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

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

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\(rac{1}{2}\) = \(rac{1}{4}\).

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

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

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} = \(rac{84 imes5^6}{6^{10}}\).

Two dice are thrown. Find the probability of getting an odd number on the first die and a multiple of 3 on the other. -Maths 9th

Description : Two dice are thrown. Find the probability of getting an odd number on the first die and a multiple of 3 on the other. -Maths 9th

Last Answer : Let A : Getting an odd number on first die; B : Getting a multiple of 3 on second die, thenA = {1, 3, 5}, B = {3, 6} ∴ P(A) = \(rac{3}{6}=rac{1}{2}\), P(B) = \(rac{2}{6}=rac{1}{3}\) ... B are independent∴ Required probability = P (A) . P (B) = \(rac{1}{2}\) x \(rac{1}{3}\) = \(rac{1}{6}\)

A die is thrown twice. What is the probability that at least one of the two throws comes up with the number 3 ? -Maths 9th

Description : A die is thrown twice. What is the probability that at least one of the two throws comes up with the number 3 ? -Maths 9th

Last Answer : (b) \(rac{11}{36}\)Let S = total ways in which two dice can be rolled ⇒ n(S) = 6 6 = 36 Let A : Event of throwing 3 with 1st dice, B : Event of throwing 3 with 2nd dice. Then, A = {(3, 1), (3, 2), (3, 3), (3, 4), ... ) - P(A ∩ B)= \(rac{6}{36}\) + \(rac{6}{36}\) - \(rac{1}{36}\) = \(rac{11}{36}\).

A die is thrown.What is the probability of getting a multiple of 3 on the upper face ? -Maths 9th

Description : A die is thrown.What is the probability of getting a multiple of 3 on the upper face ? -Maths 9th

Last Answer : Multiple of 3 on a die = 3, 6 ∴ P (a multiple of 3) = 2/6 = 1/3.

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

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...

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 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

When two dice are thrown, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 9. A) 1/2 B) 1/5 C) 2/5 D) 4/2

Description : When two dice are thrown, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 9. A) 1/2 B) 1/5 C) 2/5 D) 4/2 

Last Answer : Answer: A) Let the event of getting a greater number on the first die be G. There are 4 ways to get a sum of 9 when two dice are rolled = {(3,6),(4,5),(5,4), (6,3)}. And there are two ways where the number on the ... Now, P(G) = P(G sum equals 9)/P(sum equals 9) = (2/36)/(4/36) = 2/4 =>1/2

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!!!

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!!!

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

The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Description : The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Last Answer : Median=n=even =n/2=20/2=10th observation =98 Mode =88

The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Description : The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Last Answer : Median=n=even =n/2=20/2=10th observation =98 Mode =88

Find the area of the trapezium PQRS with height PQ given in the figure given below. -Maths 9th

Description : Find the area of the trapezium PQRS with height PQ given in the figure given below. -Maths 9th

Last Answer : The area of trapezium =

Find the area of the trapezium PQRS with height PQ given in the figure given below. -Maths 9th

Description : Find the area of the trapezium PQRS with height PQ given in the figure given below. -Maths 9th

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Consider two postulates given below: -Maths 9th

Description : Consider two postulates given below: -Maths 9th

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Given below are the seats won by different political -Maths 9th

Description : Given below are the seats won by different political -Maths 9th

Last Answer : Solution :- (i) (ii) Party A won the maximum number of seats. i.e., 75.

The marks obtained (out of 100) by a class of 80 students are given below: -Maths 9th

Description : The marks obtained (out of 100) by a class of 80 students are given below: -Maths 9th

Last Answer : In the given frequency distribution, the class intervals are not of equal width. Therefore, we would make modification in the lengths of the rectangle in the histogram so that the areas of rectangle ... draw rectangles with lengths as given in the last column. The histogram of data is given below:

The percentage of marks obtained by a student in monthly unit tests are given below. -Maths 9th

Description : The percentage of marks obtained by a student in monthly unit tests are given below. -Maths 9th

Last Answer : (i) Number of tests in which the student scored more than 70% marks = 3 ∴ P(more than 70% marks) = 3/6 = 1/2 (ii) Number of tests in which the student scored less than 70% marks = 3 ∴ P(less ... ) Number of tests in which the student scored at least 60% marks = 5 ∴ P(at least 60% marks) = 5/6

Find six rational numbers between 3 and 4. -Maths 9th

Description : Find six rational numbers between 3 and 4. -Maths 9th

Last Answer : Solution :-

Find six rational numbers between 5/7 and 6/7. -Maths 9th

Description : Find six rational numbers between 5/7 and 6/7. -Maths 9th

Last Answer : Solution :-

A Police man fires six bullets on a decoit. The probability that the decoit will be killed by one bullet is 0.6. What is the probability that the -Maths 9th

Description : A Police man fires six bullets on a decoit. The probability that the decoit will be killed by one bullet is 0.6. What is the probability that the -Maths 9th

Last Answer : Let Ai be the event that the decoit is killed by the ith bullet (1 ≤ i ≤ 6). Then \(\bar{A}_i\) is the event that the decoit is not killed, ∴ P (Ai) = 0.6 and P(\(\bar{A}_i\)) = 1 − 0.6 = 0.4∴ Probability ... decoit p = 0.6 ⇒ q = 1 − p = 1 − 0.6 = 0.4Required probability = qqqqqq = (q)6 = (0.4)6.

A management institute has six senior professors and four junior professors. Three professors are selected at random -Maths 9th

Description : A management institute has six senior professors and four junior professors. Three professors are selected at random -Maths 9th

Last Answer : (a) \(rac{5}{6}\)P(At least one junior professor is selected) = P(Selecting 1 Junior) P(Selecting 2 Seniors) + P(Selecting 2 Junior) P(Selecting 1 Senior) + P(Selecting all 3 Juniors)∴ Required probability = \(rac{^4C_1 imes^ ... }{30}\) = \(rac{15+9+1}{30}\) = \(rac{25}{30}\) = \(rac{5}{6}\).

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Description : Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Last Answer : (c) \(rac{1}{10}\)Let S be the sample space.Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon= 6C3 = \(rac{6 imes5 imes4}{3 imes2}\) = 20.Let A : Event that the ... Required probability = \(rac{n(A)}{n(S)}\) = \(rac{2}{20}\) = \(rac{1}{10}\).

What could be an actual geopolitical issue for China in a deglobalising post-pandemic world in relation to other countries? Please, see the noted details below.

Description : What could be an actual geopolitical issue for China in a deglobalising post-pandemic world in relation to other countries? Please, see the noted details below.

Last Answer : I can think of several: 1. Raw material supply chain for manufacturing in China 2. Intellectual property, creativity, invention in a post-Covid Asia where travel and visa are restricted. 3. ... to product political change 5. The use of government-sponsored assassination as a tool of diplomacy.

In a throw of a die, find the probability of getting an even number. -Maths 9th

Description : In a throw of a die, find the probability of getting an even number. -Maths 9th

Last Answer : Total even number on a die = 3 P (getting an even numbers) = 3/6 = 1/2

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

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

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

In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Description : In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Last Answer : Required probability = 1 – P(4) – P(5) =1- 1 / 6 - 1 / 6 = 4 / 6 = 2 / 3

In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Description : In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Last Answer : Required probability = 1 – P(4) – P(5) =1- 1 / 6 - 1 / 6 = 4 / 6 = 2 / 3

Two unbiased dice are rolled. Find the probability of getting a multiple of 2 on one die and a multiple of 3 on the other die ? -Maths 9th

Description : Two unbiased dice are rolled. Find the probability of getting a multiple of 2 on one die and a multiple of 3 on the other die ? -Maths 9th

Last Answer : When two unbiased dice are rolled, the possible out comes are∴ n(S) = 36 Let A : getting a multiple of 2 on one die and a multiple of 3 on the other die. ⇒ A = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), ( ... (3, 6), (6, 2), (6, 4)} ⇒ n(A) = 11∴ P(A) = \(rac{n(A)}{n(S)} =rac{11}{36}.\)

One bag contains 3 black and 4 white balls and the other bag contains 4 black and 3 white balls. A die is rolled. -Maths 9th

Description : One bag contains 3 black and 4 white balls and the other bag contains 4 black and 3 white balls. A die is rolled. -Maths 9th

Last Answer : Let A : Getting 2 or 5 B : Getting white ball from first bag C : Getting white ball from second bag.∴ P(A) = \(rac{2}{6}\) = \(rac{1}{3}\) ⇒ P(\(\bar{A}\)) = 1 - P(A) = 1 - \(rac{1}{3}\) = \(rac{2}{3}\)∴ Required ... \(rac{4}{7}\) + \(rac{2}{3}\) x \(rac{3}{7}\) = \(rac{4+6}{21}\) = \(rac{10}{21}.\)

What is the probability of getting a 4 or a 6 when a die is thrown together? a) 2/3 b) 1/3 c) 3/6 d) 4/6

Description : What is the probability of getting a 4 or a 6 when a die is thrown together? a) 2/3 b) 1/3 c) 3/6 d) 4/6

Last Answer : Answer: B) Taking the individual probabilities of each number, getting a 4 is 1/6 and so is getting a 6. Applying the formula of compound probability, Probability of getting a 4 or a 6, P(4 or 6) = P(4) + P(6) – P(4 and 6) ==> 1/6 + 1/6 – 0 2/6 = 1/3

Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

In testing a pile by load test, pile platform is loaded with one and half times the design load and a maximum settlement is noted. The load is gradually removed and the consequent rebound is measured. For a safe pile, the net settlement (i.e. total settlement minus rebound) per tonne of test load should not exceed (A) 10 mm (B) 15 mm (C) 20 mm (D) 25 mm

Description : In testing a pile by load test, pile platform is loaded with one and half times the design load and a maximum settlement is noted. The load is gradually removed and the consequent rebound is measured. For a safe pile, the net ... load should not exceed (A) 10 mm (B) 15 mm (C) 20 mm (D) 25 mm

Last Answer : Answer: Option D

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

Heat load in a cooling tower (A) Means the amount of heat thrown away (KCal/hr.) by the cooling tower (B) Is equal to the number of kg, of water circulated times the cooling range (C) Both (A) & (B) (D) Neither (A) nor (B)

Description : Heat load in a cooling tower (A) Means the amount of heat thrown away (KCal/hr.) by the cooling tower (B) Is equal to the number of kg, of water circulated times the cooling range (C) Both (A) & (B) (D) Neither (A) nor (B)

Last Answer : (C) Both (A) & (B)

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

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

The number of solutions satisfying the given equation -Maths 9th

Description : The number of solutions satisfying the given equation -Maths 9th

Last Answer : (d) 3Taking log of both the sides to base 3, we have,\(\big[(log_3\,x)^2-rac{9}{2}log_3\,x+5\big]\) log3x = log333/2 = \(rac{3}{2}\) (∵ log33 = 1)⇒ 2(log3x)3 - 9(log3x)2 + 10 log3x - 3 = 0 ⇒ ... log3x = 3, 2 log3x = 1 ⇒ x = 31, x = 33, x2 = 31⇒ \(x\) = (3, 27, √3)∴ There are three solutions.