answer:The necessary and sufficient condition for the complex number z=a+bi (a, b in mathbb{R}) to be a purely imaginary number is "a=0 and b neq 0". Therefore, a=0 is a necessary but not sufficient condition for the complex number z=a+bi (a, b in mathbb{R}) to be a purely imaginary number. The correct choice is B. boxed{text{B}}

answer:First, by Vieta's formulas: - a + b = 6 - ab = 8 Compute a^2 + b^2: [ a^2 + b^2 = (a+b)^2 - 2ab = 6^2 - 2 times 8 = 36 - 16 = 20 ] Next compute a^5 b^3 + a^3 b^5: [ a^5 b^3 + a^3 b^5 = a^3 b^3 (a^2 + b^2) = (ab)^3 ((a+b)^2 - 2ab) ] [ = 8^3 cdot (6^2 - 2 times 8) = 512 cdot (36 - 16) = 512 cdot 20 = 10240 ] Summing all the terms: [ a^2 + a^5 b^3 + a^3 b^5 + b^2 = 20 + 10240 = boxed{10260} ]

answer:To find the variance D(X) of a discrete random variable X which is Poisson-distributed with parameter lambda, we can use the formula for the variance: [ D(X) = Mleft(X^2right) - [M(X)]^2 ] 1. **Expectation ( M(X) )**: For a Poisson-distributed random variable X with parameter lambda, the expected value M(X) is known to be lambda. This can be derived from the properties of the Poisson distribution or seen in previous problem references (e.g., problem 207). 2. **Rewriting the variance formula**: Given M(X) = lambda, we rewrite the variance formula: [ D(X) = Mleft(X^2right) - lambda^2 ] 3. **Expectation of ( X^2 )**: To find M(X^2), we use the definition of expectation for X^2: [ Mleft(X^2right) = sum_{k=0}^{infty} k^2 cdot P(X = k) ] Given that (X) follows a Poisson distribution with parameter (lambda), we have: [ P(X = k) = frac{lambda^k e^{-lambda}}{k!} ] Thus, [ Mleft(X^2right) = sum_{k=0}^{infty} k^2 cdot frac{lambda^k e^{-lambda}}{k!} ] 4. **Simplifying the sum**: The term k^2 can be rearranged as k cdot k. We rewrite and separate the series: [ Mleft(X^2right) = sum_{k=1}^{infty} k cdot frac{k lambda^k e^{-lambda}}{k!} = sum_{k=1}^{infty} k cdot frac{lambda^k e^{-lambda}}{(k-1)! (k)} ] [ = lambda sum_{k=1}^{infty} k cdot frac{lambda^{k-1} e^{-lambda}}{(k-1)!} ] Change of index by letting ( m = k-1 ): [ Mleft(X^2right) = lambda sum_{m=0}^{infty} (m+1) cdot frac{lambda^m e^{-lambda}}{m!} ] Split the sum into two parts: [ Mleft(X^2right) = lambda left[ sum_{m=0}^{infty} m cdot frac{lambda^m e^{-lambda}}{m!} + sum_{m=0}^{infty} frac{lambda^m e^{-lambda}}{m!} right] ] 5. **Evaluating the sums**: The first part is the series for the expectation (which is lambda): [ sum_{m=0}^{infty} m cdot frac{lambda^m e^{-lambda}}{m!} = lambda ] The second part is the series for the Poisson distribution sum: [ sum_{m=0}^{infty} frac{lambda^m e^{-lambda}}{m!} = e^{-lambda} cdot e^{lambda} = 1 ] 6. **Substituting back to find (boldsymbol{M(X^2)})**: [ M(X^2) = lambda left( lambda + 1 right) = lambda^2 + lambda ] 7. **Calculating the variance (boldsymbol{D(X)})**: Now substituting ( M(X^2) ) back into the variance formula: [ D(X) = left( lambda^2 + lambda right) - lambda^2 = lambda ] **Conclusion**: The variance of a Poisson-distributed random variable with parameter (lambda) is: [ boxed{lambda} ]

answer:Let's calculate the distance covered by the first train from station P to the meeting point. The first train starts at 7 a.m. and meets the second train at 12 p.m., which means it has been traveling for 5 hours at a speed of 20 kmph. Distance covered by the first train = Speed × Time Distance covered by the first train = 20 kmph × 5 hours = 100 km Since the two trains meet each other and the total distance between the two stations is 200 km, the second train must have covered the remaining distance. Distance covered by the second train = Total distance - Distance covered by the first train Distance covered by the second train = 200 km - 100 km = 100 km Now, let's calculate the time taken by the second train to cover this 100 km distance at a speed of 25 kmph. Time taken by the second train = Distance / Speed Time taken by the second train = 100 km / 25 kmph = 4 hours The trains meet at 12 p.m., so we need to subtract the time taken by the second train from this meeting time to find out when it started from station Q. Starting time of the second train = Meeting time - Time taken by the second train Starting time of the second train = 12 p.m. - 4 hours = 8 a.m. Therefore, the second train started from station Q at boxed{8} a.m.