answer:Given a quadrilateral (ABCD) with points (P, Q, R, S) on the extensions of sides (AB, BC, CD, DA) respectively, such that (AP = AB), (BQ = BC), (CR = CD), and (DS = DA). We need to prove that the ratio of the areas of quadrilaterals (ABCD) and (PQRS) is (1:5). 1. **Consider the diagonal (AC) and analyze the triangles:** The problem hints at drawing the diagonal (AC) and splitting the figure into triangles. By drawing diagonal (AC), quadrilateral (ABCD) is divided into two triangles: (triangle ABC) and (triangle ACD). 2. **Consider the areas of (triangle ADC) and (triangle ARS):** - Note that triangle (ADC) lies within triangle (ARS) since (AR = AC + CD). - The areas of these triangles are proportional to the squares of the corresponding sides. Since (AD = DS) and extension (DR = DC), the area of (triangle ADR) is identical to that of (triangle ADC). Similarly, for the triangles related by the points (P) and (Q): [ text{Area} (triangle ADC) = text{Area} (triangle ADR) = text{Area} (triangle ARS) ] 3. **Apply the same argument to the other diagonal (BD):** When drawing a diagonal (BD), we get triangle pairs (triangle ABD) and (triangle BQR), and (triangle BCD) and (triangle CRP): [ text{Area} (triangle ABD) = text{Area} (triangle ABP) = text{Area} (triangle AQP) ] [ text{Area} (triangle BCD) = text{Area} (triangle BCQ) = text{Area} (triangle CPQ) ] 4. **Summing areas in (ABCD) and comparing to (PQRS):** Since each triangle with a pair of congruent sides doubles the area of the triangle forming quadrilateral (P, Q, R, S), the additive result of the areas ensures every triangle count is effectively doubled. 5. **Form the area ratio for (ABCD) and (PQRS):** Given that each of these smaller identically constructed triangles in regions outside the perimeter (resulting from larger encompassing triangles formed by the construction rules), the additional area occupied leads up the respective contributions: [ text{Area}(PQRS) = 5 times text{Area}(ABCD) ] Thus, the ratio of the areas of quadrilaterals (ABCD) and (PQRS) is: [ frac{text{Area}(ABCD)}{text{Area}(PQRS)} = frac{1}{5} ] Hence, the ratio of the areas of quadrilaterals (ABCD) and (PQRS) is indeed (1:5). [ boxed{1:5} ]

answer:1. Firstly, we are given a two-dimensional Gaussian vector (xi, eta) with mathrm{E}[xi] = mathrm{E}[eta] = 0, mathrm{D}[xi] = mathrm{D}[eta] = 1, and correlation coefficient rho. The goal is to establish the equality: [ mathrm{P}(xi > a, eta > b) - mathrm{P}(xi > a) mathrm{P}(eta > b) = int_{0}^{rho} phi(a, b, r) , dr ] where a, b in mathbb{R}, and phi(a, b, r) is the density of a bivariate normal distribution with mean (0, 0) and covariance matrix [ begin{pmatrix} 1 & r r & 1 end{pmatrix}. ] 2. We will use Hoeffding's formula which states: [ operatorname{cov}(f(xi), f(eta)) = int_{mathbb{R}^2} [mathrm{P}(xi > a, eta > b) - mathrm{P}(xi > a) mathrm{P}(eta > b)] f'(a) g'(b) , da , db. ] 3. Assume rho geq 0. If rho < 0, we can replace eta with -eta and utilize the relation: [ operatorname{cov}(I(xi > a), I(eta > b)) = -operatorname{cov}(I(xi > a), I(-eta > -b)). ] 4. To compute mathrm{P}(xi > a, eta > b), we use Hermite polynomials. We have: [ I(xi > a) = sum_{n} H_n(xi) frac{mathrm{E}[H_n(xi) I(xi > a)]}{n!} ] and [ mathrm{E}[H_n(xi) I(xi > a)] = (-1)^n int_{a}^{infty} varphi^{(n)}(x) , dx = (-1)^n Psi^{(n)}(a), ] where Psi(x) = mathrm{P}(xi > x) and varphi(x) = -Psi'(x). 5. Therefore, by using the task II.13.62(b), we have: [ mathrm{E}[I(xi > a) mid eta] = sum_n frac{Psi^{(n)}(a)}{n!} (-1)^n rho^n H_n(eta). ] 6. Consequently, [ begin{aligned} mathrm{P}(xi > a, eta > b) &= mathrm{E}[I(xi > a) I(eta > b)] &= mathrm{E} left[ sum_n frac{Psi^{(n)}(a)}{n!} (-1)^n rho^n H_n(eta) sum_k frac{Psi^{(k)}(b)}{k!} (-1)^k H_k(eta) right] &= sum_n frac{Psi^{(n)}(a) Psi^{(n)}(b)}{n!} rho^n. end{aligned} ] 7. Thus, the density function (xi, eta) is: [ phi(a, b, rho) = sum_n frac{Psi^{(n+1)}(a) Psi^{(n+1)}(b)}{n!} rho^n. ] 8. Therefore, [ mathrm{P}(xi > a, eta > b) = Psi(a) Psi(b) + int_{0}^{rho} phi(a, b, r) , dr, ] which is the desired equality. 9. For proving the given inequality, we use Hoeffding's formula: [ begin{aligned} operatorname{cov}(f(xi), g(eta)) &= int_{mathbb{R}^2} [mathrm{P}(xi > a, eta > b) - mathrm{P}(xi > a) mathrm{P}(eta > b)] f'(a) g'(b) , da , db &= int_{mathbb{R}^2} left( int_{0}^{rho} f'(a) g'(b) phi(a, b, r) , dr right) , da , db &= int_{0}^{rho} mathrm{E}[ f'(xi_r) g'(eta_r) ] , dr. quad blacksquare end{aligned} ] Conclusively, we have shown both the desired equality and the covariance expression using Hoeffding's formula. (boxed{mathrm{P}(xi > a, eta > b) - mathrm{P}(xi > a) mathrm{P}(eta > b) = int_{0}^{rho} phi(a, b, r) , dr})

answer:Let's denote the initial number of machines as ( M ). The initial group of machines can finish the job in 40 days, so their work rate is ( frac{1}{40} ) jobs per day. If 4 more machines are added, making the total number of machines ( M + 4 ), they can finish the job in one-fourth less time than the initial group. One-fourth less time than 40 days is ( 40 - frac{1}{4} times 40 = 40 - 10 = 30 ) days. The work rate of ( M + 4 ) machines is ( frac{1}{30} ) jobs per day. Since work rate is directly proportional to the number of machines, we can set up the following proportion: [ frac{M}{40} = frac{M + 4}{30} ] Now we can solve for ( M ): [ 30M = 40(M + 4) ] [ 30M = 40M + 160 ] [ 160 = 40M - 30M ] [ 160 = 10M ] [ M = frac{160}{10} ] [ M = 16 ] So, initially, there were boxed{16} machines working on the job.

answer:Solution: Since the area S of a triangle is given by S = frac{1}{2}ah, then S' = frac{1}{2} times frac{1}{3}a times 3h = frac{1}{2}ah = S, thus, the answer is: unchanged. According to the formula for the area of a triangle S = frac{1}{2}ah, we know that if the base of the triangle is reduced by a factor of 3 and its height is increased by a factor of 3, the change in area is frac{1}{3} times 3 = 1; this is how the problem is solved. This problem examines the area of a triangle. The key to solving this problem is to use the formula for the area of a triangle S = frac{1}{2}ah and the rules of product change to solve the problem. Therefore, the change in the area is boxed{text{unchanged}}.