Three coins were tossed 30 times simultaneously. -Maths 9th

Three coins were tossed 30 times simultaneously. -Maths 9th
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Frequency disribution of above data in tabular form is given as:
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Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Last Answer : Total number of chances = 23 + 72 + 77 + 28 = 200 Number of chances of coming 2 heads = 72 therefore P( coming 2 heads)= 514 / 642 = 9 / 25

Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. -Maths 9th

Description : Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. -Maths 9th

Last Answer : Total number of outcomes = 360 Let the number of times ‘No Tail’ appeared be x Then, number of times ‘2 Tails’ appeared =3x Number of times ‘1 Tail’ appeared =2x Now, x + 2x + 3x =360 ⇒ 6x =360 ⇒ x= 60 P(of getting two tails)=(3 x 60) / 360 =1 / 2

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Last Answer : It is given that coin is tossed 200 times Total number of trials = 200 Number of events for getting less than three tails = 68 + 82 + 30 = 180 Probability of getting less than 3 tails =180 / 200 =9 / 10

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Last Answer : Total number of chances = 23 + 72 + 77 + 28 = 200 Number of chances of coming 2 heads = 72 therefore P( coming 2 heads)= 514 / 642 = 9 / 25

Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. -Maths 9th

Description : Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. -Maths 9th

Last Answer : Total number of outcomes = 360 Let the number of times ‘No Tail’ appeared be x Then, number of times ‘2 Tails’ appeared =3x Number of times ‘1 Tail’ appeared =2x Now, x + 2x + 3x =360 ⇒ 6x =360 ⇒ x= 60 P(of getting two tails)=(3 x 60) / 360 =1 / 2

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Last Answer : It is given that coin is tossed 200 times Total number of trials = 200 Number of events for getting less than three tails = 68 + 82 + 30 = 180 Probability of getting less than 3 tails =180 / 200 =9 / 10

Two coins are tossed simultaneously 500 times. -Maths 9th

Description : Two coins are tossed simultaneously 500 times. -Maths 9th

Last Answer : Since, frequency of one or more than one head = 100 + 270 = 370 Therefore, P (one or more heads) = 370/500 = 37/50

Three coins are tossed simultaneously -Maths 9th

Description : Three coins are tossed simultaneously -Maths 9th

Last Answer : Frequency of more than one tail = 135 + 85 = 220 ∴ P (more than one tail) = 220/500 = 11/25

Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails. -Maths 9th

Description : Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails. -Maths 9th

Last Answer : (d) \(rac{1}{2}\)Let S be the sample space. Then, S = {HHH, HHT, HTH, HTT, THH, THT,TTH, TTT} ⇒ n(S) = 8 Let A : Event of getting more heads than number of tails. Then, A = {HHH, HHT, HTH, THH} ⇒ n(A) = 4∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{8}\) = \(rac{1}{2}.\)

Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

Two coins are tossed 1000 times and the outcomes are recorded as below : -Maths 9th

Description : Two coins are tossed 1000 times and the outcomes are recorded as below : -Maths 9th

Last Answer : Required probability = P(0 heads) + P(1 head) = 250/1000 + 550 / 1000 = 800/ 1000 =4 / 5 =0.8

Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

Two coins are tossed 1000 times and the outcomes are recorded as below : -Maths 9th

Description : Two coins are tossed 1000 times and the outcomes are recorded as below : -Maths 9th

Last Answer : Required probability = P(0 heads) + P(1 head) = 250/1000 + 550 / 1000 = 800/ 1000 =4 / 5 =0.8

Two coins are tossed 1000 times and the outcomes are recorded as below: -Maths 9th

Description : Two coins are tossed 1000 times and the outcomes are recorded as below: -Maths 9th

Last Answer : P (at most one head) = P (0 head) + P (1 head) = 250/1000 + 550/1000 = 800/1000 = 4/5

Let a pair of fair coins be tossed. Here S = {HH, HT, TH, TT}. Consider the events A = {heads on the first coin} = {HH, HT}, -Maths 9th

Description : Let a pair of fair coins be tossed. Here S = {HH, HT, TH, TT}. Consider the events A = {heads on the first coin} = {HH, HT}, -Maths 9th

Last Answer : ThenP (A) = P (B) = P (C) = \(rac{2}{4}\) = \(rac{1}{2}\) andP (A ∩ B) = P ({HH}) = \(rac{1}{4}\), P (A ∩ C) = P ({HT}) = \(rac{1}{4}\)P ( ... C)Thus condition (i) is satisfied, i.e., the events are pairwise independent. But condition (ii) is not satisfied and so the three events are not independent

If two coins are tossed once, what is the probability of getting at least one head ? -Maths 9th

Description : If two coins are tossed once, what is the probability of getting at least one head ? -Maths 9th

Last Answer : When two coins are tossed once, there are four possible outcomes, i.e., S = {HH, HT, TH, TT} ∴ Total number of outcomes = n(S) = 4 Let A : Event of getting at least one head ⇒ A = {HH, HT, TH} ⇒ n(A) = 3∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{3}{4}.\)

When 2 coins are tossed simultaneously, write all possible outcomes.

