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

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

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

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 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 : 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

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

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 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:

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) = \(rac{n(A)}{n(S)}\) = \(rac{4}{8}\) = \(rac{1}{2}.\)

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.

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 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 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 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.

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) = \(rac{n(A)}{n(S)}\) = \(rac{3}{4}.\)

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 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 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

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

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : NEED ANSWER

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : (c) Class marks i.e., the mid-point of the classes are abscissa of the points, which we plot for frequency polygon.

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:

When a wide band of frequencies is transmitted simultaneously, each frequency will vary in the amount of fading. What is this variable fading called?

Description : When a wide band of frequencies is transmitted simultaneously, each frequency will vary in the amount of fading. What is this variable fading called?

Last Answer : Selective fading.

Three audio waves with 100,200 and 300 volts amplitude respectively, simultaneously modulate a 450 volts carrier. What is the total percent of modulation of the AM wave? A. 69 % B. 115.5 % C. 50% D. 83%

Description : Three audio waves with 100,200 and 300 volts amplitude respectively, simultaneously modulate a 450 volts carrier. What is the total percent of modulation of the AM wave? A. 69 % B. 115.5 % C. 50% D. 83%

Last Answer : D. 83%

Average cost of 10 pens is rupees 200 & average cost of 10 pencils is 100 rupees find the average cost of the all pens and pencils​ -Maths 9th

Description : Average cost of 10 pens is rupees 200 & average cost of 10 pencils is 100 rupees find the average cost of the all pens and pencils​ -Maths 9th

Last Answer : 10 pens of average cost = 200 => 10×200 = 2000 rupees 10 pencils average cost = 100 rupees => 10×100=1000 Pens + Pencils = 20 Total average 2000+1000 = 3000 Rupees 150 is average for average cost of each pen and pencil ✏

The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Description : The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Last Answer : Here's ur answer..

The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Description : The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Last Answer : Here's ur answer..

Over the past 200 working days, -Maths 9th

Description : Over the past 200 working days, -Maths 9th

Last Answer : (i) P (no defective part) = 50/200 = 0.25 (ii) P (at least one defective part) = 1 - P (no defective part) = 1 - 0.25 = 0.75 (iii) P (not more than 5 defective parts) = P (no defective part) + P (1 ... (5 defective parts) = 50/200 + 32/200 + 22/200 + 18/200 + 12/200 + 12/200 = 146/200 = 0.73

The remainder, when x^(200) is divided by x^2 – 3^x + 2 is -Maths 9th

Description : The remainder, when x^(200) is divided by x^2 – 3^x + 2 is -Maths 9th

Last Answer : answer:

Write the following decimals in the place value table. (a) 0.29 (b) 2.08 (c) 19.60 (d) 148.32 (e) 200.812 -Maths 9th

Description : Write the following decimals in the place value table. (a) 0.29 (b) 2.08 (c) 19.60 (d) 148.32 (e) 200.812 -Maths 9th

Last Answer : (a) 0.29 = 0.2 + 0.09 = 2 / 10 + 9 / 100 (b) 2.08 = 2 + 0.08 = 2 + 8 / 100 (c) 19.60 = 19 + 0.60 = 10 + 9 + 6 / 10 (d) 148.32 = 148 + 0.3 + 0. ... / 100 (e) 200.812 = 200 + 0.8 + 0.01 + 0.002 =200 + 8 / 10 + 1 / 100 + 2 / 1000HundredsTensOnesTenthsHundredthsThousandths000290002080019600148320200812

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: