💡Okay, so I have this problem where a point (x, y) is randomly picked from a rectangular region with vertices at (0,0), (3012,0), (3012, 3013), and (0,3013). I need to find the probability that x > 3y. Hmm, probability questions can sometimes be tricky, but let me break it down step by step.First, I know that when dealing with probabilities involving areas, especially in uniform distributions over regions, the probability is usually the area of the region where the condition is satisfied divided by the total area of the region. So, in this case, the total area is the area of the rectangle, and the area where x > 3y is some region within that rectangle.Let me visualize the rectangle. It has a width of 3012 units along the x-axis and a height of 3013 units along the y-axis. So, it's a pretty large rectangle, but the exact dimensions might not matter because when we take the ratio, some terms might cancel out.Now, the condition is x > 3y. I can rewrite this inequality as y < x/3. This is the equation of a straight line, y = x/3, which is a line passing through the origin with a slope of 1/3. So, the region where y < x/3 is below this line.I need to figure out the area within the rectangle where y < x/3. Since the rectangle extends from (0,0) to (3012, 3013), I should find where the line y = x/3 intersects the rectangle. Let me find the intersection points. The line y = x/3 starts at (0,0). As x increases, y increases proportionally. The rectangle's top side is at y = 3013, so let me see where the line y = x/3 intersects this top side. Setting y = 3013 in the equation, we get 3013 = x/3, so x = 3013 * 3 = 9039. But wait, the rectangle only goes up to x = 3012, so the line doesn't intersect the top side of the rectangle. Instead, it must intersect the right side of the rectangle, which is at x = 3012.So, plugging x = 3012 into y = x/3, we get y = 3012 / 3 = 1004. Therefore, the line intersects the right side of the rectangle at (3012, 1004). So, the region where y < x/3 within the rectangle is a triangle with vertices at (0,0), (3012, 0), and (3012, 1004).Now, I need to calculate the area of this triangle. The formula for the area of a triangle is (base * height) / 2. In this case, the base can be considered as the distance along the x-axis from (0,0) to (3012,0), which is 3012 units. The height is the y-coordinate where the line intersects the right side, which is 1004 units.So, the area of the triangle is (3012 * 1004) / 2. Let me compute that. But before I do, I notice that when I divide by 2, it might simplify with the other terms later on.Next, I need the area of the entire rectangle to find the probability. The area of the rectangle is simply its width multiplied by its height, which is 3012 * 3013.So, the probability that a randomly picked point (x, y) satisfies x > 3y is the area of the triangle divided by the area of the rectangle. That is:Probability = ( (3012 * 1004) / 2 ) / (3012 * 3013 )I can simplify this expression. The 3012 in the numerator and denominator cancels out, leaving:Probability = (1004 / 2) / 3013Simplifying 1004 / 2 gives 502, so:Probability = 502 / 3013Wait, but 502 and 3013, do they have any common factors? Let me check. 502 is 2 * 251, and 3013 divided by 251 is approximately 12, but 251 * 12 is 3012, which is one less than 3013. So, 3013 is a prime number? Let me verify.3013 divided by 13 is 231.769, which isn't an integer. Divided by 7 is about 430.428, not an integer. Divided by 3 is 1004.333, not an integer. So, it seems 3013 is a prime number. Therefore, 502 and 3013 don't have any common factors besides 1, so the fraction 502/3013 is already in its simplest form.Wait, but earlier I thought the area of the triangle was (3012 * 1004)/2, which is 3012 * 502. Then, when divided by 3012 * 3013, the 3012 