How had India kept her relationship with the world in ancient times? -Maths 9th

How had India kept her relationship with the world in ancient times? -Maths 9th
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INDIA's contact with the world have continued through the ages but her relationships through the land route are much older than her maritime contacts.The various passes across the mountains in the north have provided passages to the ancient travellers, while the oceans restricted such interaction for a long time. These routes have contributed in the exchange of ideas and commodities since ancient times.The ideas of the Upanishads and the Ramayana, the stories  of Panchtantra, the Indian numerals and the decimal system thus could reach many parts of the world.
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How had India kept her relationship with the world in ancient times? -Geography 9th

Description : How had India kept her relationship with the world in ancient times? -Geography 9th

Last Answer : How had India kept her relationship with the world in ancient times? ... These routes have contributed in the exchange of ideas and commodities since ancient times. The ideas of the Upanishads ... of Panchtantra, the Indian numerals and the decimal system thus could reach many parts of the world.

A tyre manufacturing company kept a record -Maths 9th

Description : A tyre manufacturing company kept a record -Maths 9th

Last Answer : The total number of trials = 1000 (i) P (tyre to be replaced before it covers 4000 km) = 20/1000 = 0.02 (ii) The frequency of a tyre that will last more than 9000 km = 325 + 445 = 770 ∴ P ( ... = 210 + 325 = 535 So, P (tyre requiring replacement between 4000 km and 14000 km) = 535/1000 = 0.535

What is the relationship between chord of a circle and a perpendicular to it from the centre? -Maths 9th

Description : What is the relationship between chord of a circle and a perpendicular to it from the centre? -Maths 9th

Last Answer : Solution :- Perpendicular line from the centre bisect the chord.

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Last Answer : Given : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged ... length of coloured tape required = 30 cm (b) The values are : Happiness, beauty, Knowledge.

Mary wants to decorate her ********* tree. She wants to place the tree on a wooden block covered with coloured paper with picture of Santa Claus on it. -Maths 9th

Description : Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with coloured paper with picture of Santa Claus on it. -Maths 9th

Last Answer : Length of box (l) = 80 cm Breadth (b) = 40 cm and height (h) = 20 cm ∴ Total surface area = 2(lb + bh + hl) = 2[80 x 40 + 40 x 20 + 20 x 80] cm² = 2[3200 + 800 + 1600] ... of one sheet = (40 cm)² = 1600 cm² ∴ No. of sheets required for the box = 11200 = 1600 = 7 sheets

On her birthday, Amisha donated 2 toffees to each -Maths 9th

Description : On her birthday, Amisha donated 2 toffees to each -Maths 9th

Last Answer : x = 2y + 15 (a) x – 2y – 15 = 0 (b) 61 = 2y + 15 ⇒ y = 46/2 = 23 children (c) Caring, kind, socially active.

For her records, a teacher asked the students about their heights. -Maths 9th

Description : For her records, a teacher asked the students about their heights. -Maths 9th

Last Answer : Yes, Things which are equal to the same thing are equal to one another. Knowledge, curiosity, truthfulness

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm. He wants to decorate the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that ... the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th

Last Answer : Let us consider the following lay out of the greeting card. Trapeziums are arranged in such a way that AB || HC || GD || FE. Also BC=CD=DE and GF=6 cm and DE = 4cm. If three parallel lines make equal ... HG+GF+BC+CD+DE = 6+6+6+4+4+4=30 cm. (b) The values are: Happiness, beauty, Knowledge.

A child consumed an ice-cream of inverted right-circular conical shape from the top and left only 12.5% of the cone for her mother. -Maths 9th

Description : A child consumed an ice-cream of inverted right-circular conical shape from the top and left only 12.5% of the cone for her mother. -Maths 9th

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What can you infer from the fact that Ancient Greeks and Egyptians kept slaves?

Description : What can you infer from the fact that Ancient Greeks and Egyptians kept slaves?

Last Answer : Feel Free to Answer

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 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

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

A dice is rolled number of times and its outcomes are recorded as below : -Maths 9th

Description : A dice is rolled number of times and its outcomes are recorded as below : -Maths 9th

Last Answer : Total number of outcomes = 250 Total number of outcomes of getting odd numbers = 35 + 50 + 53 = 138 P (getting an odd number) = 138 / 250 = 69 / 125

A die is thrown six times and number on it is noted as given below : -Maths 9th

Description : A die is thrown six times and number on it is noted as given below : -Maths 9th

Last Answer : Here, in 6 trials, each number occur once and total prime numbers i.e., 2, 3, 5 occur one time each Hence, the number of prime occur = 3 Probability of getting a prime = 3/6 =1/2

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

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

A die was rolled 100 times and the number of times, 6 came up was noted. -Maths 9th

Description : A die was rolled 100 times and the number of times, 6 came up was noted. -Maths 9th

Last Answer : Here, total number of trials = 100 Let x be the number of times occuring 6. We know, Probability of an ever = Frequency of the event occuring / Total number of trials ⇒ x / 100 = 2 / 5 [∵ Probability is given] ⇒ x = 40

Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Description : Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Last Answer : Let the abscissa of the point be x, According to the question, Ordinate (y) = 3 x Abscissa ⇒ y=3x When x = 1, then y = 3 x 1 = 3 and when x = 2, then y = 3 x 2 = 6. Here, ... the line AB. Hence, y = 3x is the required equation such that each point on its graph has an ordinate 3 times its abscissa.

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

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

A dice is rolled number of times and its outcomes are recorded as below : -Maths 9th

Description : A dice is rolled number of times and its outcomes are recorded as below : -Maths 9th

Last Answer : Total number of outcomes = 250 Total number of outcomes of getting odd numbers = 35 + 50 + 53 = 138 P (getting an odd number) = 138 / 250 = 69 / 125

A die is thrown six times and number on it is noted as given below : -Maths 9th

Description : A die is thrown six times and number on it is noted as given below : -Maths 9th

Last Answer : Here, in 6 trials, each number occur once and total prime numbers i.e., 2, 3, 5 occur one time each Hence, the number of prime occur = 3 Probability of getting a prime = 3/6 =1/2

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

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

A die was rolled 100 times and the number of times, 6 came up was noted. -Maths 9th

Description : A die was rolled 100 times and the number of times, 6 came up was noted. -Maths 9th

Last Answer : Here, total number of trials = 100 Let x be the number of times occuring 6. We know, Probability of an ever = Frequency of the event occuring / Total number of trials ⇒ x / 100 = 2 / 5 [∵ Probability is given] ⇒ x = 40

Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Description : Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Last Answer : Let the abscissa of the point be x, According to the question, Ordinate (y) = 3 x Abscissa ⇒ y=3x When x = 1, then y = 3 x 1 = 3 and when x = 2, then y = 3 x 2 = 6. Here, ... the line AB. Hence, y = 3x is the required equation such that each point on its graph has an ordinate 3 times its abscissa.

The weight of a man is four times the weight of a child. Write an equation in two variables for this situation. -Maths 9th

Description : The weight of a man is four times the weight of a child. Write an equation in two variables for this situation. -Maths 9th

Last Answer : Solution :-

Determine the point on the graph of the linear equation x + y = 6, whose ordinate is 2 times its abscissa. -Maths 9th

Description : Determine the point on the graph of the linear equation x + y = 6, whose ordinate is 2 times its abscissa. -Maths 9th

Last Answer : Solution :-

How many times area is changed, when sides of a triangle are doubled. -Maths 9th

Description : How many times area is changed, when sides of a triangle are doubled. -Maths 9th

Last Answer : Four times area is changed, when sides of a triangle are doubled.

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

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

Last Answer : Frequency disribution of above data in tabular form is given as:

The following table gives the life times of 400 neon lamps: -Maths 9th

Description : The following table gives the life times of 400 neon lamps: -Maths 9th

Last Answer : Sol. (i) (ii) Numbers of lamps having life 700 or more hours 74 + 62 + 48 = 184

In tossing a coin 100 times head appears 56 times. -Maths 9th

Description : In tossing a coin 100 times head appears 56 times. -Maths 9th

Last Answer : P (head) = 56/100 = 0.56.

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

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

A die was rolled 100 times and the number of times, -Maths 9th

Description : A die was rolled 100 times and the number of times, -Maths 9th

Last Answer : This answer was deleted by our moderators...

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

In cricket match, a batsman hits a boundary 12 times -Maths 9th

Description : In cricket match, a batsman hits a boundary 12 times -Maths 9th

Last Answer : Let E be the event 'the batsman hits a boundary'. Then, E bar is the event 'the batsman does hit a boundary'. ∴ P(E) = 12/48 = 1/4 ⇒ P(E bar) = 1 - P(E) = 1 - 1/4 = 3/4

linear equation such that each point on its graph has an ordinate one more than 3×1÷2 times its abscissa -Maths 9th

Description : linear equation such that each point on its graph has an ordinate one more than 3×1÷2 times its abscissa -Maths 9th

Last Answer : This answer was deleted by our moderators...

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)

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 =\(rac{20}{216}\) = \(rac{5}{24}\).

A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is -Maths 9th

Description : A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is -Maths 9th

Last Answer : (c) \(rac{84 imes5^6}{6^{10}}\)In the first nine throws we should have three sixes and six non-sixes and a six in the tenth throw and thereafter whatever face appears, it doesn't matter. ∴ Required probability = 9C3 \(\bigg(rac{1} ... x 1 x 1 ............x 1 {10 times} = \(rac{84 imes5^6}{6^{10}}\).

A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. -Maths 9th

Description : A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. -Maths 9th

Last Answer : Let radius of iron rod = r ∴∴ Height = 8r ∴∴ Volume of iron rod =π×(r)2×8r⇒8πr3=π×(r)2×8r⇒8πr3 ⇒⇒ Radius of spherical ball =r2=r2 Volume of spherical ball =43π(r2)3=43π(r2)3 Let n balls are casted ∴n×43π(r38)=8πr3∴n×43π(r38)=8πr3 ⇒n6=8⇒n=48

A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th

Description : A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th

Last Answer : Height of cone = 3 times height of cylinder = 3 3 = 9 cm Volume of cylinder = volume of cone r2 = 8 8 r = 8 cm l2 = h2 + r2 = (9)2 + (8)2 l = = 12 cm C.S.A (cone) = = 301.71 cm2

Find the condition that one root of ax^2 + bx + c = 0 may be four times the other. -Maths 9th

Description : Find the condition that one root of ax^2 + bx + c = 0 may be four times the other. -Maths 9th

Last Answer : answer: