The probability of student A passing examination is 3/7 and of student B passing is 5/7 Assuming the two events “A passes”, -Maths 9th

The probability of student A passing examination is 3/7 and of student B passing is 5/7 Assuming the two events “A passes”, -Maths 9th
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p1 = P(A) = \(rac{3}{7}\), p2 = P(B) = \(rac{5}{7}\) ∴ q1 = P(\(\bar{A}\)) = 1 - P(A) = 1 - \(rac{3}{7}\) = \(rac{4}{7}\). q2 = P(\(\bar{B}\)) = 1 - P(B) = 1 - \(rac{5}{7}\) = \(rac{2}{7}\)(i) P (only A passes) = p1 q2 = \(rac{3}{7}\) x \(rac{2}{7}\) = \(rac{6}{49}.\) P (only B passes) = q1 p2 = \(rac{4}{7}\) x \(rac{5}{7}\) = \(rac{20}{49}\)(ii) P (only one of them passes) = p1 q2 + q1 p2 = \(rac{3}{7}\) x \(rac{2}{7}\) + \(rac{4}{7}\) x \(rac{5}{7}\) = \(rac{26}{49}.\)
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Related questions

In an examination there are 3 multiple choice questions and each question has 4 choices. If a student randomly selects answer for all -Maths 9th

Description : In an examination there are 3 multiple choice questions and each question has 4 choices. If a student randomly selects answer for all -Maths 9th

Last Answer : Probability of selecting a correct choice for a question = \(rac{1}{4}\)(∵ Out of 4 choices only one is correct)∴ Probability of answering all the three questions correctly = \(rac{1}{4}\)x \(rac{1}{4}\)x\ ... of not answering all the three questions correctly = 1 - \(rac{1}{64}\) = \(rac{63}{64}\).

In an examination, 900 students appeared. Out of these students; 56 % got first division, 27 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed. a) 225 b) 153 c) 245 d) 148 e) 298

Description : In an examination, 900 students appeared. Out of these students; 56 % got first division, 27 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed. a) 225 b) 153 c) 245 d) 148 e) 298

Last Answer : Answer: B The number of students with first division = 56 % of 900  = 56/100 × 900 = 50400/100  = 504 And, the number of students with second division = 27 % of 900  = 27/100 × 900  =24300/100  = 243 Therefore, the number of students who just passed = 900 – (504 + 243)  = 900- 747  = 153

If the three independent events E1, E2 and E3, the probability that only E1 occurs is α, E2 occurs -Maths 9th

Description : If the three independent events E1, E2 and E3, the probability that only E1 occurs is α, E2 occurs -Maths 9th

Last Answer : (c) 6Let \(x\), y, z be the probabilities of happening of events E1, E2 and E3 respectively. Then, \(\alpha\) = P (occurrence of E1 only) = x (1 - y) (1 - z) \(\beta\) = P (occurrence of E2 only) = (1 - \(x\)) y (1 - ... (x\) = 6z (From (i) and (ii))⇒ \(rac{x}{z}\) = 6 ⇒ \(rac{P(E_1)}{P(E_3)}\) = 6.

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Description : A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Last Answer : (d) \(rac{31}{66}\)Total number of balls in the bag = 12 (5 Green + 7 Red) Let S be the sample space of drawing 2 balls out of 12 balls.Thenn(S) = 12C2 = \(rac{12 imes11}{2}\) = 66∴ Let A : Event of drawing two red balls⇒ ... \(rac{n(B)}{n(S)}\) = \(rac{21}{66}\) + \(rac{10}{66}\) = \(rac{31}{66}\).

A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white. -Maths 9th

Description : A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white. -Maths 9th

Last Answer : Let A, B, C, D denote the events of not getting a white ball in first, second, third and fourth draw respectively. Since the balls are drawn with replacement, therefore, A, B, C, D are independent events such that P (A) = P (B) ... x \(rac{11}{16}\) x \(rac{11}{16}\) = \(\big(rac{11}{16}\big)^4.\)

Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers ? -Maths 9th

Description : Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers ? -Maths 9th

Last Answer : The chosen consists of players (6 + 5). ∴ Number of ways of selecting 11 players out of 15 players = n(S) = 15C11 Let A : Event of choosing 6 batsmen of 8 batsmen and 5 bowlers of 7 bowlers Then, n(A) = 8C6 ... 7 imes6}{2}\) x \(rac{4 imes3 imes2 imes1}{15 imes14 imes13 imes12}\) = \(rac{28}{65}.\)