cancels, leaving 502/3013. So, that seems correct.But just to double-check, maybe I made a mistake in calculating the area. Let me recalculate:Area of triangle = (base * height) / 2 = (3012 * 1004) / 2 = 3012 * 502.Area of rectangle = 3012 * 3013.So, the ratio is (3012 * 502) / (3012 * 3013) = 502 / 3013.Yes, that seems correct.Alternatively, I can think about the ratio without calculating the areas. Since the line y = x/3 divides the rectangle into two regions, and the area below the line is a triangle, the probability is the area of that triangle over the area of the rectangle.Another way to think about it is that for each x from 0 to 3012, the maximum y that satisfies y < x/3 is y = x/3. So, integrating y from 0 to x/3 for x from 0 to 3012 would give the area, but that's essentially the same as calculating the area of the triangle.So, I think my answer is correct. The probability is 502/3013.Wait, but 502 is half of 1004, and 1004 is exactly one-third of 3012, right? Because 3012 divided by 3 is 1004. So, that makes sense because the line y = x/3 intersects the rectangle at (3012, 1004), which is one-third of the way up the y-axis from the origin.So, the triangle's height is one-third of the rectangle's height, but the base is the full width of the rectangle. So, the area of the triangle is (base * height)/2 = (3012 * 1004)/2, which is half of 3012 * 1004, and since 1004 is 3012/3, it's (3012 * (3012/3))/2 = (3012²)/(6). Then, the area of the rectangle is 3012 * 3013, so the probability is (3012² / 6) / (3012 * 3013) = (3012 / 6) / 3013 = (502) / 3013.Yes, that confirms it again. So, I think I'm confident that the probability is 502/3013.Just to make sure, let me plug in some smaller numbers to test the concept. Suppose the rectangle was from (0,0) to (3,0) to (3,4) to (0,4). Then, the line y = x/3 would intersect the right side at (3,1). The area of the triangle would be (3*1)/2 = 1.5. The area of the rectangle is 3*4=12. So, the probability would be 1.5/12 = 1/8. Alternatively, 1.5 is 3*1/2, so 3*1/2 divided by 3*4 is (1/2)/4 = 1/8. That makes sense.In this case, 1.5/12 simplifies to 1/8, which is the same as (1/2)/4. So, in my original problem, it's (1004/2)/3013 = 502/3013.Therefore, I think my answer is correct.

💡First, I recognize that the average of the original data set ( a_1, a_2, a_3, a_4, a_5 ) is 8. This means that the sum of these five numbers is ( 5 times 8 = 40 ).Next, I consider the transformed data set: ( a_1 + 10 ), ( a_2 - 10 ), ( a_3 + 10 ), ( a_4 - 10 ), ( a_5 + 10 ). To find the average of this new set, I need to calculate the sum of these transformed values and then divide by 5.Calculating the sum of the transformed values:[(a_1 + 10) + (a_2 - 10) + (a_3 + 10) + (a_4 - 10) + (a_5 + 10) = (a_1 + a_2 + a_3 + a_4 + a_5) + (10 - 10 + 10 - 10 + 10)][= 40 + 10 = 50]Finally, I find the average of the transformed data set by dividing the total sum by the number of data points:[text{Average} = frac{50}{5} = 10]

💡Okay, so I have this sequence where a₁ is 1, a₂ is 3, a₃ is 6, and for n greater than 3, each term is defined by the recurrence relation aₙ = 3aₙ₋₁ - aₙ₋₂ - 2aₙ₋₃. I need to prove that for all natural numbers n > 3, aₙ is greater than 3 times 2 to the power of (n-2). Hmm, that sounds like an inequality proof, probably using mathematical induction.First, let me write down the given information clearly:- a₁ = 1- a₂ = 3- a₃ = 6- For n > 3, aₙ = 3aₙ₋₁ - aₙ₋₂ - 2aₙ₋₃And I need to show that for all n > 3, aₙ > 3 × 2^{n-2}.Alright, so induction is a good approach here. Induction usually involves two steps: the base case and the inductive step. The base case would be to verify the statement for the first few values of n, and then the inductive step would assume it's true for some k and prove it for k+1.But