As the number of tosses of a coin increases, -Maths 9th

As the number of tosses of a coin increases, -Maths 9th
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1 Answer

Answer :

No. As the number of tosses of a coin increases, the ratio of the number of heads to the total number of tosses will be near to 1/2, not exactly 1/2.
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What is the average number of coin tosses required to flip both sides at least once?

Description : What is the average number of coin tosses required to flip both sides at least once?

Last Answer : Two.

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 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

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.

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.

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)

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

A and B throw a coin alternately till one of them gets a ‘head’ and wins the game. Find their respective probabilities of winning . -Maths 9th

Description : A and B throw a coin alternately till one of them gets a ‘head’ and wins the game. Find their respective probabilities of winning . -Maths 9th

Last Answer : Let A : Event of A getting a head ⇒ \(\bar{A}\) : Event of A not getting a head ∴ P(A) = \(rac{1}{2}\) and P(\(\bar{A}\)) = 1 - \(rac{1}{2}\) = \(rac{1}{2}\)Similarly, B : Event of B ... exclusive events, as either of them will win, P(B winning the game first) = 1 - \(rac{2}{3}\) = \(rac{1}{3}\).

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

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When a pitcher tosses a baseball to the catcher across the home plates what is the projectile?

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Mickey is fishing from his row boat at his favorite spot in the lake. He has a large rock in the bottom of the boat for stability and ballast. Mickey decides to call it quits. To lighten his load, he tosses the rock overboard before rowing to shore. Does the water level of the lake: w) go up x) go down y) remain the same

Description : Mickey is fishing from his row boat at his favorite spot in the lake. He has a large rock in the bottom of the boat for stability and ballast. Mickey decides to call it quits. To lighten his load, he tosses the ... to shore. Does the water level of the lake: w) go up x) go down y) remain the same

Last Answer : ANSWER: X -- GO DOWN

The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th

Description : The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th

Last Answer : Let r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So r1 = 7cm r2 = 14 cm Now, Required ratio = (initial surface area)/(Surface area after pumping air into ... = (7/14)2 = (1/2)2 = ¼ Therefore, the ratio between the surface areas is 1:4.

The radius of a spherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Description : The radius of a spherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Last Answer : Surface area of a spherical balloon whose radius is 6 cm. = 4π × 6 × 6 cm2 Surface area of a spherical balloon whose radius is 12 cm. = 4π × 12 × 12 cm2 ∴ Ration of surface areas = 4π × 6 × 6 / 4π × 12 × 12 = 1 / 4 = 1 : 4

The radius of a spherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Description : The radius of a spherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Last Answer : Surface area of a spherical balloon whose radius is 6 cm. = 4π × 6 × 6 cm2 Surface area of a spherical balloon whose radius is 12 cm. = 4π × 12 × 12 cm2 ∴ Ration of surface areas = 4π × 6 × 6 / 4π × 12 × 12 = 1 / 4 = 1 : 4

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Description : The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Last Answer : NEED ANSWER

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Description : The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Last Answer : Find the ratios of the surface areas of the balloon in the two cases.

The radius of a spherical balloon increases -Maths 9th

Description : The radius of a spherical balloon increases -Maths 9th

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Description : 1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? -Maths 9th

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Important question on Number game -Maths 9th

Description : Important question on Number game -Maths 9th

Last Answer : Let x2 + 19x + 92 = k2 Multiplying both sides by 4 we get, Factors of 7 are 7 and 1 as it is a prime number. 2k-m = 7 and 2k+m = 1 ------------ (1) 2k-2 = 7 and 2k+m = 7 ------ ... (2), k = 2 Hence the value of k = 2 Therefore, there are 2 integers for which x2 + 19x + 92 is a perfect square.

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!!!

Express the following number in standard form 4730000000 -Maths 9th

Description : Express the following number in standard form 4730000000 -Maths 9th

Last Answer : The standard form of 4,730,000,000 is 4.73 * 10^9 Explanation :- = 4730000000. * 10^0 = 4.73 * 10^9

If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Description : If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Last Answer : We know that product of two rational number is always a rational number. Hence if a is a rational number then a2 = a x a is a rational number, a3 = 4:2 x a is a rational number. ∴ an = an-1 x a is a rational number.

