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

1 Answer

Answer :

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

Related questions

Description : Now-a-days yellow lamps are frequently used as street lights. Which of the following gases is used in these lamps? (1) Sodium (2) Neon (3) Hydrogen (4) Nitrogen

Last Answer : (1) Sodium Explanation: A sodium-vapor lamp is a gasdischarge lamp that uses sodium in an excited state to produce light. There are two varieties of such lamps: low pressure and high. ... High-pressure sodium lamps have a broader spectrum of light but poorer colour rendering than other types.

Description : What are the materials used in Neon sign tube lamps for different colors?

Last Answer : Following are the materials used in neon sign tubes for different colors. For, Red - Neon gas. Reddish orange - Neon gas + Argon gas. Blue - Vapour of mercury. Golden - Neon gas + ... tube. By depositing fluorescent powder on the inner surface of the tube varying colors in intensity can be made.

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Last Answer : (B) Used in color discharge tube

Description : Now-a-days yellow lamps are frequently used as street lights. Which of the following gases is used in these lamps ? (1) Sodium (2) Neon (3) Hydrogen (4) Nitrogen

Last Answer : Sodium

Description :  In HPSV lamps, traces of neon gas   (a) Helps in stabilizing the arc (b) Changes the colour of light (c) Prevents the vaporization of filament (d) Assists during starting to develop enough heat to vaporize sodium

Last Answer :  In HPSV lamps, traces of neon gas Assists during starting to develop enough heat to vaporize sodium

Description : The following table gives the pocket money (in Rs) given to children per day by their parents : Represent the data in the form of a histogram. -Maths 9th

Last Answer : The required histogram is as below :

Description : Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them. -Maths 9th

Last Answer : Clearly, the mean score of two sections A and B is same

Description : The following table gives the pocket money (in Rs) given to children per day by their parents : Represent the data in the form of a histogram. -Maths 9th

Last Answer : The required histogram is as below :

Description : Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them. -Maths 9th

Last Answer : Clearly, the mean score of two sections A and B is same

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

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

Description : A neon flashlight cell with an emf of 1.5V gives a current of 15A when connected directly to an ammeter of resistance 0.04?. Internal resistance of the cell is a. 0.0004? b. 0.06? c. 0.10? d. 0.13?

Last Answer : b. 0.06?

Description : In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Last Answer : Frequency distribution table

Description : In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Last Answer : Here, class size = 135 - 305 = 10 ∴ Lower limit of first class interval is 305 - 10 / 2 = 300 Upper limit of first class interval is 305 + 10 / 2 = 310 Thus, first class interval is 300 - 310 Required histogram and frequency polygon is given on the graph paper .

Description : The table shows the marks obtained by a student in unit tests out of 50 : -Maths 9th

Last Answer : Here the marks are out of 50 , so we find its percentage (i.e. out of 100)

Description : Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table : -Maths 9th

Last Answer : Total number of books = 700 (i) P(no defective books) = 400 / 700 = 4 / 7 (ii) P(more than 0 but less than 4 defective books) = 269 / 700 (iii) P(more than 4 defective books) = 13 / 700

Description : The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Last Answer : (a) P (later half of the year) = 23 / 40 (b) P (month having 31 days) = 26 / 40 = 13 / 20 (c) P(month having 30 days) = 10 / 40 = 1 / 4

Description : Plot the points (x, y) given by the following table. -Maths 9th

Last Answer : On plotting the given points on the graph, we get the points P(2,4), Q(4,2) R (-3, 0), S (-2, 5), T (3, – 3)and O (0 0).

Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Last Answer : Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

Description : In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Last Answer : Frequency distribution table

Description : In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Last Answer : Here, class size = 135 - 305 = 10 ∴ Lower limit of first class interval is 305 - 10 / 2 = 300 Upper limit of first class interval is 305 + 10 / 2 = 310 Thus, first class interval is 300 - 310 Required histogram and frequency polygon is given on the graph paper .

Description : The table shows the marks obtained by a student in unit tests out of 50 : -Maths 9th

Last Answer : Here the marks are out of 50 , so we find its percentage (i.e. out of 100)

Description : Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table : -Maths 9th

Last Answer : Total number of books = 700 (i) P(no defective books) = 400 / 700 = 4 / 7 (ii) P(more than 0 but less than 4 defective books) = 269 / 700 (iii) P(more than 4 defective books) = 13 / 700

Description : The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Last Answer : (a) P (later half of the year) = 23 / 40 (b) P (month having 31 days) = 26 / 40 = 13 / 20 (c) P(month having 30 days) = 10 / 40 = 1 / 4

Description : Plot the points (x, y) given by the following table. -Maths 9th

Last Answer : On plotting the given points on the graph, we get the points P(2,4), Q(4,2) R (-3, 0), S (-2, 5), T (3, – 3)and O (0 0).

Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Last Answer : Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

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Last Answer : NEED ANSWER

Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Last Answer : NEED ANSWER

Description : A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data -Maths 9th

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Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

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Description : The cost of a table is 100 more than half the cost of the chair write the statement linear equation in two varible -Maths 9th

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

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Last Answer : Total number of ways in which 10 person can sit around a circular table = 9! (∵ We shall keep one place fixed and the rest of the 9 places will be filled in (9 8 7 6 5 4 3 2 1) ways asthere is ... probability = \(rac{2 imes8!}{9!}\) = \(rac{2 imes8!}{9 imes8!}\) = \(rac{2}{9}.\)

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Last Answer : ∴ Required number of ways of seating = 9!

Description : In how many ways can 7 men and 7 women sit on a round table such that no two women sit together? -Maths 9th

Last Answer : Now there are 7 places vacant between these 7 men. ∴ 7 women can seat themselves in these 7 places in 7 ! ways. ∴ Total number of required arrangement where no two women sit together = 6!

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Last Answer : If each gentleman has to have a lady by his side, the seating arrangement can be done as shown below: This can be done in 5! (Gentlemen) × 3! (Ladies) = 720 ways.

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Description : Eighteen quests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and -Maths 9th

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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

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Last Answer : Helium

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Last Answer : A. Mercury and Bromine (Answer)

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

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

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

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

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

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

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

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