answer:Let's denote Ram's original weight as ( R ) and Shyam's original weight as ( S ). According to the problem, the ratio of their weights is ( 2 : 5 ), so we can write: [ R : S = 2 : 5 ] This means that ( R = 2k ) and ( S = 5k ) for some constant ( k ). Now, Ram's weight is increased by 10%, so his new weight is ( R + 0.10R = 1.10R ). The total original weight of Ram and Shyam is ( R + S = 2k + 5k = 7k ). The total new weight of Ram and Shyam is 82.8 kg, which is a 15% increase from their original total weight. So, we can write: [ 1.15(7k) = 82.8 ] Now, let's solve for ( k ): [ 8.05k = 82.8 ] [ k = frac{82.8}{8.05} ] [ k = 10.28 ] Now we can find Ram's original weight ( R ) and Shyam's original weight ( S ): [ R = 2k = 2 times 10.28 = 20.56 text{ kg} ] [ S = 5k = 5 times 10.28 = 51.4 text{ kg} ] Since the total new weight is 82.8 kg and Ram's new weight is ( 1.10R ), we can find Shyam's new weight ( S' ): [ S' = 82.8 - 1.10R ] [ S' = 82.8 - 1.10 times 20.56 ] [ S' = 82.8 - 22.616 ] [ S' = 60.184 text{ kg} ] Now we can calculate the percentage increase in Shyam's weight: [ text{Percentage increase} = frac{text{New weight} - text{Original weight}}{text{Original weight}} times 100 ] [ text{Percentage increase} = frac{60.184 - 51.4}{51.4} times 100 ] [ text{Percentage increase} = frac{8.784}{51.4} times 100 ] [ text{Percentage increase} approx 17.09% ] So, Shyam's weight had to be increased by approximately boxed{17.09%} .

answer:To solve for the expected number of coins showing heads, we'll use the probability and expected values again, considering the new conditions: 1. **Probability Analysis for each Coin:** - Probability a coin lands heads on the first toss (and is tossed again): frac{1}{2}. Probability it shows heads again on this second toss is frac{1}{2} times frac{1}{2} = frac{1}{4}. - Probability a coin lands tails on the first toss and is tossed again: frac{1}{2}. Probability it then shows heads on the second toss and is tossed again: frac{1}{2} times frac{1}{2} = frac{1}{4}. Probability it shows heads on this third toss is frac{1}{4} times frac{1}{2} = frac{1}{8}. - Probability a coin lands tails on the first two tosses and is tossed a third time: frac{1}{2} times frac{1}{2} = frac{1}{4}. Probability it shows heads on this third toss is frac{1}{4} times frac{1}{2} = frac{1}{8}. 2. **Total Probability of a Coin Showing Heads:** - Compiling all scenarios where the coin can end up heads: frac{1}{4} (heads on first and second toss) + frac{1}{8} (second toss heads, third toss heads) + frac{1}{8} (heads on the third toss after two tails) = frac{1}{4} + frac{1}{8} + frac{1}{8} = frac{1}{2}. 3. **Expected Number of Heads:** - With each coin now having a frac{1}{2} probability of being heads after all possible tosses and scenarios: [ E(text{number of heads}) = 80 times frac{1}{2} = 40 ] Concluding, the expected number of coins that show heads after up to three tosses and with the included conditions is 40. The final answer is boxed{textbf{(B)} 40}

answer:1. **Calculate the total pages read in the first four days**: Jordan read an average of 42 pages per day for the first four days. Therefore, the total number of pages he read in these four days is: [ 4 times 42 = 168 text{ pages} ] 2. **Calculate the total pages read in the next two days**: For the following two days, he read an average of 38 pages per day. Thus, the total number of pages he read during these days is: [ 2 times 38 = 76 text{ pages} ] 3. **Add the pages read in the first six days**: The total number of pages read in the first six days is the sum of the pages read in the first four days and the next two days: [ 168 + 76 = 244 text{ pages} ] 4. **Include the pages read on the last day**: On the seventh day, Jordan read 20 more pages. Therefore, the total number of pages in the book, including the last day, is: [ 244 + 20 = 264 text{ pages} ] 5. **Conclusion**: The total number of pages in the book that Jordan read is 264. The final answer is boxed{B}

