answer:**1. Verifying Continuity and Self-Adjointness of ( A ):** Let ( f, g in L^2[a, b] ). Consider the operator ( A ) defined by: [ A f(s) = int_a^b K(s, t) f(t) , dt ] **Continuity:** To show that ( A ) is continuous, we need to show that for any ( f, g in L^2[a, b] ): [ begin{aligned} int_a^b left| A f(s) - A g(s) right|^2 , ds & leq int_a^b left| int_a^b K(s, t) (f(t) - g(t)) , dt right|^2 , ds & leq int_a^b left( int_a^b |K(s, t)||f(t) - g(t)| , dt right)^2 , ds & leq (b-a) |K|_infty^2 int_a^b |f(t) - g(t)|^2 , dt. end{aligned} ] This implies that ( A ) is continuous since: [ left| A f - A g right|_{L^2} leq (b-a) |K|_infty left| f - g right|_{L^2}. ] **Self-Adjointness:** To prove self-adjointness, consider: [ begin{aligned} (A f, g) &= int_a^b A f(s) overline{g(s)} , ds &= int_a^b left( int_a^b K(s, t) f(t) , dt right) overline{g(s)} , ds &= int_{[a, b]^2} K(s, t) f(t) overline{g(s)} , ds , dt &= int_a^b f(t) left( int_a^b K(s, t) overline{g(s)} , ds right) , dt &= int_a^b f(t) overline{A g(t)} , dt = (f, A g). end{aligned} ] Hence, ( A ) is self-adjoint. **Conclusion:** ( A ) is a continuous self-adjoint operator. **2. Applying Hilbert-Schmidt Theorem and Establishing Mercer's Theorem:** By the Hilbert-Schmidt theorem (considering the compactness of ( A ) in ( L^2 )), we find a complete orthonormal system of complex-valued functions ( {phi_n} ) in ( L^2[a, b] ) which satisfy the integral equation: [ int_a^b K(s, t) phi_n(t) , dt = lambda_n phi_n(s) ] where ( lambda_n in mathbb{R} ) with ( |lambda_1| geq |lambda_2| geq cdots geq 0 ) and ( lambda_n to 0 ) as ( n to infty ). Thus, for ( f in L^2[a, b] ), [ A f = sum_{n=1}^infty lambda_n (f, phi_n) phi_n ] in ( L^2 ). Furthermore, due to the non-negativity of ( K(s, t) ): [ lambda_n = (A phi_n, phi_n) = int_{[a, b]^2} K(s, t) phi_n(t) overline{phi_n(s)} , ds , dt geq 0. ] **Continuity of Eigenfunctions:** If ( lambda_n > 0 ), using Lebesgue's dominated convergence theorem, we obtain that ( phi_n(s) ) are continuous in ( s ). **Real-Valued Eigenfunctions:** If ( phi_n ) is a solution, then ( operatorname{Re} phi_n ) and ( operatorname{Im} phi_n ) are also solutions. We consider a system ( A_i ) consisting of real and imaginary parts of ( phi_n ) classified by distinct eigenvalues ( mu_i ). Using the Gram-Schmidt orthogonalization process, we can form a complete orthonormal set of real-valued eigenfunctions ( {f_{ni}} ). Then, [ bigcup_{i=1}^infty {f_{1i}, ldots, f_{ni}} ] is a complete orthonormal system of real-valued functions in ( L^2[a, b] ), forming ( {varphi_n} ). **Mercer's Representation:** For any ( t ), [ K_n(s, t) = K(s, t) - sum_{i=1}^n lambda_i phi_i(s) phi_i(t) ] is continuous. By the properties of the integral operators ( A_n ), [ int_{[a, b]^2} K_n(s, t) f(s) overline{f(t)} , ds , dt = sum_{i=n+1}^infty lambda_i | (f, phi_i) |^2 ] implying that ( K_n(s, s) geq 0 ). **Summation of Series:** Using Lebesgue's dominated convergence theorem and applying Dini's theorem for uniform convergence, we conclude that: [ K(s, t) = sum_{n=1}^infty lambda_n phi_n(s) phi_n(t) ] with the series converging absolutely and uniformly on ( [a, b] times [a, b] ). This proves Mercer's theorem. **3. Gaussian Process Representation:** Given a centered Gaussian process ( X = (X_t, t in [a, b]) ) with covariance function ( K(s, t) ), [ X_t = sum_{n=1}^infty xi_n sqrt{lambda_n} phi_n(t), ] where ( {xi_n} ) are i.i.d. ( mathscr{N}(0, 1) ) random variables converging almost surely and in ( L^2 ). **Covariance Calculation:** [ begin{aligned} mathrm{E} X_s X_t &= lim_{n to infty} mathrm{E} left( sum_{i leq n} sqrt{lambda_i} phi_i(s) xi_i right) left( sum_{i leq n} sqrt{lambda_i} phi_i(t) xi_i right) &= lim_{n to infty} sum_{i leq n} lambda_i phi_i(s) phi_i(t) &= K(s, t). end{aligned} ] Thus, the process ( X_t ) has the covariance function ( K(s, t) ). Conclusion: [ boxed{X_t = sum_{n=1}^infty xi_n sqrt{lambda_n} phi_n(t)} ]

answer:1. **Identify the Variables and Given Information:** Let us denote the number of physicists by ( n_P ) and the number of chemists by ( n_C ). According to the problem, there are ( n ) physicists and ( n ) chemists, thus: [ n_P = n quad text{and} quad n_C = n ] 2. **Understanding the Liars:** It is given that the number of liar chemists ((C_L)) is equal to the number of liar physicists ((P_L)), so: [ C_L = P_L ] 3. **Analyzing Responses:** Each person sitting around the table responded that their right neighbor is a chemist. We need to consider the implications of this in terms of the truth-tellers and liars. 4. **Determine if There is an Odd or Even Number of Each Type:** We know that there are in total (2n) individuals seated at the table, and the number of liars (be it physicists or chemists) is evenly split according to the problem statement: [ n_L = P_L + C_L ] where (n_L) is the total number of liars. 5. **Using the Response Analysis:** The key point is that everyone claims their right neighbor is a chemist. For the statement to hold consistently under the given conditions, we need to delve deeper into the truth-teller/liar dynamics: - A truth-telling physicist will truthfully say the right neighbor is a chemist. - A liar physicist will lie and say the right neighbor is a chemist as well, implying the person to their right is anything but a chemist. - Similar logic applies for chemists who always lies or tells the truth. 6. **Combination and Circular Arrangement:** Consider individuals arranged in a circle, we know: [ sum (text{right neighbor is chemist}) = 2n ] Clearly, relations imply (n_P = n_C = n). Consistency enforces even numbers of physicists and chemists across (2n), mandating (n mod 2 = 0). 7. **Final Consideration on Parity:** Since the problem explicitly stated every (2n) individual's response matched consistently to avoid paradoxical contradiction: Even split implies (n) must be even for these inputs to remain logically consistent. [ boxed{n text{ is even}} ]

answer:1. The equation of the circle (O) is given by (x^2 + y^2 = 4). 2. The line (l) passing through point (P(0, 1)) intersects the circle at points (A) and (B). To find the line equation, assume it is in the form (y = kx + 1). 3. Substitute (y = kx + 1) into the circle equation (x^2 + y^2 = 4): [ x^2 + (kx + 1)^2 = 4 ] [ x^2 + k^2x^2 + 2kx + 1 = 4 ] [ (1 + k^2)x^2 + 2kx - 3 = 0 ] 4. Since the line (l) intersects the circle at two points (A) and (B), the quadratic equation has two solutions. Denote these points as (A(x_1, y_1)) and (B(x_2, y_2)). 5. The line (l) also intersects the (x)-axis at point (Q). For (Q), set (y = 0) in (y = kx + 1): [ 0 = kx + 1 ] [ Q left( -frac{1}{k}, 0 right) ] 6. Using the problem’s condition, let (overrightarrow{QA} = lambda overrightarrow{PA}) and (overrightarrow{QB} = mu overrightarrow{PB}). 7. Express these conditions algebraically: [ x_1 + frac{1}{k} = lambda x_1 ] [ x_2 + frac{1}{k} = mu x_2 ] 8. Solving these equations for (lambda) and (mu): [ lambda = 1 + frac{1}{kx_1} ] [ mu = 1 + frac{1}{kx_2} ] 9. Sum these expressions and simplify: [ lambda + mu = 2 + frac{1}{kx_1} + frac{1}{kx_2} ] 10. Use Vieta’s formulas to express (x_1) and (x_2) from the quadratic equation ( (1+k^2)x^2 + 2kx - 3 = 0): [ x_1 + x_2 = frac{-2k}{1 + k^2} ] [ x_1 x_2 = frac{-3}{1 + k^2} ] 11. Substitute Vieta’s formulas into the expression for (lambda + mu): [ lambda + mu = 2 + frac{x_1 + x_2}{k x_1 x_2} ] [ lambda + mu = 2 + frac{frac{-2k}{1 + k^2}}{k cdot frac{-3}{1 + k^2}} ] Simplifying the above: [ lambda + mu = 2 + frac{-2k}{frac{-3k}{1 + k^2}} ] [ lambda + mu = 2 + frac{2}{3} ] [ lambda + mu = frac{8}{3} ] 12. Conclusion: Therefore, the sum (lambda + mu) is a constant value: [ boxed{frac{8}{3}} ]

answer:Alani earns 45 for 3 hours of baby-sitting, so her rate is 45 / 3 hours = 15 per hour. If she baby-sits for 5 hours at the rate of 15 per hour, she would earn 5 hours * 15/hour = boxed{75} .