before jumping into induction, maybe I should compute a few more terms of the sequence to see if the inequality holds for the initial terms beyond n=3. That might give me some intuition.Let's compute a₄:a₄ = 3a₃ - a₂ - 2a₁ = 3×6 - 3 - 2×1 = 18 - 3 - 2 = 13.Now, let's check the inequality for n=4:3 × 2^{4-2} = 3 × 2² = 3×4 = 12.So, a₄ = 13 > 12. Okay, that works.Next, a₅:a₅ = 3a₄ - a₃ - 2a₂ = 3×13 - 6 - 2×3 = 39 - 6 - 6 = 27.Check the inequality for n=5:3 × 2^{5-2} = 3 × 2³ = 3×8 = 24.a₅ = 27 > 24. Good.a₆:a₆ = 3a₅ - a₄ - 2a₃ = 3×27 - 13 - 2×6 = 81 - 13 - 12 = 56.Inequality for n=6:3 × 2^{6-2} = 3 × 16 = 48.a₆ = 56 > 48. Nice, it's holding up.a₇:a₇ = 3a₆ - a₅ - 2a₄ = 3×56 - 27 - 2×13 = 168 - 27 - 26 = 115.Inequality for n=7:3 × 2^{7-2} = 3 × 32 = 96.115 > 96. Still good.a₈:a₈ = 3a₇ - a₆ - 2a₅ = 3×115 - 56 - 2×27 = 345 - 56 - 54 = 235.Inequality for n=8:3 × 2^{8-2} = 3 × 64 = 192.235 > 192. Yep, it's working.So, up to n=8, the inequality holds. That gives me some confidence. Now, let's think about how to approach the proof.Since it's a linear recurrence relation, maybe solving the recurrence would help, but that might be complicated. Alternatively, using induction seems feasible.Let me outline the induction approach:1. **Base Case:** Verify the inequality for n=4, n=5, n=6, etc., as needed. Since the recurrence relation starts at n=4, I might need to check a few base cases.2. **Inductive Hypothesis:** Assume that for some k ≥ 4, the inequality holds for n=k, n=k-1, and n=k-2. That is, a_k > 3×2^{k-2}, a_{k-1} > 3×2^{k-3}, and a_{k-2} > 3×2^{k-4}.3. **Inductive Step:** Using the recurrence relation, express a_{k+1} in terms of a_k, a_{k-1}, and a_{k-2}, and then apply the inductive hypothesis to show that a_{k+1} > 3×2^{(k+1)-2} = 3×2^{k-1}.Wait, but the recurrence is a_{k+1} = 3a_k - a_{k-1} - 2a_{k-2}. So, if I can express a_{k+1} in terms of the previous terms and then use the inequalities from the inductive hypothesis, I might be able to derive the desired inequality.Let me write that out:Assume for n = k, k-1, k-2, we have a_n > 3×2^{n-2}.Then, for n = k+1:a_{k+1} = 3a_k - a_{k-1} - 2a_{k-2}.Using the inductive hypothesis:a_k > 3×2^{k-2},a_{k-1} > 3×2^{k-3},a_{k-2} > 3×2^{k-4}.So, substituting these into the expression for a_{k+1}:a_{k+1} > 3*(3×2^{k-2}) - (3×2^{k-3}) - 2*(3×2^{k-4}).Let me compute the right-hand side:3*(3×2^{k-2}) = 9×2^{k-2},- (3×2^{k-3}) = -3×2^{k-3},-2*(3×2^{k-4}) = -6×2^{k-4}.So, combining these:9×2^{k-2} - 3×2^{k-3} - 6×2^{k-4}.Let me factor out 2^{k-4} to simplify:2^{k-4}*(9×2^{2} - 3×2^{1} - 6×1) = 2^{k-4}*(9×4 - 3×2 - 6) = 2^{k-4}*(36 - 6 - 6) = 2^{k-4}*24.24 is equal to 3×8, which is 3×2³. So, 24 = 3×2³.Therefore, 2^{k-4}*24 = 3×2³×2^{k-4} = 3×2^{k-1}.So, putting it all together:a_{k+1} > 3×2^{k-1}.Which is exactly what we needed to show for n = k+1.Therefore, by the principle of mathematical induction, the inequality aₙ > 3×2^{n-2} holds for all natural numbers n > 3.Wait, but I need to make sure that the inductive step is valid. I assumed the inequality holds for n = k, k-1, k-2, and then showed it holds for n = k+1. So, as long as the base cases are verified, the induction should hold.Earlier, I computed up to a₈ and checked the inequality for n=4 to n=8, so the base cases are covered.Therefore, the proof is complete.

💡First, calculate the total current age of all 7 people by multiplying the average age by the number of people.Next, determine the total age reduction for all 7 people by multiplying the youngest person's age by the number of people.Subtract the total age reduction from the current total age to find the total age of the people when the youngest was born.Finally, calculate the average age when the youngest was born by dividing the total age at that time by the number of older people.

💡Okay, so I need to find the sum of the seven smallest distinct positive integer multiples of 13. Hmm, let's break this down step by step. First, what exactly are multiples of 13? Well, multiples of a number are just that number multiplied by any integer. So, the multiples of 13 would be 13×1, 13×2, 13×3, and so on.Since we're looking for the smallest seven, I guess that means I need to consider the first seven positive integers multiplied by 13. That would be 13×1, 13×2, up to 13×7. Let me write those out to make sure I have them right:1. 13×1 = 132. 13×2 = 263. 13×3 = 394. 13×4 = 525. 13×5 = 656. 13×6 = 787. 13×7 = 91Okay, so those are the seven smallest distinct positive integer multiples of 13. Now, I need to find their sum. That means I have to add all these numbers together: 13 + 26 + 39 + 52 + 65 + 78 + 91.Adding them one by one might be a bit tedious, but maybe there's a smarter way to do this. I remember that when you have a series of numbers that form an arithmetic sequence, you can use a formula to find the sum without adding each term individually. An arithmetic sequence is a sequence where each term increases by a constant difference. In this case, each multiple of 13 is 13 more than the previous one, so yes, it's an arithmetic sequence.The formula for the sum of the first n terms of an arithmetic sequence is:[ S_n = frac{n}{2} times (a_1 + a_n) ]Where:- ( S_n ) is the sum of the first n terms,- ( n ) is the number of terms,- ( a_1 ) is the first term,- ( a_n ) is the nth term.In this problem, n is 7 because we're adding the first seven multiples. The first term ( a_1 ) is 13, and the seventh term ( a_7 ) is 91. Plugging these into the formula:[ S_7 = frac{7}{2} times (13 + 91) ]First, let's compute the sum inside the parentheses:13 + 91 = 104Now, multiply that by 7/2:[ S_7 = frac{7}{2} times 104 ]Calculating that:7 divided by 2 is 3.5, and 3.5 multiplied by 104 is... let's see, 104 × 3 = 312, and 104 × 0.5 = 52, so 312 + 52 = 364.So, the sum of the first seven multiples of 13 is 364.Wait, just to double-check, maybe I can add them the long way to make sure I didn't make a mistake. Let's add them step by step:13 + 26 = 3939 + 39 = 7878 + 52 = 130130 + 65 = 195195 + 78 = 273273 + 91 = 364Yes, that matches the result I got using the formula. So, both methods give me the same answer, which is reassuring.Alternatively, I could factor out the 13 from each term and then add the remaining numbers. Let's try that approach too:13(1 + 2 + 3 + 4 + 5 + 6 + 7)First, calculate the sum inside the parentheses:1 + 2 = 33 + 3 = 66 + 4 = 1010 + 5 = 1515 + 6 = 2121 + 7 = 28So, the sum inside is 28. Then, multiply that by 13:13 × 28Let me compute that:13 × 20 = 26013 × 8 = 104260 + 104 = 364Again, I get 364. This consistency across different methods makes me confident that the answer is correct.Just to recap, the three methods I used were:1. Using the arithmetic series sum formula.2. Adding each multiple one by one.3. Factoring out the common multiple and summing the coefficients.All three methods led me to the same result, so I feel pretty sure that 364 is the correct sum of the seven smallest distinct positive integer multiples of 13.I guess another way to think about it is that since each multiple is 13 times a number from 1 to 7, the sum is just 13 times the sum of numbers from 1 to 7. The sum from 1 to n is given by n(n + 1)/2, so for n=7, that's 7×8/2 = 28. Then, 13×28 is 364. Yep, that's consistent too.I don't think I made any calculation errors because all the different approaches gave me the same answer. It's always good to verify using multiple methods, especially when dealing with arithmetic and series. It helps catch any possible mistakes that might happen if you're doing the calculations quickly or if you're just starting out with these concepts.So, in conclusion, after carefully considering different ways to approach the problem and verifying each step, I'm confident that the sum of the seven smallest distinct positive integer multiples of 13 is 364.

💡First, I need to determine the breadth of the rectangular plot. I know the total cost of fencing is Rs. 5300 and the cost per metre is Rs. 26.50. By dividing the total cost by the cost per metre, I can find the total perimeter of the plot.Next, I'll use the formula for the perimeter of a rectangle, which is (2 times (text{length} + text{breadth})). Given that the length is 60 metres, I can set up an equation to solve for the breadth.Once I have the breadth, I can easily find out how much more the length is than the breadth by subtracting the breadth from the length.