answer:To find the minimum value of the integral (int_{-1}^1 {f(x)-(a|x|+b)}^2 , dx), we need to first express the function (f(x)) and then set up the integral. 1. **Define the function (f(x)):** The function (f(x)) is given by the piecewise linear function connecting the points ((-1, 0)), ((0, 1)), and ((1, 4)). Therefore, we can write: [ f(x) = begin{cases} 1 + x & text{for } -1 leq x < 0, 1 + 3x & text{for } 0 leq x leq 1. end{cases} ] 2. **Set up the integral:** We need to minimize the integral: [ int_{-1}^1 {f(x) - (a|x| + b)}^2 , dx. ] This can be split into two integrals: [ int_{-1}^0 {(1 + x) - (a(-x) + b)}^2 , dx + int_0^1 {(1 + 3x) - (ax + b)}^2 , dx. ] 3. **Simplify the integrands:** For (x in [-1, 0)): [ (1 + x) - (a(-x) + b) = 1 + x - (-ax + b) = 1 + x + ax - b = (1 - b) + (1 + a)x. ] For (x in [0, 1]): [ (1 + 3x) - (ax + b) = 1 + 3x - (ax + b) = 1 - b + (3 - a)x. ] 4. **Set up the integrals with the simplified expressions:** [ int_{-1}^0 {(1 - b) + (1 + a)x}^2 , dx + int_0^1 {(1 - b) + (3 - a)x}^2 , dx. ] 5. **Expand the integrands:** [ int_{-1}^0 {(1 - b)^2 + 2(1 - b)(1 + a)x + (1 + a)^2 x^2} , dx + int_0^1 {(1 - b)^2 + 2(1 - b)(3 - a)x + (3 - a)^2 x^2} , dx. ] 6. **Integrate each term:** [ int_{-1}^0 (1 - b)^2 , dx + int_{-1}^0 2(1 - b)(1 + a)x , dx + int_{-1}^0 (1 + a)^2 x^2 , dx + int_0^1 (1 - b)^2 , dx + int_0^1 2(1 - b)(3 - a)x , dx + int_0^1 (3 - a)^2 x^2 , dx. ] 7. **Evaluate the integrals:** [ (1 - b)^2 int_{-1}^0 dx + 2(1 - b)(1 + a) int_{-1}^0 x , dx + (1 + a)^2 int_{-1}^0 x^2 , dx + (1 - b)^2 int_0^1 dx + 2(1 - b)(3 - a) int_0^1 x , dx + (3 - a)^2 int_0^1 x^2 , dx. ] [ (1 - b)^2 [x]_{-1}^0 + 2(1 - b)(1 + a) left[frac{x^2}{2}right]_{-1}^0 + (1 + a)^2 left[frac{x^3}{3}right]_{-1}^0 + (1 - b)^2 [x]_0^1 + 2(1 - b)(3 - a) left[frac{x^2}{2}right]_0^1 + (3 - a)^2 left[frac{x^3}{3}right]_0^1. ] [ (1 - b)^2 (0 - (-1)) + 2(1 - b)(1 + a) left(0 - frac{1}{2}right) + (1 + a)^2 left(0 - left(-frac{1}{3}right)right) + (1 - b)^2 (1 - 0) + 2(1 - b)(3 - a) left(frac{1}{2} - 0right) + (3 - a)^2 left(frac{1}{3} - 0right). ] [ (1 - b)^2 + 2(1 - b)(1 + a) left(-frac{1}{2}right) + (1 + a)^2 left(frac{1}{3}right) + (1 - b)^2 + 2(1 - b)(3 - a) left(frac{1}{2}right) + (3 - a)^2 left(frac{1}{3}right). ] 8. **Combine like terms and simplify:** [ 2(1 - b)^2 - (1 - b)(1 + a) + frac{(1 + a)^2}{3} + (1 - b)(3 - a) + frac{(3 - a)^2}{3}. ] 9. **Differentiate with respect to (a) and (b) and set to zero to find critical points:** [ frac{partial}{partial a} left(2(1 - b)^2 - (1 - b)(1 + a) + frac{(1 + a)^2}{3} + (1 - b)(3 - a) + frac{(3 - a)^2}{3}right) = 0, ] [ frac{partial}{partial b} left(2(1 - b)^2 - (1 - b)(1 + a) + frac{(1 + a)^2}{3} + (1 - b)(3 - a) + frac{(3 - a)^2}{3}right) = 0. ] Solving these equations, we get: [ 2a + 3b = 5, ] [ a + 2b = 3. ] 10. **Solve the system of equations:** [ begin{cases} 2a + 3b = 5, a + 2b = 3. end{cases} ] Multiply the second equation by 2: [ 2a + 4b = 6. ] Subtract the first equation from this result: [ (2a + 4b) - (2a + 3b) = 6 - 5, ] [ b = 1. ] Substitute (b = 1) into the second equation: [ a + 2(1) = 3, ] [ a = 1. ] 11. **Substitute (a = 1) and (b = 1) back into the integral:** [ int_{-1}^0 (1 + x - (1(-x) + 1))^2 , dx + int_0^1 (1 + 3x - (1x + 1))^2 , dx, ] [ int_{-1}^0 (x)^2 , dx + int_0^1 (2x)^2 , dx, ] [ int_{-1}^0 x^2 , dx + int_0^1 4x^2 , dx, ] [ left[frac{x^3}{3}right]_{-1}^0 + 4left[frac{x^3}{3}right]_0^1, ] [ left(0 - left(-frac{1}{3}right)right) + 4left(frac{1}{3} - 0right), ] [ frac{1}{3} + frac{4}{3} = frac{5}{3}. ] The final answer is (boxed{frac{8}{3}}).