answer:1. **Calculate the cost price of the candy bars:** The troop buys candy bars at a rate of 3 bars for 1.50. Therefore, the cost per bar is: [ frac{1.50 text{ dollars}}{3 text{ bars}} = 0.5 text{ dollars per bar} ] Since they buy 1500 bars, the total cost is: [ 1500 text{ bars} times 0.5 text{ dollars per bar} = 750 text{ dollars} ] 2. **Calculate the selling price of the candy bars:** The troop sells the candy bars at a rate of 3 bars for 2. Therefore, the selling price per bar is: [ frac{2 text{ dollars}}{3 text{ bars}} = 0.667 text{ dollars per bar} ] Since they sell 1500 bars, the total revenue is: [ 1500 text{ bars} times 0.667 text{ dollars per bar} = 1000.5 text{ dollars} ] 3. **Calculate the profit:** The profit is the difference between the total revenue and the total cost: [ text{Profit} = text{Total Revenue} - text{Total Cost} = 1000.5 text{ dollars} - 750 text{ dollars} = 250.5 text{ dollars} ] Thus, the scout troop's profit is 250.5 dollars. The final answer is boxed{textbf{(C)} 250.5}

answer:Consider a repeating decimal x = 0.overline{def}. Convert to fraction: x = 0.overline{def} 1000x = def.overline{def} 999x = 1000x - x = def x = frac{def}{999} If def (as a number) is not divisible by factors of 999 that are not 3, the fraction is in its lowest terms. We are given that d+e+f is not divisible by 3, hence def is not divisible by 3. The factors of 999 are 3^3 and 37, but the condition d+e+f not divisible by 3 ensures the fraction frac{def}{999} can't be reduced by 3. We need to consider 37: - Total possible values for def from 001 to 999 are 999. - Values divisible by 37 are 27 multiples. - Thus, numbers not divisible by 37 are 999 - 27 = 972. Since each valid def not divisible by 37 yields a unique numerator: boxed{972}

answer:Given the line equation x = 2y + 5, we can substitute the x and y coordinates of the first point (m, n) into the equation to find the y-coordinate (n). So, we have: m = 2n + 5 We don't have the exact values of m and n, but we can use the information about the second point (m + 5, n + 2.5) to find n. Since the second point lies on the same line, it must also satisfy the line equation. Therefore, we can substitute the x and y coordinates of the second point into the line equation: m + 5 = 2(n + 2.5) + 5 Now, let's solve for n: m + 5 = 2n + 5 + 5 m + 5 = 2n + 10 Since we know from the first point that m = 2n + 5, we can substitute this into the equation: 2n + 5 + 5 = 2n + 10 2n + 10 = 2n + 10 This equation is true for all values of n, which means that any value of n that satisfies the first point's equation (m = 2n + 5) will also satisfy the second point's equation. Therefore, to find the y-coordinate (n) of the first point, we can rearrange the first point's equation: m = 2n + 5 m - 5 = 2n (m - 5)/2 = n Since we don't have the exact value of m, we cannot determine the exact value of n. However, the y-coordinate of the first point (n) can be expressed in terms of m as (m - boxed{5)/2} .

answer:To determine the probability of only one student solving the problem, we need to consider the following mutually exclusive events: - Student A solves the problem, but students B and C do not. - Student B solves the problem, but students A and C do not. - Student C solves the problem, but students A and B do not. Given that the events of each student solving the problem are independent of each other, we can calculate the required probability as follows: For student A solving and others not solving: - Probability that A solves the problem: frac{1}{2}. - Probability that B does not solve the problem: 1 - frac{1}{3} = frac{2}{3}. - Probability that C does not solve the problem: 1 - frac{1}{4} = frac{3}{4}. - Combined probability for this scenario: frac{1}{2} times frac{2}{3} times frac{3}{4}. Similarly, we can calculate for B and C: - For B: frac{1}{2} times frac{1}{3} times frac{3}{4}. - For C: frac{1}{2} times frac{2}{3} times frac{1}{4}. Adding all the scenarios together, the required probability is: P(text{only one solves}) = left(frac{1}{2} times frac{2}{3} times frac{3}{4}right) + left(frac{1}{2} times frac{1}{3} times frac{3}{4}right) + left(frac{1}{2} times frac{2}{3} times frac{1}{4}right). Carrying out the calculation by multiplying the fractions and then performing the sum: P(text{only one solves}) = left(frac{1}{2} times frac{2}{3} times frac{3}{4}right) + left(frac{1}{2} times frac{1}{3} times frac{3}{4}right) + left(frac{1}{2} times frac{2}{3} times frac{1}{4}right) = left(frac{6}{24}right) + left(frac{3}{24}right) + left(frac{2}{24}right) = frac{11}{24}. Therefore, the probability that only one of the three students solves the problem is boxed{frac{11}{24}}.