A bag contains 7 red and 5 green balls. The probability of drawing all four balls asred balls, when four balls are drawn at random is -Maths 9th

Description : A bag contains 7 red and 5 green balls. The probability of drawing all four balls asred balls, when four balls are drawn at random is -Maths 9th

Last Answer : (b) \(rac{7}{99}\)There are (7 + 5) = 12 balls in the bag. 4 balls can be drawn at random from 12 balls in 12C4 ways. ∴ n(S) = 12C4 = \(rac{|\underline{7}}{|\underline3|\underline4}\) = \(rac{7 imes6 imes5}{3 ... ) = 35∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{35}{495}\) = \(rac{7}{99}\).

There are 5 red, 4 white and 3 blue marbles in a bag. They are drawn one by one and arranged in a row. Assuming that all the 12 marbles -Maths 9th

Description : There are 5 red, 4 white and 3 blue marbles in a bag. They are drawn one by one and arranged in a row. Assuming that all the 12 marbles -Maths 9th

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Description : Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3). -Maths 9th

Last Answer : For three points to be collinear, the area of the triangle formed by the three points should be zero. ∴ Area of D formed by the given three points= \(rac{1}{2}\) [a((c + a) - (a + b)) + b((a + b) - (b + c)) + c((b ... ba - bc + cb - ca] = 0.Hence (a, b + c), (b, c + a) and (c, a + b) are collinear.

Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7. -Maths 9th

Description : Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7. -Maths 9th

Last Answer : (c) \(rac{7}{12}\)Total number of ways in which 2 dice are rolled = 6 6 = 36 ⇒ n(S) = 36 Let A : Event of rolling an even number of 1st dice B : Event of rolling a total of 7 ⇒ A = {(2, 1), (2, 2) , (2, 6), (4 ... (rac{18}{36}\) + \(rac{6}{36}\) - \(rac{3}{36}\) = \(rac{21}{36}\) = \(rac{7}{12}\).

The probability of A, B, C solving a problem are 1/3, 2/7 and 3/8 respectively. -Maths 9th

Description : The probability of A, B, C solving a problem are 1/3, 2/7 and 3/8 respectively. -Maths 9th

Last Answer : Let E1, E2, E3 be the eventsthat the problem issolved by A, B, C respectively and let p1, p2, p3 be corresponding probabilities. Then,p1 = P(E1) = \(rac{1}{3}\), p2 = P(E2) = \(rac{2}{7}\), p3 = P(E3) = \(rac{3}{8}\) ... }{8}\) = \(rac{25}{168}\) + \(rac{5}{42}\) + \(rac{5}{28}\) = \(rac{25}{56}.\)

If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9 is -Maths 9th

Description : If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9 is -Maths 9th

Last Answer : (d) 226 × 52C26 | 104C26Since there are 52 distinct cards in a deck and each distinct card is 2 in number.∴2 decks will also contain only 52 distinct cards, two each.∴ Probability that the player gets all distinct cards = \(rac{^{52}C_{26} imes2^{26}}{^{104}C_{26}}\).

On planet A , each alien has 3 eyes and on planet B , each alien has 4 eyes. A group of aliens from planets A and B landed on earth. They had 30 eyes in all. Assuming number of aliens from planets A and B respectively , express the situation in the form of linear equation in two variables. -Maths 9th

Description : On planet A , each alien has 3 eyes and on planet B , each alien has 4 eyes. A group of aliens from planets A and B landed on earth. They had 30 eyes in all. Assuming number of ... planets A and B respectively , express the situation in the form of linear equation in two variables. -Maths 9th

Last Answer : answer:

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. -Maths 9th

Description : A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. -Maths 9th

Last Answer : (c) \(rac{11}{3^5}\)Probability of guessing a correct answer = \(rac{1}{3}\)Probability of guessing an incorrect answer = \(rac{2}{3}\)∴ Probability of guessing 4 or more correct answers = 5C4 \(\big(rac{1}{3}\big)^4\)\(\big(rac{2 ... )^5\) = 5 x \(rac{2}{3^5}\) + \(rac{1}{3^5}\) = \(rac{11}{3^5}\).

