answer:1. To determine the probability, denote k as the number of combinations of five numbers in which every pair of differences is at least 5. The probability p we are seeking is: [ p = frac{k}{binom{90}{5}} ] 2. Consider the sequence (1 leq a_1 < a_2 < a_3 < a_4 < a_5 leq 90), where (a_{i+1} - a_i geq 5) for (i = 1, 2, 3, 4). 3. Transform (a_i) to a new sequence (b_i) by setting: [ b_i = a_i - 4(i-1) quad text{for } i = 1, 2, ldots, 5 ] 4. The new sequence (b_i) satisfies: [ 1 leq b_1 < b_2 < b_3 < b_4 < b_5 leq 90 - 4 times (5 - 1) = 74 ] This transformation ensures that every sequence of (a_i) yielding differences of at least 5 corresponds bijectively to a sequence (b_i) from 1 to 74. 5. Conversely, each sequence (1 leq b_1 < b_2 < b_3 < b_4 < b_5 leq 74) can be mapped back to an (a_i) sequence by: [ a_i = b_i + 4(i-1) ] 6. The number of valid sequences of (b_i) is: [ binom{74}{5} ] 7. Therefore, the probability (p) is given by: [ p = frac{binom{74}{5}}{binom{90}{5}} ] 8. Compute the binomial coefficients: [ binom{74}{5} = frac{74 times 73 times 72 times 71 times 70}{5 times 4 times 3 times 2 times 1} = 17259390 ] [ binom{90}{5} = frac{90 times 89 times 88 times 87 times 86}{5 times 4 times 3 times 2 times 1} = 43842110 ] 9. Thus, the probability: [ p = frac{17259390}{43842110} approx 0.367 ] # Conclusion: [ boxed{0.367} ]

answer:The formula for the projection of a vector mathbf{u} onto vector mathbf{w} is: [ operatorname{proj}_{mathbf{w}} mathbf{u} = frac{mathbf{u} cdot mathbf{w}}{|mathbf{w}|^2} mathbf{w} ] Given operatorname{proj}_{mathbf{w}} mathbf{v} = begin{pmatrix} 6 -3 end{pmatrix}, we can compute operatorname{proj}_{mathbf{w}} (2mathbf{v}) using: [ operatorname{proj}_{mathbf{w}} (2mathbf{v}) = 2 cdot operatorname{proj}_{mathbf{w}} mathbf{v} = 2 cdot begin{pmatrix} 6 -3 end{pmatrix} = begin{pmatrix} 12 -6 end{pmatrix} ] Also, the projection of mathbf{w} onto itself is simply mathbf{w}, hence: [ operatorname{proj}_{mathbf{w}} (3mathbf{w}) = 3 cdot mathbf{w} ] Finally, using the linearity of projection: [ operatorname{proj}_{mathbf{w}} (2mathbf{v} + 3mathbf{w}) = operatorname{proj}_{mathbf{w}} (2mathbf{v}) + operatorname{proj}_{mathbf{w}} (3mathbf{w}) = begin{pmatrix} 12 -6 end{pmatrix} + 3mathbf{w} ] If we consider operatorname{proj}_{mathbf{w}} mathbf{w} to be mathbf{w} (under the assumption |mathbf{w}|=1 for simple replacement), then: [ boxed{begin{pmatrix} 12 -6 end{pmatrix} + 3 mathbf{w}} ] Conclusion: Without further details on mathbf{w}, the final answer depends on the actual vector mathbf{w} but is otherwise correctly derived under linear operations.

answer:Option C Since E(X)=x_1+x_2=. Thus, x_2=4-2x_1, and D(X)=2times+2times=. Since x_1<x_2, thus, x_1+x_2=boxed{3}.

answer:Given vector (bar{a} = (1, -2, 5)), we need to find coordinates of the vector (bar{b}), which lies in the plane (xOy) and is perpendicular to the vector (bar{a}), and has a magnitude (|vec{b}| = 2 sqrt{5}). 1. **Check Coordinates in the Plane (xOy):** Since vector (bar{b}) lies in the plane (xOy), it has coordinates of the form ((x, y, 0)). 2. **Magnitude Condition:** Given the magnitude of (bar{b}): [ |vec{b}| = sqrt{x^2 + y^2} = 2sqrt{5} ] Therefore, we have: [ sqrt{x^2 + y^2} = 2sqrt{5} ] Squaring both sides: [ x^2 + y^2 = 20 ] 3. **Orthogonality Condition:** Since (bar{b}) is perpendicular to (bar{a}), the dot product of (bar{a}) and (bar{b}) must be zero. [ bar{a} cdot bar{b} = 1x + (-2)y + 5 cdot 0 = x - 2y = 0 ] Thus: [ x - 2y = 0 ] 4. **Solve the System of Equations:** We now solve the system: [ left{ begin{array}{l} x^2 + y^2 = 20 x - 2y = 0 end{array} right. ] From the second equation, (x - 2y = 0), we get: [ x = 2y ] Substitute (x = 2y) into the first equation: [ (2y)^2 + y^2 = 20 ] [ 4y^2 + y^2 = 20 ] [ 5y^2 = 20 ] [ y^2 = 4 ] [ y = pm 2 ] When (y = 2): [ x = 2y = 2 times 2 = 4 ] So, one vector (bar{b}) is ((4, 2, 0)). When (y = -2): [ x = 2y = 2 times (-2) = -4 ] So, another vector (bar{b}) is ((-4, -2, 0)). # Conclusion: The vectors that satisfy the given conditions are ((4, 2, 0)) and ((-4, -2, 0)). [ boxed{(4, 2, 0) text{ or } (-4, -2, 0)} ]