The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th
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Answer :

Since the class marks are equally spaced. ∴ Class size = 114 – 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / 2 respectively. ∴ we have h = 10 ∴ Lower limit of first class interval = 104 - 10 / 2 = 99 Upper limit of first class interval = 104 + 10 / 2 = 109 ∴ First class interval is 99 - 109 Hence, the class intervals are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.
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Related questions

The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

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Description : What is the median of 112 135 124 115 119 90 112 134 113 114 115.?

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Description : Following is the frequency distribution of total marks obtained by the students of different section of class-IX. -Maths 9th

Last Answer : Since class intervals of the given frequency distribution are not of equal width. We would make modifications in the lengths of the rectangles in the histogram, so that the areas of rectangles are proportional ... lengths as given in the last column .The histogram of the data is given below.

Following is the frequency distribution of total marks obtained by the students of different section of class-IX. -Maths 9th

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Last Answer : Since class intervals of the given frequency distribution are not of equal width. We would make modifications in the lengths of the rectangles in the histogram, so that the areas of rectangles are proportional ... lengths as given in the last column .The histogram of the data is given below.

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Last Answer : NEED ANSWER

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Last Answer : NEED ANSWER

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Last Answer : NEED ANSWER

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Last Answer : NEED ANSWER

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Description : The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

Last Answer : (c) Let x and y be the upper and lower class limit of frequency distribution. Given, width of the class = 5 ⇒ x-y= 5 (i) Also, given lower class (y) = 10 On putting y = 10 in Eq. (i), we get ... 20-25, 25-30 and 30-35. Thus, the highest class is 30-35, Hence, the upper limit of this class is 35.

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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

Last Answer : (b) We arrange the given data into groups like 13-22,23-32 103-112. (since, our data is from 14 to 112). The class width in this case is 9. Now, the given data can be arranged in tabular form as follows. Hence, the number of classes in distribution will be 10.

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Description : 187.5, 182, 171, ?, 132.5, 105, 72 a) 145.5 b) 155.5 c) 154.5 d) 144.5 e) None of these

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

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 .

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

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 .

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 cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.

The curved surface area of a cylinder is 154 cm2. -Maths 9th

Description : The curved surface area of a cylinder is 154 cm2. -Maths 9th

Last Answer : Since curved surface area of a cyclinder = 154 cm2 [given] Total surface area of cyclinder = 3 curved surface area 2πrh + 2πr2 = 3 154 154 + 2πr2 = 462 2πr2 = 462 - 154 = 308 r2 = 308 7 / 2 22 = 49 ... 154 / 44 = 3.5 cm ∴ Volume of cyclinder = πr2h = 22 / 7 7 7 3.5 = 539 cm3

A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.

The curved surface area of a cylinder is 154 cm2. -Maths 9th

Description : The curved surface area of a cylinder is 154 cm2. -Maths 9th

Last Answer : Since curved surface area of a cyclinder = 154 cm2 [given] Total surface area of cyclinder = 3 curved surface area 2πrh + 2πr2 = 3 154 154 + 2πr2 = 462 2πr2 = 462 - 154 = 308 r2 = 308 7 / 2 22 = 49 ... 154 / 44 = 3.5 cm ∴ Volume of cyclinder = πr2h = 22 / 7 7 7 3.5 = 539 cm3

Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th

Description : Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th

Last Answer : Let 'r' be the radius of sphere Surface area of sphere = 4 πr2 ⇒ 154 = 4 πr2 ⇒ 154 = 4 x 22/7 x r2 ⇒ r 2 = 154 x 7/4 x 22 = 49/4 ⇒ r = 7/2 cm = 3.5 cm

Find the volume of a sphere whose surface area is 154 cm sq. -Maths 9th

Description : Find the volume of a sphere whose surface area is 154 cm sq. -Maths 9th

Last Answer : Let r cm be the radius of sphere. Surface area of the sphere = 4 πr2 ⇒ 154 = 4 πr2 ⇒ 4 x 22/7 x r2 = 154 r 2 = 154 x 7/4 x 22 = 72/22 ⇒ r = 7/2 Volume of sphere = 4/3 πr3 = 4/3 x 22/7 x 7/2 x 7/2 x 7/2 cm3 = 539/3 cm3 = 179.2/3 cm3

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:

In an examination, Salim secured `52%` marks and got 114 marks more than the pass marks and A zam secured `39%` marks and got 36 marks more than the p

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

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

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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

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The frequency distribution has been represented graphically as follows. -Maths 9th

Description : The frequency distribution has been represented graphically as follows. -Maths 9th

Last Answer : NEED ANSWER

To draw a histogram to represent the following frequency distribution. -Maths 9th

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