answer:Since it takes June 4 minutes to travel 1 mile to Julia's house, her rate is frac{1 text{ mile}}{4 text{ minutes}}. To find the time required to travel to Carl's house, multiply the distance by her travel time per mile: [ text{Time} = 7 text{ miles} times 4 text{ minutes per mile} = 28 text{ minutes} ] Thus, it would take June boxed{28} minutes to travel to Carl's house.

answer:If a certain number x, when divided by 61, gives a remainder of 24, it means that x can be expressed as: x = 61k + 24 where k is some integer. Now, we want to find the remainder when x is divided by 5. To do this, we can substitute x into the expression and then take the modulo with respect to 5: (61k + 24) mod 5 Since 61 is congruent to 1 modulo 5 (because 61 divided by 5 gives a remainder of 1), we can simplify the expression: (1k + 24) mod 5 Now, we just need to find the remainder when 24 is divided by 5: 24 mod 5 = 4 Therefore, the remainder when x is divided by 5 is boxed{4} .

answer:First, we solve the inequality x^{2}-5x-6geqslant 0 from set P. By factoring, we get (x+1)(x-6)geqslant 0. Solving for x, we find that xleqslant -1 or xgeqslant 6. Thus, the set P can be represented as P={x|xleqslant -1 text{ or } xgeqslant 6}. Given that the universal set is U=mathbb{R}, We find the complement of P in U to be ∁_UP=(-1,6). Therefore, the answer is boxed{(-1,6)}. To solve this problem, we determined the solution set of the inequality in set P, defined set P, and then found the elements in the universal set U that do not belong to set P, thus obtaining the complement of set P. This problem tests the understanding of complements and their operations, and employs the concept of transformation. When finding complements, it is important to consider the range of the universal set.

answer:Since the function is f(x)= dfrac {1}{2}+log _{2} dfrac {x}{1-x}, We have f(x)+f(1-x)=1+log _{2} dfrac {x}{1-x}+log _{2} dfrac {1-x}{x}=1+log _{2}1=1, Given S_{n}=f( dfrac {1}{n})+f( dfrac {2}{n})+f( dfrac {3}{n})+…+f( dfrac {n-1}{n}), We have 2S_{n}=f( dfrac {1}{n})+f( dfrac {n-1}{n})+f( dfrac {2}{n})+f( dfrac {n-2}{n})+…+f( dfrac {n-1}{n})+f( dfrac {1}{n}) =1×n=n, Hence, S_{n}= dfrac {n}{2}. So, the answer is: boxed{ dfrac {n}{2} }. The function f(x)= dfrac {1}{2}+log _{2} dfrac {x}{1-x} leads to f(x)+f(1-x)=1+log _{2} dfrac {x}{1-x}+log _{2} dfrac {1-x}{x}=1+log _{2}1=1, and then by using the "reverse order addition method", we can obtain the answer. This problem tests the "reverse order addition method", properties of functions, and summation of sequences, requiring both reasoning and computational skills, making it a moderate-level question.