Description : When 2 coins are tossed simultaneously, write all possible outcomes.

Last Answer : When 2 coins are tossed simultaneously, write all possible outcomes.

If we tossed simultaneously two coins. Find the probability of exactly one tail.

Description : If we tossed simultaneously two coins. Find the probability of exactly one tail.

Last Answer : If we toss two coins simultaneously,there are four possible outcomes HEAD-HEAD  TAIL-TAIL HEAD-TAIL  TAIL-HEAD  so probability of getting exactly one tail=2/4=1/2

A fair coin is tossed three times. Let A, B and C be defined as follows: -Maths 9th

Description : A fair coin is tossed three times. Let A, B and C be defined as follows: -Maths 9th

Last Answer : The sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HHH, HHT, HTH, HTT}, B = {HHH, HHT, THH, THT} and C = {HHT, THH} Also, A ∩ B = {HHH, HHT}, B ∩ C = {HHT, THH}, C ∩ A = {HHT}P (A ... (C), i.e., if the events are pairwise independent and (ii) P (A ∩ B ∩ C) = P (A) . P (B) . P (C)

Three coins are tossed 100 times, and three heads one head occurred 14 times and head did not occur 23 times. Find the probability of getting more tha

Description : Three coins are tossed 100 times, and three heads one head occurred 14 times and head did not occur 23 times. Find the probability of getting more tha

Last Answer : Three coins are tossed 100 times, and three heads one head occurred 14 times and head did not ... Find the probability of getting more than one head.

A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

A coin is tossed thrice and all eight outcomes are assumed equally likely. Find whether the events E -Maths 9th

Description : A coin is tossed thrice and all eight outcomes are assumed equally likely. Find whether the events E -Maths 9th

Last Answer : When a coin is tossed three times, the sample space is given by S = [HHH, HHT, HTH, THT, THH, HTT, TTH, TTT] E = {HHH, HTT, THT, TTH}, F = {TTT, HTH, THH, HHT}E ∩ F = ϕP(E) = \(rac{4}{8}\) = \(rac{1}{2}\ ... rac{1}{2}\) x \(rac{1}{2}\) x \(rac{1}{4}\) ≠ P(E ∩ F) ∴ E and F are not independent events.

Two dice are thrown simultaneously 500 times. -Maths 9th

Description : Two dice are thrown simultaneously 500 times. -Maths 9th

Last Answer : (i) P (getting a sum more than 10) = P (getting a sum of 11) + P (getting a sum of 12) = 28/500 + 15/500 = 28 + 15/500 = 43/500 = 0.869 = 0.09 (ii) P (getting a sum less than or equal to 5) = P ( ... + P (getting a sum of 10) + P (getting a sum of 11) = 53/500 + 46/500 + 28/500 = 127/500 = 0.254

Two coins are tossed. Find the number of outcomes of getting one head.

Description : Two coins are tossed. Find the number of outcomes of getting one head.

Last Answer : Two coins are tossed. Find the number of outcomes of getting one head.

Hari has some two rupee and five rupee coins .The total amount with him is rs. 43. Express the given information as a linear equation in two variables. -Maths 9th

Description : Hari has some two rupee and five rupee coins .The total amount with him is rs. 43. Express the given information as a linear equation in two variables. -Maths 9th

Last Answer : answer:

A bag contains 2n + 1 coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. -Maths 9th

Description : A bag contains 2n + 1 coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. -Maths 9th

Last Answer : (a) 10As (n + 1) coins are fair P (Tossing a tail) = \(rac{rac{n+1}{2}}{2n+1}\) = \(rac{n+1}{2(2n+1)}\)∴ P (Tossing a head) = 1 - \(rac{n+1}{2(2n+1)}\) = \(rac{4n+2-n-1}{2(2n+1)}\) = \(rac{3n+1}{4n+2}\)Given, \(rac{3n+1}{4n+2}\) = \(rac{31}{42}\)⇒ 126n + 42 = 124n + 62 ⇒ 2n = 20 ⇒ n = 10.

Two dice are rolled simultaneously. The probability of getting a multiple of 2 on one dice and a multiple of 3 on the other is -Maths 9th

Description : Two dice are rolled simultaneously. The probability of getting a multiple of 2 on one dice and a multiple of 3 on the other is -Maths 9th

Last Answer : (c) \(rac{11}{36}\)Total number of outcomes when two identical dice are rolled, n(S) = 6 6 = 36 Let A : Event of rolling a multiple of 2 on one die and a multiple of 3 on the other die ⇒ A = {(2, 3), (2, 6), (4, 3), (4, ... , 4), (3, 6)} ⇒ n(A) = 11 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{11}{36}\).

A coin and six faced die, both unbiased are thrown simultaneously. -Maths 9th

Description : A coin and six faced die, both unbiased are thrown simultaneously. -Maths 9th

Last Answer : (c) \(rac{1}{4}\)Let A : Event of getting a tail on the coin B : Event of getting an even number on the die. Then, P(A) = \(rac{1}{2}\)P(B) = \(rac{3}{6}\) = \(rac{1}{2}\) as B = {2,4,6}A and B being independent events ... die)= P(A ∩ B) = P(A) P(B) = \(rac{1}{2}\)x\(rac{1}{2}\) = \(rac{1}{4}\).

Find the range of values of x which satisfy x^2 + 6x – 27 > 0, –x^2 + 3x + 4 > 0 simultaneously. -Maths 9th

Description : Find the range of values of x which satisfy x^2 + 6x – 27 > 0, –x^2 + 3x + 4 > 0 simultaneously. -Maths 9th

Last Answer : answer:

The set of values of x for which the inequalities x^2 – 3x – 10 < 0, 10x – x^2 – 16 > 0 hold simultaneously is -Maths 9th

Description : The set of values of x for which the inequalities x^2 – 3x – 10 < 0, 10x – x^2 – 16 > 0 hold simultaneously is -Maths 9th

Last Answer : answer:

A coin is tossed 500 times. Head occurs 343 times and tail occurs 157 times. Find the probability of each event.

Description : A coin is tossed 500 times. Head occurs 343 times and tail occurs 157 times. Find the probability of each event.

Last Answer : A coin is tossed 500 times. Head occurs 343 times and tail occurs 157 times. Find the probability of each event.

A coin is tossed 20 times and head occurred 12 times. How many times did tail occur?

Description : A coin is tossed 20 times and head occurred 12 times. How many times did tail occur?

Last Answer : A coin is tossed 20 times and head occurred 12 times. How many times did tail occur?

A single coin is tossed 7 times. What is the probability of getting at least one tail? a) 127/128 b) 128/127 c) 2/128 d) 4/128

Description : A single coin is tossed 7 times. What is the probability of getting at least one tail? a) 127/128 b) 128/127 c) 2/128 d) 4/128

Last Answer : Answer: A) Consider solving this using complement. Probability of getting no tail = P(all heads) = 1/128 P(at least one tail) = 1 – P(all heads) = 1 – 1/128 = 127/128

the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th

Description : the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th

Last Answer : T.S.A = 3*154 = 462 cm² C.S.A = 154 cm² C.S.A = 2πrh T.S.A = 2πr(r+h) Now, In T.S.A = 2πrr + 2πrh 462 = 2πrr + 2πrh 462 = 2*22/7*r*r + 154 462 - 154 = 2*22/7*r*r 308*7/2*22 = r*r 49 = r*r R = 7 cm ... 7*h 154/44 = h 7/2 =h H = 3.5 cm or 7/2 cm Now volume = πrrh = 22/7 * 7* 7 *7/2 = 11*49 = 539 cm³

A can hit a target three times in five shots, B two times in five shots and C three times in four shots. They fire a volley. -Maths 9th

Description : A can hit a target three times in five shots, B two times in five shots and C three times in four shots. They fire a volley. -Maths 9th

Last Answer : P(A) = \(rac{3}{5}\), P(B) = \(rac{2}{5}\), P(C) = \(rac{3}{4}\). In order that two shots may hit the target the following three mutually exclusive events are possible,(i) A and B hit the target and not C. (ii) B ... {100}\) + \(rac{12}{100}\) + \(rac{27}{100}\) = \(rac{45}{100}\) = \(rac{9}{20}.\)

A die is rolled three times. The probability of getting a larger number than the previous number each time is: -Maths 9th

Description : A die is rolled three times. The probability of getting a larger number than the previous number each time is: -Maths 9th

Last Answer : (b) \(rac{5}{24}\)Total number of ways three die can be rolled = 6 6 6 = 216 A larger number than the previous number can be got in the three throws as (1, 2, 3), (1, 2, 4), (1, 2, 5) ( ... , 5, 6). ∴ Total number of favourable cases = 20∴ Required probability =\(rac{20}{216}\) = \(rac{5}{24}\).