There are two groups, X and Y wrote an examination. The probability of X's pass is 3/5 and the probability of Y's pass is 5/7. What is the probability that only one of them is passed out? a) 15/16 b) 16/35 c) 12/43 d) 18/35

Description : There are two groups, X and Y wrote an examination. The probability of X's pass is 3/5 and the probability of Y's pass is 5/7. What is the probability that only one of them is passed out? a) 15/16 b) 16/35 c) 12/43 d) 18/35

Last Answer : Answer: B) Let X be the event of the group X pass Let Y be the event of the group Y pass Then, X'= Event of the group X's fail and Y'= event of the group Y's fail. Therefore, p(X) = 3/5 and p(Y) = 5/7, P(X') = 1 - P( ... )] + p[(Y And X')] = p(X) * p(Y') + P(Y) *P(X') = (6/35 + 10/35) = 16/35

In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Description : In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 Number of students having blood group ‘B’ = 12 Required probability =12 / 40 = 3 / 10

In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Description : In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 Number of students having blood group ‘B’ = 12 Required probability =12 / 40 = 3 / 10

In a medical examination of students of a class, -Maths 9th

Description : In a medical examination of students of a class, -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 P (a student has blood group B) = 12/40 = 3/10

In a public examination , Raghav scored twice twice Sunitha’s score. If r and s represent the scores of Raghav and Sunitha respectively , Write a linear equation in r and s representing the above situation ? -Maths 9th

Description : In a public examination , Raghav scored twice twice Sunitha’s score. If r and s represent the scores of Raghav and Sunitha respectively , Write a linear equation in r and s representing the above situation ? -Maths 9th

Last Answer : Let raghav and sunita score be r and S respectively. Sunita score=2times raghav score Therefore, s=2r

The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Description : The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Last Answer : x̅ = 1 / n (Σxi) ⇒ 55 = x1 + x2 + .......+ ⇒ x7 / 7 ⇒ x1 + x2 + ...... + x7 = 55 × 7 = 385 x1 + x2 + ...... + x6 = 52 + 54 + 55 + 53 + 56 + 54 = 324 ∴ x7 = 385 - 324 = 61 kg ∴ weight of the seventh student is 61 kg.

The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Description : The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Last Answer : x̅ = 1 / n (Σxi) ⇒ 55 = x1 + x2 + .......+ ⇒ x7 / 7 ⇒ x1 + x2 + ...... + x7 = 55 × 7 = 385 x1 + x2 + ...... + x6 = 52 + 54 + 55 + 53 + 56 + 54 = 324 ∴ x7 = 385 - 324 = 61 kg ∴ weight of the seventh student is 61 kg.

Check whether the graph of the equation y = 3x + 5 passes through the origin or not. -Maths 9th

Description : Check whether the graph of the equation y = 3x + 5 passes through the origin or not. -Maths 9th

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Find the value of k if the line on 2x + y = k passes through the point (3,5). -Maths 9th

Description : Find the value of k if the line on 2x + y = k passes through the point (3,5). -Maths 9th

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Find the slope and inclination of the line which passes through the points (1, 2) and (5, 6) ? -Maths 9th

Description : Find the slope and inclination of the line which passes through the points (1, 2) and (5, 6) ? -Maths 9th

Last Answer : Let A ≡ (x, y), B ≡ (2, 1), C ≡ (3, -2) Area of ΔABC = \(rac{1}{2}\) |{x1 (y2 - y3) + x2(y3 - y1) + x3(y1 - y2)}|= \(rac{1}{2}\) | \(x\)(1 + 2) + 2(-2 - y) + 3(y - 1) | = \(rac{1}{2 ... \(rac{3}{2}\)∴ Co-ordinates of A are \(\bigg(rac{7}{2},rac{13}{2}\bigg)\) or \(\bigg(rac{-3}{2},rac{3}{2}\bigg)\)

Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Description : Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Last Answer : (a) 45º 3x + y - 7 = 0 ⇒ y = -3x + 7 ⇒ Slope (m1) = -3 x + 2y + 9 = 0 ⇒ y = \(rac{-x}{2}\) - \(rac{9}{2}\) ⇒ Slope (m2) = \(-rac{1}{2}\)If θ is the angle between the given lines, then tan θ = \(\ ... \bigg|rac{-rac{5}{2}}{1+rac{3}{2}}\bigg|\)= \(\bigg|rac{-rac{5}{2}}{rac{5}{2}}\bigg|\) = 1∴ θ = 45°.

The line L is given by x/5 + y/b = 1 passes through the point (13, 32). The line K is parallel to L and has the equation -Maths 9th

Description : The line L is given by x/5 + y/b = 1 passes through the point (13, 32). The line K is parallel to L and has the equation -Maths 9th

Last Answer : (a) 45º The equations of the given lines are: A\(x\) + By = A + B ⇒ By = -A\(x\) + (A + B) ⇒ y = \(-rac{A}{B}x\) + \(rac{(A+B)}{B}\) ....(i)and A(\(x\) - y) + B(\(x\) ... (ii) = m2 = \(rac{(A+B)}{B-A}\)Let θ be the angle between both the lines, then∴ tan θ = 1 ⇒ θ = tan-1 (1) = 45°.

Find the equation of the straight line with a positive gradient which passes through the point (–5, 0) -Maths 9th

Description : Find the equation of the straight line with a positive gradient which passes through the point (–5, 0) -Maths 9th

Last Answer : (d) Both (a) and (c)Since the line passes through A(a, 0) and B(0, b), it makes intercepts a and b on x-axis and y-axis respectively. Let the equation of this line in the intercept from be \(rac{x}{a}\) + \(rac{y}{a}\) ... \(rac{x}{-12}\) + \(rac{y}{-5}\) = 1⇒ 5x + 12y = 60 and 5x + 12y + 60 = 0.

A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Description : A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Last Answer : (b) \(rac{\sqrt{17}}{2}\)Equation of the line through the points (5, 0) and (0, 3) y - 0 = \(rac{3-0}{0-5}\) (x - 5)⇒ y = \(rac{-3}{5}\)(x - 5)⇒ 5y + 3x - 15 = 0 ∴ Distance of perpendicular from ... (rac{|20+12-15|}{\sqrt{25+9}{}}\) = \(rac{17}{\sqrt{34}}\) units. = \(rac{\sqrt{17}}{2}\) units.

Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -Maths 9th

Description : Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -Maths 9th

Last Answer : Solution :- x= -4

Find the value of a, if the line passing through (–5, –8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a). -Maths 9th

Description : Find the value of a, if the line passing through (–5, –8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a). -Maths 9th

Last Answer : Let C(x, y) be the centre of the circle passing through the points P(6, -6), Q(3, -7) and R(3, 3) Then, PC = QC = RC(Being radius of the same circle) PC2 = QC2 ⇒ (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2 ⇒ x2 - ... i), we get 3\(x\) + (-2) - 7 = 0 ⇒ 3\(x\) = 9 ⇒ \(x\) = 3 ∴ The centre is (3, -2).

Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th

Description : Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th

Last Answer : There are two ways to prove it. 1st way: Area of triangle formed by the given points = 0 if they are collinear.∴ \(rac{1}{2}\) [\(x\)(2 - (y + 1) ) + 1((y + 1) - 1) + 0(1 - 2)] = 0⇒ \(rac{1}{2}\) [2\(x\) - \(x\)y - \( ... y - \(x\)y - 1 + \(x\) ⇒ x + y = \(x\)y ⇒ \(rac{1}{x}\) + \(rac{1}{y}\) = 1.

Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : Given : E and F are mid points of 2 chords AB and CD respectively. Line EF passes through centre. To prove : AB||CD ∠ OFC = ∠ OEA = 90° as line drawn through the centre to bisect the ... EF as traversal for lines AB and CD as alternate interior angles on same side are equal. Therefore, AB || CD

Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : Given : E and F are mid points of 2 chords AB and CD respectively. Line EF passes through centre. To prove : AB||CD ∠ OFC = ∠ OEA = 90° as line drawn through the centre to bisect the ... EF as traversal for lines AB and CD as alternate interior angles on same side are equal. Therefore, AB || CD

The graph of a linear equation in two variables always passes through three quadrants of the graph paper. True/false -Maths 9th

Description : The graph of a linear equation in two variables always passes through three quadrants of the graph paper. True/false -Maths 9th

Last Answer : answer:

The graph of the linear equation y = x passes through the point. -Maths 9th

Description : The graph of the linear equation y = x passes through the point. -Maths 9th

Last Answer : (c) The linear equation y = x has same value of x and y-coordinates are same. Therefore, the point (1,1) must lie on the line y = x.

P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

The graph of the linear equation y = x passes through the point. -Maths 9th

Description : The graph of the linear equation y = x passes through the point. -Maths 9th

Last Answer : (c) The linear equation y = x has same value of x and y-coordinates are same. Therefore, the point (1,1) must lie on the line y = x.

P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

2x + y = 3 passes from origin. Is this statement true or false? -Maths 9th

Description : 2x + y = 3 passes from origin. Is this statement true or false? -Maths 9th

Last Answer : Solution :-

The slope of a line perpendicular to the line which passes through the points (–k, h) and (b, – f ) is -Maths 9th

Description : The slope of a line perpendicular to the line which passes through the points (–k, h) and (b, – f ) is -Maths 9th

Last Answer : (b) (2, 2)The line 3x + 4y - 24 = 0 cuts the axis at A. To obtain the co-ordinates of A put y = 0, as on x-axis, y = 0. ∴ A ≡ (8, 0) ...(i) Also, on y-axis, x = 0, therefore B ≡ (0, 6 ... 8+10},rac{6 imes0+8 imes6+10 imes0}{6+8+10}\bigg)\)= \(\bigg(rac{48}{24},rac{48}{24}\bigg)\) = (2, 2).

A line passes through the point of intersection of the lines 100x + 50y – 1 = 0 and 75x + 25y + 3 = 0 and makes equal intercepts on the axes. -Maths 9th

Description : A line passes through the point of intersection of the lines 100x + 50y – 1 = 0 and 75x + 25y + 3 = 0 and makes equal intercepts on the axes. -Maths 9th

Last Answer : (d) x + 2y = 2Let the required equation make intercept on x-axis = 2a ⇒ intercept made on y-axis = a ∴ Eqn of the given line in the intercept from:\(rac{x}{2a}+rac{y}{a}=1\) ...(i)Since the line ... 1 ⇒ a = 1.∴ Required equation of line : \(rac{x}{2 imes1}+rac{y}{1}=1\) ⇒ x + 2y = 2.

A straight line passes through the points (a, 0) and (0, b). The length of the line segment contained between the axes is 13 and the product of -Maths 9th

Description : A straight line passes through the points (a, 0) and (0, b). The length of the line segment contained between the axes is 13 and the product of -Maths 9th

Last Answer : (d) \(rac{23}{\sqrt{17}}\)The given lines are:L : \(rac{x}{5}+rac{y}{b}=1\) ....(i)K : \(rac{x}{c}+rac{y}{3}=1.\) ...(ii)Since line L passes through (13, 32),\(rac{13}{ ... between parallel lines ax + by + c1 = 0 and ax + by c2 = 0 is d = \(rac{|c_2-c_1|}{\sqrt{a^2+b^2}}\bigg)\)

What is the equation of the straight line which passes through (3, 4) and the sum of whose x-intercept and y-intercept is 14 ? -Maths 9th

Description : What is the equation of the straight line which passes through (3, 4) and the sum of whose x-intercept and y-intercept is 14 ? -Maths 9th

Last Answer : (a) 4x + 3y = 24 Let the x-intercept = a. Then, y-intercept = 14 - a ∴ Eqn of the straight line is \(rac{x}{a}\) + \(rac{y}{14-a}\) = 1Since it passes through (3, 4), so\(rac{3}{a}\) + \(rac{4}{14-a}\) = 1⇒ 3(14 - ... = 1 ⇒ x + y = 7or \(rac{x}{6}\) + \(rac{y}{8}\) = 1 ⇒ 8x + 6y = 48 ⇒ 4x + 3y = 24.

Find the probability that a two digit number formed by the digit 1, 2, 3, 4 and 5 is divisible by 4. -Maths 9th

Description : Find the probability that a two digit number formed by the digit 1, 2, 3, 4 and 5 is divisible by 4. -Maths 9th

Last Answer : The two digit numbers can be formed by putting any of 5 digits at the one 's place and also one of the 5 digits at ten's place. So, Total number of 2-digit numbers that can be formed using these 5-digits = 5 5 = ... 52}, i.e, 5 in number. ∴ Required probability = \(rac{5}{25}\) = \(rac{1}{5}.\)