The following data on the number of girls to the nearest -Maths 9th

The following data on the number of girls to the nearest -Maths 9th
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(i) (ii) The gender equity exists most in scheduled tribe and least in urban areas. Yes, gender equity leads to economic growth which, in turn, helps in the development of a  society.
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Related questions

The following data on the number of girls to -Maths 9th

Description : The following data on the number of girls to -Maths 9th

Last Answer : Solution :- (i) (ii) Number of girls to the nearest ten per thousand boys are maximum in scheduled tribes whereas they are minimun in urban areas.

The ratio of girls and boys in a class is 1: 3. Set up an equation between the students of a class and boys and then draw its graph. Also find the number of boys in a class of 40 students from the graph. -Maths 9th

Description : The ratio of girls and boys in a class is 1: 3. Set up an equation between the students of a class and boys and then draw its graph. Also find the number of boys in a class of 40 students from the graph. -Maths 9th

Last Answer : Total number of boys and girl = 40, Ratio = 1 : 3 Number of girls be A and Number of boys be B. Ratio of number of girls and boys is 1 : 3, so Therefore 3A=B To find number of boys we ... the number 30 represents the number of girls. 40 as total on the line A = 10, which is the common equation.

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5 m drawn in a park. -Maths 9th

Description : Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5 m drawn in a park. -Maths 9th

Last Answer : Solution :- Let R, S and M represent the position of Reshma, Salma and Mandeep respectively. Clearly △RSM is an isosceles triangle as RS = SM = 6m Join OS which intersect RM at A. In △ROS and △MOS OR = OM ( ... . ∴ RM = 2RA RM = 2 x 4.8 = 9.6m Hence, distance between Reshma and Mandeep is 9.6m.

A class consists of 50 students out of which 30 are girls. -Maths 9th

Description : A class consists of 50 students out of which 30 are girls. -Maths 9th

Last Answer : Mean marks scored by girls ( x̅1 ) = 73 Number of girls (n1) = 30 Mean marks scored by boys ( x̅2 ) = 71 Number of boys (n2) = 50 - 30 = 20 Mean score of the whole class ( x̅12 ) = n1 x̅1 + n2 x̅2 /n1 + n2 = 30 x 73 + 20 x 71/30 + 20 = 2190 + 1420/50 = 3610/50 x̅ 2 = 72.2

A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. -Maths 9th

Description : A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. -Maths 9th

Last Answer : Let P (A) = Probability of selecting a fair complexioned person. ThenP(A) = \(rac{20}{80}\) = \(rac{1}{4}\)Let P(B) = Probability of selecting a rich person. Then P(B) = \(rac{10}{80}\) = \(rac{1}{8}\)Let P (C) = ... ) = \(rac{1}{4}\)x \(rac{1}{8}\)x \(rac{5}{16}\) = \(rac{5}{512}\) = 0.009.

Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ? -Maths 9th

Description : Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ? -Maths 9th

Last Answer : Total number of ways of arranging 4 boys and 3 girls, i.e., 7 people in a queue (row) = n(S) = 7! Let A : Event in which the 4 boys and 3 girls occupy alternate position. This is possible when the ... {4 imes3 imes2 imes1 imes3 imes2 imes1}{7 imes6 imes5 imes4 imes3 imes2 imes1}\) = \(rac{1}{35}.\)

6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is -Maths 9th

Description : 6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is -Maths 9th

Last Answer : (b) \(rac{1}{462}\)Let S be the sample space. Then, n(S) = Number of ways in which 6 boys and 6 girls can sit in a row = 12! Let E : Event of 6 girls and 6 boys sitting alternately. Then, the ... )= \(rac{2 imes6 imes5 imes4 imes3 imes2 imes1}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{462}\).

6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th

Description : 6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th

Last Answer : (d) \(rac{1}{132}\)Let S be the sample space for arranging 6 boys and 6 girls in a row. Then, n(S) = 12! If all 6 boys are to sit together, then consider the 6 boys as one entity. Now the ... ) = \(rac{7 imes6 imes5 imes4 imes3 imes2}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{132}\).

A group of 2n boys and 2n girls is divided at random into two equal batches. -Maths 9th

Description : A group of 2n boys and 2n girls is divided at random into two equal batches. -Maths 9th

Last Answer : (c) \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)Total number of boys and girls = 2n + 2n = 4n Since, there are two equal batches, each batch has 2n members ∴ Let S (Sample space) : Selecting one batch out of 2 ⇒ S : ... )2∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)

There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together? -Maths 9th

Description : There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together? -Maths 9th

Last Answer : Have the 55 boys stand in a line. This can be done in 5!5! ways. For the moment, add a boy at each end, who will be removed when we're done. Now send the 33 girls, one at a time, to ... that each girl can either wedge herself between two boys or else stand next to a boy at one end or the other.

There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls -Maths 9th

Description : There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls -Maths 9th

Last Answer : Since the chairs are numbered, so they are distinguishable. Therefore 3 boys can be arranged on 3 alternate chairs in 3! ways. 3 girls can be arrenged in 3! ways Also, the girls can be seated before the boys. Total number of required ways = 3! × 3! + 3! × 3! = 2 × (3!)2

The heights of 50 students, measured to the nearest centimetre, -Maths 9th

Description : The heights of 50 students, measured to the nearest centimetre, -Maths 9th

Last Answer : Frequency distribution of above data in tabular form is given as: (ii) One conclusion we can draw from the above table is that more than 50% of students are shorter than 165 cm.

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

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

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 :

The following data given the weight (in grams) of 30 oranges picked from a basket: -Maths 9th

Description : The following data given the weight (in grams) of 30 oranges picked from a basket: -Maths 9th

Last Answer : Here , class width = 20 class mark = 70 Half of the class width =20 /2 =10 Upper limit of first class interval = 70 + 10 = 80 Lower limit of first class interval = 70 - 10 = 60 Thus, class interval becomes 60 ... than 180 g = 1 + 1 = 2 (b) Number of oranges weights less than 100 g = 3 + 10 = 13

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

The following data given the weight (in grams) of 30 oranges picked from a basket: -Maths 9th

Description : The following data given the weight (in grams) of 30 oranges picked from a basket: -Maths 9th

Last Answer : Here , class width = 20 class mark = 70 Half of the class width =20 /2 =10 Upper limit of first class interval = 70 + 10 = 80 Lower limit of first class interval = 70 - 10 = 60 Thus, class interval becomes 60 ... than 180 g = 1 + 1 = 2 (b) Number of oranges weights less than 100 g = 3 + 10 = 13

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

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 :

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : In conclusion, the range of data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, and 20 is 26.

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

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

Last Answer : NEED ANSWER

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

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

If each observation of the data is increased by 5, then their mean -Maths 9th

Description : If each observation of the data is increased by 5, then their mean -Maths 9th

Last Answer : NEED ANSWER

The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Description : The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Last Answer : NEED ANSWER

The mode of given data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15,17 and 15 is -Maths 9th

Description : The mode of given data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15,17 and 15 is -Maths 9th

Last Answer : We first arrange the given data in ascending order as follows 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20 From above, we see that 15 occurs most frequently i.e., 5 times. Hence, the mode of the given data is 15.

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : (d) In a given data, maximum value = 32 and minimum value = 6 We know, range of the data = maximum value – minimum value = 32 – 6 = 26 Hence, the range of the given data is 26.

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

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

Last Answer : (c) Here, we arrange the given data into groups like 210-230, 230-250 390-410. (since, our data is from 210 to 406). The class width in this case is 20. Now, the given data can be arrange in tabular form as follows

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

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.

If each observation of the data is increased by 5, then their mean -Maths 9th

Description : If each observation of the data is increased by 5, then their mean -Maths 9th

Last Answer : Simplify the question

The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Description : The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Last Answer : According to question find the median of the data

The mode of given data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15,17 and 15 is -Maths 9th

Description : The mode of given data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15,17 and 15 is -Maths 9th

Last Answer : (b) We first arrange the given data in ascending order as follows 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20 From above, we see that 15 occurs most frequently i.e., 5 times. Hence, the mode of the given data is 15.

Let cost of a pen and a pencil be “x” and “y” respectively. A girl pays ₹16 for 2 pens and 3 pencils. Write the given data in the form of a linear equation in two variables. Also represent it graphically. -Maths 9th

Description : Let cost of a pen and a pencil be “x” and “y” respectively. A girl pays ₹16 for 2 pens and 3 pencils. Write the given data in the form of a linear equation in two variables. Also represent it graphically. -Maths 9th

Last Answer : Solution :-

Half the perimeter of a rectangular garden in 36m. Write a linear equation which satisfies this data. Draw the graph for the same. -Maths 9th

Description : Half the perimeter of a rectangular garden in 36m. Write a linear equation which satisfies this data. Draw the graph for the same. -Maths 9th

Last Answer : Solution :-

The mean of the data: -Maths 9th

Description : The mean of the data: -Maths 9th

Last Answer : It is correct as the 2nd data is obtained by multiplying each observation of data by 2, therefore, the mean will be 2 times the mean of the data, i.e., 10.

If the mode of the data 5, 8, 4,5,5,8, 4, 7, 8, x is 5, then find the value of x. -Maths 9th

Description : If the mode of the data 5, 8, 4,5,5,8, 4, 7, 8, x is 5, then find the value of x. -Maths 9th

Last Answer : The value of x = 5.

If the mean of the following data is 20.2, find the value of p: -Maths 9th

Description : If the mean of the following data is 20.2, find the value of p: -Maths 9th

Last Answer : x̅ = ∑fx/∑f ∴ 20.2 = 610 + 20p/30 + p ⇒ 20.2(30 + p) = 610 + 20p ⇒ 606 + 20.2p = 610 + 20p ⇒ 20.2p - 20p = 610 - 606 ⇒ 0.2p = 4 ⇒ p = 4/0.2 = 40/2 ⇒ p = 20

Prepare a continuous grouped frequency distribution from the following data: -Maths 9th

Description : Prepare a continuous grouped frequency distribution from the following data: -Maths 9th

Last Answer : If m is mid-point of a class and h is the class size, lower and upper limits of the class intervals are m - h/2 and m + h/2 respectively. Class size (h) = 15 - 5 = 10 So, the class interval formed ... 2) - (5 + 10/2) i.e., 0 - 10 Continuing in the same manner, the continuous classes formed are:

What percent of the total number of girls in the college do not know Hindi? (rounded off to nearest integer) (1) 38 (2) 46 (3) 48 (4) 36 (5) 43

Description : What percent of the total number of girls in the college do not know Hindi? (rounded off to nearest integer) (1) 38 (2) 46 (3) 48 (4) 36 (5) 43

Last Answer : (5) 43

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

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

Last Answer : As we can see from the figure, that two circles have two points in common. Two circles cannot intersect each other at more than two points. Let us assume that two circles cut each other at ... circle can pass. So, two circles if intersect each other will intersect at maximum two points.

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.