Insert a rational and an irrational number between 2 and 3. -Maths 9th

Description : Insert a rational and an irrational number between 2 and 3. -Maths 9th

Last Answer : A rational number between 2 and 3 = 2 + 3 / 2 = 2.5 An irrational number between 2 and 3 is √5 .

Give an example to show that the product of a rational number and an irrational number may be a rational number . -Maths 9th

Description : Give an example to show that the product of a rational number and an irrational number may be a rational number . -Maths 9th

Last Answer : A rational number 0 multiplied by an irrational number gives the irrational number 0.

Give two examples to show that the product of two irrational numbers may be a rational number . -Maths 9th

Description : Give two examples to show that the product of two irrational numbers may be a rational number . -Maths 9th

Last Answer : Take a = (2+ √3) and b =(2 - √3 ); a and b are irrational numbers, but their product = 4-3 = 1, is a rational number. Take c = √3 and d = -√3; c and d are irrational numbers. but their product = -3, is a rational number.

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Description : Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Last Answer : (i) Frequency distribution table (ii) From the above frequency distribution table, we observe that number of children in the class - interval 15 - 20 is 2. So, 2 children view television for 15 hours or more than 15 hours a week .

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 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

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

Insert a rational number and an irrational number between the following : -Maths 9th

Description : Insert a rational number and an irrational number between the following : -Maths 9th

Last Answer : We know that, there are infinitely many rational and irrational values between any two numbers. (i) A rational number between 2 and 3 is 2.1. To find an irrational number between 2 and 3. Find a ... and 6.375738 is 6.3753. An irrational number between 6.375289 and 6.375738 is 6.375414114111

If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation. -Maths 9th

Description : If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation. -Maths 9th

Last Answer : (b) By property, if we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation remains the same i.e., the solution of the linear equation is remains unchanged.

Important question on Number game -Maths 9th

Description : Important question on Number game -Maths 9th

Last Answer : Let x2 + 19x + 92 = k2 Multiplying both sides by 4 we get, Factors of 7 are 7 and 1 as it is a prime number. 2k-m = 7 and 2k+m = 1 ------------ (1) 2k-2 = 7 and 2k+m = 7 ------ ... (2), k = 2 Hence the value of k = 2 Therefore, there are 2 integers for which x2 + 19x + 92 is a perfect square.

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!!!

Express the following number in standard form 4730000000 -Maths 9th

Description : Express the following number in standard form 4730000000 -Maths 9th

Last Answer : The standard form of 4,730,000,000 is 4.73 * 10^9 Explanation :- = 4730000000. * 10^0 = 4.73 * 10^9

If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Description : If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Last Answer : We know that product of two rational number is always a rational number. Hence if a is a rational number then a2 = a x a is a rational number, a3 = 4:2 x a is a rational number. ∴ an = an-1 x a is a rational number.

Insert a rational and an irrational number between 2 and 3. -Maths 9th

Description : Insert a rational and an irrational number between 2 and 3. -Maths 9th

Last Answer : A rational number between 2 and 3 = 2 + 3 / 2 = 2.5 An irrational number between 2 and 3 is √5 .

Give an example to show that the product of a rational number and an irrational number may be a rational number . -Maths 9th

Description : Give an example to show that the product of a rational number and an irrational number may be a rational number . -Maths 9th

Last Answer : A rational number 0 multiplied by an irrational number gives the irrational number 0.

Give two examples to show that the product of two irrational numbers may be a rational number . -Maths 9th

Description : Give two examples to show that the product of two irrational numbers may be a rational number . -Maths 9th

Last Answer : Take a = (2+ √3) and b =(2 - √3 ); a and b are irrational numbers, but their product = 4-3 = 1, is a rational number. Take c = √3 and d = -√3; c and d are irrational numbers. but their product = -3, is a rational number.

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Description : Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Last Answer : (i) Frequency distribution table (ii) From the above frequency distribution table, we observe that number of children in the class - interval 15 - 20 is 2. So, 2 children view television for 15 hours or more than 15 hours a week .

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 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