answer:First, we need to find the slopes of both lines.The slope of line l_{1}: 3x+4y+1=0 can be found by rearranging the equation into slope-intercept form (y = mx + b).begin{align}3x + 4y + 1 &= 0 4y &= -3x - 1 y &= -frac{3}{4}x - frac{1}{4}end{align}So the slope of line l_{1} is -frac{3}{4}.Similarly, for line l_{2}: 4x-3y+2=0begin{align}4x - 3y + 2 &= 0 -3y &= -4x + 2 y &= frac{4}{3}x - frac{2}{3}end{align}So the slope of line l_{2} is frac{4}{3}.Two lines are perpendicular if and only if the product of their slopes is equal to -1. In this case:-frac{3}{4} times frac{4}{3} = -1Therefore, line l_{1} and line l_{2} are perpendicular.The answer is: boxed{B}.

answer:To calculate 3-left(-2right), we follow the steps:1. Recognize that subtracting a negative number is equivalent to adding its positive counterpart. Thus, 3-left(-2right) becomes 3+2.2. Add the two positive numbers: 3+2=5.Therefore, the calculation simplifies to:3-left(-2right) = 3+2 = 5Hence, the correct answer is boxed{D}.

answer:Solution. Let n_{1} points fall on the first die, n_{2} - on the second. The space of elementary events is the set of pairs (n_{1}, n_{2}):Omega=left{left(n_{1}, n_{2}right): n_{1}, n_{2}=1,2,3,4,5,6right}Event A is of the formA=left{left(n_{1}, n_{2}right): n_{1}, n_{2}=1,2,3,4 ; n_{1}+n_{2} leq 5right} .The set Omega contains 36 elements (see Table 1.1), the set A-10 elements ((1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2), (4,1)). According to formula (1.7.5) we getP(A)=frac{10}{36}=frac{5}{18}## Problems1. Two dice are tossed. Find the probability that the sum of the points on both dice is greater than 6 (event A)?2. In a lottery, 100 tickets are raffled. The prize falls on 10 tickets. Someone bought 3 tickets. What is the probability that at least one of them will win?3. In a box, there are 6 blue and 9 red balls. Three balls are drawn from the box. Find the probability that two of them will be blue.4. On eight identical cards, the numbers 2, 4,6,7,8,11,12 and 13 are written. Two cards are drawn at random. Determine the probability that the fraction formed from the two obtained numbers is reducible.5. Out of ten tickets, two are winning. Find the probability that among the five tickets taken at random: a) one is winning; b) both are winning.## Answerstext { 1. } frac{21}{36}=frac{7}{12} cdot 2 cdot frac{C_{100}^{3}-C_{90}^{3}}{C_{100}^{3}}=1-frac{C_{90}^{3}}{C_{100}^{3}}=1-frac{90 cdot 89 cdot 88}{100 cdot 99 cdot 98} approx 0.2735 cdot 3 cdot frac{27}{91} cdot 4 cdot frac{5}{14} text {. }5. a) frac{5}{9}; b) frac{2}{9}.## Questions1. What is called the space of elementary events?2. What is called an elementary event?3. What is called an event?4. What is called an impossible event?5. What is called a certain event?6. What is called a random event?7. How are opposite events defined?8. What are called incompatible events?9. What is called a complete group of events?10. What is called an algebra of events?11. How is the probability of an event defined?12. What are the axioms of probability?13. What is called a probability space?14. What is the sum of the probabilities of opposite events?## § 1.8. Addition and Multiplication of Probabilities## Theorem of Addition of Probabilities of Two EventsThe probability of the sum of two events is equal to the sum of the probabilities of these events minus the probability of their joint occurrence:P(A+B)=P(A)+P(B)-P(A B)Theorem of Addition of Probabilities of Two Incompatible EventsThe probability of the sum of two incompatible events is equal to the sum of the probabilities of these events:P(A+B)=P(A)+P(B)Remark 1. Formula (1.8.2) is derived from formula (1.8.1) when A and B are incompatible events; in this case A B is an impossible event and P(A B)=0.Theorem of Addition of Probabilities of n Incompatible EventsThe probability of the sum of n incompatible events A_{1}, A_{2}, ldots, A_{n} is equal to the sum of the probabilities of these events:Pleft(A_{1}+A_{2}+ldots+A_{n}right)=Pleft(A_{1}right)+Pleft(A_{2}right)+ldots+Pleft(A_{n}right)The sum of the probabilities of events A_{1}, A_{2}, ldots, A_{n}, forming a complete group, is equal to one:Pleft(A_{1}right)+Pleft(A_{2}right)+ldots+Pleft(A_{n}right)=1The sum of the probabilities of opposite events is equal to one:P(A)+P(bar{A})=1If we denoteP(A)=p, P(bar{A})=qthen formula (1.8.5) takes the formp+q=1The probability of event B given that event A has occurred is called the conditional probability of event B and is denoted as: P(B / A), or P_{A}(B).## Theorem of Multiplication of ProbabilitiesThe probability of the product of two events is equal to the product of the probability of one of them and the conditional probability of the other given that the first event has occurred:P(A B)=P(A) P(B / A), P(A B)=P(B) P(A / B)Event B is independent of event A ifP(B / A)=P(B)i.e., the probability of event B does not depend on whether event A has occurred.In this case, event A is also independent of event B, i.e., the property of independence of events is mutual.Note that if A and B are independent, then bar{A} and B, A and bar{B}, bar{A} and bar{B} are also independent.Theorem of Multiplication of Probabilities of Two Independent EventsThe probability of the product of two independent events is equal to the product of their probabilities:P(A B)=P(A) P(B)Theorem of Multiplication of Probabilities of n EventsThe probability of the product of n events is equal to the product of one of them and the conditional probabilities of all the others, calculated under the assumption that all previous events have occurred:Pleft(A_{1} A_{2} ldots A_{n}right)=Pleft(A_{1}right) Pleft(A_{2} / A_{1}right) Pleft(A_{3} / A_{1} A_{2}right) ldots Pleft(A_{n} / A_{1} A_{2} ldots A_{n-1}right)In particular, for three events A, B, C formula (1.8.11) takes the formP(A B C)=P(A) P(B / A) P(C / A B)Events A_{1}, A_{2}, ldots, A_{n} are called independent in the aggregate, or independent, if they are pairwise independent, and also each of them is independent of the product of k of the others (k=2,3, ldots, n-1).Remark 2. Pairwise independence of events does not imply their independence in the aggregate.Remark 3. If events A_{1}, A_{2}, ldots, A_{n} are independent, then the opposite events bar{A}_{1}, bar{A}_{2}, ldots, bar{A}_{n} are also independent.## Theorem of Multiplication of Probabilities of n Independent EventsIf events A_{1}, A_{2}, ldots, A_{n} are independent, then the probability of their product is equal to the product of the probabilities of these events:Pleft(A_{1} A_{2} ldots A_{n}right)=Pleft(A_{1}right) Pleft(A_{2}right) ldots Pleft(A_{n}right)Remark 4. Equality (1.8.13) expresses the necessary and sufficient condition for the independence of events A_{1}, A_{2}, ldots, A_{n}.For three independent events A, B, C formula (1.8.13) takes the formP(A B C)=P(A) P(B) P(C)The calculation of the probability of the sum of events can be reduced to the calculation of the probability of the product of opposite events by the formulaPleft(A_{1}

answer:To find the value of k for which the line y=-2x+4 intersects the line y=kx at a point on the line y=x+2, we follow these steps:1. Set up the system of equations for the intersection of the lines y=-2x+4 and y=kx: [ left{ begin{array}{l} y = -2x + 4 y = kx end{array} right. ]2. Solve the system for x and y in terms of k: From y = -2x + 4 and y = kx, setting the two expressions for y equal gives -2x + 4 = kx. Solving for x yields: [ x = frac{4}{k + 2} ] Substituting this back into y = kx gives: [ y = kleft(frac{4}{k + 2}right) = frac{4k}{k + 2} ]3. Since the intersection point lies on the line y = x + 2, substitute x and y into this equation: [ frac{4k}{k + 2} = frac{4}{k + 2} + 2 ]4. Solve the equation for k: Simplify the equation by multiplying both sides by (k + 2) to eliminate the denominator: [ 4k = 4 + 2(k + 2) ] This simplifies to: [ 4k = 4 + 2k + 4 ] Further simplification gives: [ 4k = 2k + 8 ] Solving for k yields: [ 2k = 8 implies k = 4 ]Therefore, the value of k is boxed{4}, which corresponds to choice boxed{A}.

answer:Solution. 1. We will provide the answer in two steps, and as preparation, we will list the ways in which 8 fruits can be arranged on 4 identical plates, with 2 fruits on each. We will call our arrangement principle that we fill the plates one after the other, and if during this process we see multiple ways to continue, we proceed by first using the type of fruit of which there is currently the least; if there are multiple types of fruits with the same least quantity, we proceed in alphabetical order of their initials. We will denote the fruits by their initial letters.The possibilities for filling the first plate:a) a, b,B) a, s z,gamma ) a, kAnd in the alpha) case, the following 2 situations can be created on the first two plates:left.left.alpha_{1}right) quad a, b, quad b, s z ; quad alpha_{2}right) quad a, b, quad b, k.The former situation clearly leads to the arrangement in the following table (I), while the latter can be completed in two ways, because there are 2 each of k and s z left: by placing only one type of fruit on each of the last two plates, or mixed: (II) and (III). (For simplicity, we will no longer insert commas between the members of a pair of letters.)The continuations of the beta) start can be obtained from (I)-(III) by swapping the b, s z letters, since their quantities in the stock are equal, these are the (IV)-(VI) arrangements, which are clearly different from the former, and since the first three are different, the latter three are also different.Applying our principles to gamma), we get the (VII)-(XI) plate fillings.(I) a b, b s z, s z k, k k;(IV) a s z, s z b, b k, k k(II) a b, b k, k k, s z s z(V) a s z, s z k, k k, b b(III) a b, b k, k s z, k s z(VII) a k, b b, k k, s z s z ;(VIII) a k, b b, k s z, k s z(IX) a k, b k, b k, s z s z(X) a k, b k, b s z, k s z(XI) a k, b s z, b s z, k k.2. In the (I), (II), (IV), (V), (VII), and (X) arrangements, there are no two plates filled identically, while in the other five, the exchange of 2-2 identical plates is not noticeable. In the former 6 arrangements, we can choose 4 ways to decide which girl gets the first plate (serving), 3 ways from the remaining girls for the second plate, 2 ways for the third, and the fourth plate is uniquely determined by the one who has not yet received. Thus, we get 4 cdot 3 cdot 2=24 distribution methods from each of the 6 cases, totaling 144.Considering the 24 distribution methods for each of the other 5 arrangements, they yield identical results in pairs, so only 12 are different, for example, if in (III) we give the third plate to Cilin and the fourth to Dora, the distribution does not change if these two (indistinguishable) servings are swapped between them.Accordingly, the 5 arrangements containing 2-2 identical servings give 5 cdot 12=60 distribution methods, and the answer to the question is: 144+60=204 different distributions are possible.Remarks. 1. Another starting point for compiling the (I)-(XI) arrangements: the 3 apples go into either 2 or 3 different plates. In the former case, the quantities to be further divided are 1,2,1,2, and with our principle, a k, b b, s z s z (VII), a k, b s z, b s z (XI); a b, b k, s z s z (II); a b, b s z, k s z (I); and (with b, s z swapped) a s z, k s z, b b (V); a s z, k b, s z b (IV); result. In the latter case, we can choose 5 ways for the still empty plate from the quantities 1,2,0,2: a b (III); a s z (VI); b b (VIII); b s z(mathrm{X}) ; s z s z (IX), and the remaining 3 fruits can be placed one by one on the still equal (not distinguished) plates through their 1-1 apple.2. We can also organize according to whether we prepare 3,2,1 or 0 servings that contain only one type of fruit; the groups receive 1,2,5 and 3 arrangements.

answer:① is incorrect, variables should be separated by ","; ② is incorrect, assignment statements cannot be input; ③ is incorrect, assignment statements cannot be output; ④ is correct.Therefore, the correct answer is boxed{text{D}}.