{1}{2} = \frac{1}{2}$

$E_1$: Bag $B_1$ is selected $B_1$: 6 W 4 B $B_2$: 4 W 6 B $B_3$: 5 W 5 B $E_2$: bag $B_2$ is selected We have to find $P(\frac{E_2}{A})$ $P(\frac{E_2}{A}) = \frac{P(E_2)P(\frac{A}{E_2})}{P(E_1)P(\frac{A}{E_1}) + P(E_2)P(\frac{A}{E_2}) + P(E_3)P(\frac{A}{E_3})} = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}} = \frac{4}{15}$

Bag $1 = \{4 W, 5 B\}$ Bag $2 = \{nW, 3B\}$ $P(\frac{W}{Bag2}) = \frac{29}{45}$ $\Rightarrow P(\frac{W}{B_1}) \times P(\frac{W}{B_2}) + P(\frac{B}{B_1}) \times P(\frac{W}{B_2}) = \frac{29}{45}$ $\frac{4}{9} \times \frac{n+1}{n+4} + \frac{5}{9} \times \frac{n}{n+4} = \frac{29}{45}$ $n = 6$

Probability that a Red ball comes from Bag I ($p$): $p(B_1/R) = \frac{p(B_1) \cdot p(R/B_1)}{p(R)}$ Substituting the values, $p(B_1/R) = \frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}}$ Simplify to find $p$: $=\frac{\frac{1}{10}}{\frac{1}{3}(1.2)} = \frac{1}{4}$ So, $p = \frac{1}{4}$. Probability that a Green ball comes from Bag III ($q$): $p(B_3/G) = \frac{p(B_3) \cdot p(G/B_3)}{p(G)}$ Substitute the values, $p(B_3/G) = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10}}$ Simplify to find $q$: $=\frac{\frac{2}{15}}{\frac{1}{3} \times 1.2} = \frac{1}{3}$ So, $q = \frac{1}{3}$. Calculation of $(\frac{1}{p} + \frac{1}{q})$: $\frac{1}{p} = 4$, $\frac{1}{q} = 3$ Therefore, $(\frac{1}{p} + \frac{1}{q}) = 4 + 3 = 7$

To find $P(X \geq 3)$, we first determine the constant $k$ using the total probability for $X$. The probability $P(X = x)$ is given by: $P(X= x) = k(x+1) \cdot 3^{-x}, x= 0,1,2,3,\dots$ The total probability must equal 1: $s = \sum_{x=0}^{\infty} k(x+1) \cdot 3^{-x}$ Calculating that series: $s = k3^0 + 2\frac{k}{3} + 3\frac{k}{3^2} + \dots$ Therefore, dividing the series by 3: $\frac{s}{3} = \frac{k}{3} + 2\frac{k}{3^2} + \dots$ Subtracting these: $s - \frac{s}{3} = k + \frac{k}{3} + \frac{k}{3^2} + \dots$ The resulting series is a geometric series: $2\frac{s}{3} = k(1 + \frac{1}{3} + \frac{1}{3^2} + \dots )$ The sum of the infinite geometric series is: $2\frac{s}{3} = k \cdot \frac{1}{1 - \frac{1}{3}} = \frac{3k}{2}$ Equating: $s = \frac{9k}{4} = 1$ Thus, solving for $k$: $k = \frac{4}{9}$ Next, compute $P(X \geq 3)$: $P(X \geq 3) = 1 - (P(X=0) + P(X=1) + P(X=2))$ Calculating these: $P(X=0) = k = \frac{4}{9}$ $P(X=1) = 2\frac{k}{3} = \frac{8}{27}$ $P(X=2) = 3\frac{k}{9} = \frac{4}{27}$ Adding these probabilities: $P(X=0) + P(X=1) + P(X=2) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} = \frac{8}{9}$ Finally, calculate $P(X \geq 3)$: $P(X \geq 3) = 1 - \frac{8}{9} = \frac{1}{9}$

$X$\n$P(X)S$\n$XP(X)$\n$\left(X_i-\mu\right)^2$ SP_i X\left(X_i-\mu\right)^2$\n$X=0$\n$\frac{{}^7 C_2}{{}^{10} C_2}$\n0\nS\left(0-\frac{3}{5}\right)^2$\n$\frac{7}{15}\left(\frac{9}

Symmetric but neither reflective nor transitive

The relation R is defined on the set A = {1, 2, 3, ..., 100} such that R = {(a, b): a = 2b + 1}. We need to find the largest integer k for which there exists a sequence of k ordered pairs from R where the second element of each pair is the first element of the next pair.

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Since |x| is always non-negative, x^2 + 5|x| + 6 is the sum of positive terms plus 6, which can never be zero. Thus, no real solution.

Boiling occurs when vapor pressure equals atmospheric pressure. Higher pressure requires higher temperature.

For the block to remain stationary relative to the wedge, the pseudo force ma must balance the component of gravity along the incline. ma cosθ = mg sinθ => a = g tanθ.

Haematite is Fe2O3 and Magnetite is Fe3O4. Bauxite is Al, Malachite is Cu.

Let z = e^(iθ). Then z/(1+z^2) = e^(iθ) / (1 + e^(2iθ)). Simplifying this using Euler formula results in a real number.

Hydrogen bonding occurs when H is bonded to highly electronegative atoms like F, O, or N.