Which one of the following is the equation whose roots are respectively three times the roots of the equation ax^2 + bx + c = 0 ? -Maths 9th

Description : Which one of the following is the equation whose roots are respectively three times the roots of the equation ax^2 + bx + c = 0 ? -Maths 9th

Last Answer : answer:

The quadratic equation whose roots are three times the roots of 3ax^2 + 3bx + c = 0 is -Maths 9th

Description : The quadratic equation whose roots are three times the roots of 3ax^2 + 3bx + c = 0 is -Maths 9th

Last Answer : answer:

30 children were asked about the number of hours -Maths 9th

Description : 30 children were asked about the number of hours -Maths 9th

Last Answer : No, the number of children who watched TV for 10 or more hours a week is 4 + 2, i.e., 6 .

Mr. Shrimant inherits 4325 gold coins and divides them among his three sons; Bharat, Parat and Marat; in a certain ratio. Out of the total coins received by each of them, Bharat sells 40 coins; Parat donates his 20 coins and Marat loses 30 coins. Now, the ratio of gold coins with them is 41:34:46, respectively. How many coins did Parat receive from his father? a) 1210 b) 1211 c) 1212 d) 1213 e) None of these

Description : Mr. Shrimant inherits 4325 gold coins and divides them among his three sons; Bharat, Parat and Marat; in a certain ratio. Out of the total coins received by each of them, Bharat sells 40 coins; Parat donates his ... did Parat receive from his father? a) 1210 b) 1211 c) 1212 d) 1213 e) None of these 

Last Answer : Answer: A 41x+40+34x+20+46x+30=4325 =>121x=4325-90 =>x=35 Required answer, number of coins received by parat =34x+20=34×35+20=1210

A cistern is filled by three faucets with uniform flow. The first two faucets operating simultaneously fill the cistern in the same during which the cistern is filled by the third faucet alone. The second faucet fills the cistern 5 hours faster than the first faucet and 4 hours slower than the third faucet. The time required by the first faucet is? A) 6 hrs B) 10 hrs C) 15 hrs D) 30 hrs

Description : A cistern is filled by three faucets with uniform flow. The first two faucets operating simultaneously fill the cistern in the same during which the cistern is filled by the third faucet alone. The second faucet fills the cistern ... by the first faucet is? A) 6 hrs B) 10 hrs C) 15 hrs D) 30 hrs

Last Answer : C Suppose, first faucet alone takes x hours to fill the cistern. Then, second and third faucets will take (x - 5) and (x - 9) hours respectively to fill the cistern. 1/x + 1/(x - 5) = 1/(x - 9) (2x - 5)(x - 9) = x(x - 5) x2 - 18x + 45 = 0 (x- 15)(x - 3) = 0 => x = 15

A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Description : A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Last Answer : Here, each side of the rhombus = 30 m. Let ABCD be the given rhombus and the diagonal, BD = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m Since, a diagonal divides the rhombus into ... Area of grass for 18 cows to graze = 864 m2 ⇒ Area of grass for 1 cow to graze = 86418 m2 = 48 m2

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Last Answer : For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm Area of the given parallelogram = Area of the given triangle ∴ Area of the parallelogram = 336 cm2 ⇒ base x height = 336 ⇒ ... be the height of the parallelogram. ⇒ h = 33628 = 12 Thus, the required height of the parallelogram = 12 cm

Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3 -Maths 9th

Description : Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3 -Maths 9th

Last Answer : The given pit has its length(l) as 8m, width (b)as 6m and depth (h)as 3 m. Volume of cuboidal pit = l×b×h = (8×6×3) = 144 (using formula) Required Volume is 144 m3 Now, Cost of digging per m3 volume = Rs 30 Cost of digging 144 m3 volume = Rs (144×30) = Rs 4320

he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th

Description : he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th

Last Answer : Say h = height of the frame of lampshade, looks like cylindrical shape r = radius Total height is h = (2.5+30+2.5) cm = 35cm and r = (20/2) cm = 10cm Use curved surface area formula to find the ... 2πrh = (2 (22/7) 10 35) cm2 = 2200 cm2 Hence, 2200 cm2 cloth is required for covering the lampshade.

The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Description : The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Last Answer : The other two sides of the triangle are 12 cm and 5 cm Explanation: Let the other two sides of triangle be x and y It's hypotenuse is 13 cm Perimeter of triangle = Sum of all sides ... When y = 12 x=17-y = 17-12 =5 So, the other two sides of the triangle are 12 cm and 5 cm

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)

Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Description : Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Last Answer : see in book okay!!!

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm