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

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

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

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.

Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Description : Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Last Answer : Arranging the data in ascending order, we have 25, 31, 32, 37, 39, 42, 43, 45, 46 Here, number of observations = 9 (odd) Median = value of (9+1 / 2)th Observation = Value of 5th Observation = 39

Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Description : Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Last Answer : Arranging the data in ascending order, we have 25, 31, 32, 37, 39, 42, 43, 45, 46 Here, number of observations = 9 (odd) Median = value of (9+1 / 2)th Observation = Value of 5th Observation = 39

Nesha scored 120 out of 150 in English, 120 out of 180 in mathematics and 160 out of 200 in Science. Find Nesha’s score as percentage: (i) in Mathematics (ii) in all the three subjects (on the whole). a) 66 2/3, 78 22/51 b) 71 23/5, 45 7/31 c) 54 13/33, 82 3/4 d) 60 3/4, 76 32/41

Description : Nesha scored 120 out of 150 in English, 120 out of 180 in mathematics and 160 out of 200 in Science. Find Nesha’s score as percentage: (i) in Mathematics (ii) in all the three subjects (on the whole). a) 66 2/3, 78 22/51 b) 71 23/5, 45 7/31 c) 54 13/33, 82 3/4 d) 60 3/4, 76 32/41 

Last Answer :  Answer: A (i) Percentage scored in Mathematics = 120/180 100  = 12000/180  =1200/18  = 66 2 /3 % (ii) Total maximum of all the three subjects = 150 + 160 + 200 = 510 and Total score in the ... percentage on the whole = (400/510 100)  = (40000/510)  = 4000/51  = 78 22/51 %

What is the mean median and mode of 52 53 54 55 55 56 57 57 57 59 60?

Description : What is the mean median and mode of 52 53 54 55 55 56 57 57 57 59 60?

Last Answer : Find the mode of the following data: 55,55, 56, 52, 57, 55, 58, 59, 60, 60

The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Description : The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Last Answer : NEED ANSWER

The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Description : The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Last Answer : It is not correct. Because the difference between two consecutive class marks should be equal to the class size. Here, difference between two consecutive marks is 0.1 and class size of 1.55-1.73 is 0.18, which are not equal.

What is the bond length of a carbon-carbon double bond? (a) 1.20 Å (b) 1.34 Å (c) 1.54 Å (d) 1.68 Å

Description : What is the bond length of a carbon-carbon double bond? (a) 1.20 Å (b) 1.34 Å (c) 1.54 Å (d) 1.68 Å

Last Answer : 1.34 Å

34.8% x 950 – 33.2% of 850 = ? a) 58.4 b) 48.4 c) 56.4 d) 68.9 e) 76.3

Description : 34.8% x 950 – 33.2% of 850 = ? a) 58.4 b) 48.4 c) 56.4 d) 68.9 e) 76.3

Last Answer : 34.8% x 950 – 33.2% of 850 = ? Sol: 0.348 x 950 – 0.332 x 850=? 330.6 – 282.2=? ?= 48.4 Answer: b)

Insert the missing number: 36/62, 39/63, 43/61, 48/64, ? (A) 51/65 (B) 56/60 (C) 54/60 (D) 33/60

Description : Insert the missing number: 36/62, 39/63, 43/61, 48/64, ? (A) 51/65 (B) 56/60 (C) 54/60 (D) 33/60

Last Answer : Answer: C Observer 36/62, 39/63, 43/61, 48/64, ? series Upper part is increase by 3, 4, 5, 6 i.e 36+3= 39, 39+4 = 43, 43+5 = 48, 48 + 6 = 54.

What is the mean median mode and range of 78 25 57 98 20 48 45 37 88 24 19?

Description : What is the mean median mode and range of 78 25 57 98 20 48 45 37 88 24 19?

Last Answer : 19, 20, 24, 25, 37, 45, 48, 57, 78, 88, 98Mean: 49Median: 45Range: 79There is no mode.

A boat covers 14 kms in upstream and 20 kms downstream in 7 hours. Also it covers 22 kms upstream and 34 kms downstream in 10 hours. Find the speed of the boat in still water and of that the stream. -Maths 9th

Description : A boat covers 14 kms in upstream and 20 kms downstream in 7 hours. Also it covers 22 kms upstream and 34 kms downstream in 10 hours. Find the speed of the boat in still water and of that the stream. -Maths 9th

Last Answer : Given, The boat covers 14 km upstream and 20km downstream . at time 7 hours also cover 22km ups. and 34km dwn in10 hours total speed = total distance/total time :.total distance = 14+20+22+34=90km and total time=7+10= ... =90/17 => 5.294km/h => 5.294km/h the speed of boat in still water is 5.29 km/h

Find the orbital dipole moment in a field of dipoles of radius 20cm and angular velocity of 2m/s(in 10 -22 order) a) 64 b) 76 c) 54 d) 78

Description : Find the orbital dipole moment in a field of dipoles of radius 20cm and angular velocity of 2m/s(in 10 -22 order) a) 64 b) 76 c) 54 d) 78

Last Answer : a) 64

The refractive index of a medium with permittivity of 2 and permeability of 3 is given by a) 3.56 b) 2.45 c) 3.21 d) 1.78

Description : The refractive index of a medium with permittivity of 2 and permeability of 3 is given by a) 3.56 b) 2.45 c) 3.21 d) 1.78

Last Answer : b) 2.45

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

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

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

10 years ago the average age of all the 25 teachers of a college was 45 years. 4 years ago, a teacher retired from the post at the age of 60 years, and was replaced by a new teacher whose age was 54 years. What is the present average age of all the teachers? a) Between 52 and 53 b) Between 53 and 54 c) Between 54 and 55 d) Between 55 and 56 e) None of the above

Description : 10 years ago the average age of all the 25 teachers of a college was 45 years. 4 years ago, a teacher retired from the post at the age of 60 years, and was replaced by a new teacher whose age was 54 years. ... 53 b) Between 53 and 54 c) Between 54 and 55 d) Between 55 and 56 e) None of the above

Last Answer : Let current year be 2015 In 2005, sum of all ages = 25 45 = 1125 In 2015, sum of all ages = 1125 + (25 10) = 1375 Now 60 year old teacher was replaced by a 54 year old teacher. So we need ... all ages in 2015 considering the replacement teacher = 1375 - 6 = 1369 Average = 1369/25 = 54.76 Answer: c)

There are three different categories of jobs A, B and C. The average salary of the students who got the job of A and B categories is Rs.32 lakh per annum. The average salary of the student who got the job of B and C category is 54 lakh per annum. And the average salary of those students who got the job of A and C is Rs.44 lakh per annum. The most appropriate (or closet) range of average salary of all the three categories (if it is known that each student gets only one category for jobs i.e, A, B and C): A) lies between 30 & 44 B) lies between 48 & 56 C) lies between 38 & 45 D) lies between 49 & 50

Description : There are three different categories of jobs A, B and C. The average salary of the students who got the job of A and B categories is Rs.32 lakh per annum. The average salary of the student who got the job of B ... 30 & 44 B) lies between 48 & 56 C) lies between 38 & 45 D) lies between 49 & 50

Last Answer : Answer: C) Let the number of students who got the jobs of A, B and C categories is a, b and c respectively, ∴Total salary of Aand B = 32 (a +b) ∴Total salary of B and C = 54 (b+ ... the minimum salary must be Rs.38 lakh and the maximum salary cannot exceed 45, which is the highest of the three..

The critical angle for two media with permittivities of 16 and 9 respectively is a) 48.59 b) 54.34 c) 60 d) 45

Description : The critical angle for two media with permittivities of 16 and 9 respectively is a) 48.59 b) 54.34 c) 60 d) 45

Last Answer : a) 48.59

Consider a disk queue with I/O requests on the following cylinders in their arriving order: 6,10,12,54,97,73,128,15,44,110,34,45 The disk head is assumed to be at cylinder 23 and moving in the direction of decreasing number of cylinders. Total number of cylinders in the disk is 150. The disk head movement using SCAN-scheduling algorithm is: (1) 172 (2) 173 (3) 227 (4) 228

Description : Consider a disk queue with I/O requests on the following cylinders in their arriving order: 6,10,12,54,97,73,128,15,44,110,34,45 The disk head is assumed to be at cylinder 23 and moving in the direction ... . The disk head movement using SCAN-scheduling algorithm is: (1) 172 (2) 173 (3) 227 (4) 228

Last Answer : (2) 173

Which of the following couldnot be an Ethernet multicast destination? A) B7:7B:6C:DE:10:00 B) 7B:AA:C1:23:45:32 C) 7C:56:21:1A:DE:F4 D) 83:32:21:21:4D:34

Description : Which of the following couldnot be an Ethernet multicast destination? A) B7:7B:6C:DE:10:00 B) 7B:AA:C1:23:45:32 C) 7C:56:21:1A:DE:F4 D) 83:32:21:21:4D:34

Last Answer : 7C:56:21:1A:DE:F4

Which of the following couldnot be an Ethernet unicast destination? A) 43:7B:6C:DE:10:00 B) 44:AA:C1:23:45:32 C) 46:56:21:1A:DE:F4 D) 48:32:21:21:4D:34

Description : Which of the following couldnot be an Ethernet unicast destination? A) 43:7B:6C:DE:10:00 B) 44:AA:C1:23:45:32 C) 46:56:21:1A:DE:F4 D) 48:32:21:21:4D:34

Last Answer : 43:7B:6C:DE:10:0

Make a stem-and-leaf plot for this set of data. Find the range,median,and mode of the data. : 47 54 70 62 59 62 82 66 62 73 55 82 59?

Description : Make a stem-and-leaf plot for this set of data. Find the range,median,and mode of the data. : 47 54 70 62 59 62 82 66 62 73 55 82 59?

Last Answer : nevermind- i figured it out

Find the median of the data 35, 28, 49, 56, 79, 81, 73 and 80.

Description : Find the median of the data 35, 28, 49, 56, 79, 81, 73 and 80.

Last Answer : Find the median of the data 35, 28, 49, 56, 79, 81, 73 and 80. A. 64.5 B. 65 C. 65.5 D. 66

Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Description : Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Last Answer : Here's ur answer Hope it has helped u..

The median of a triangle divides it into two -Maths 9th

Description : The median of a triangle divides it into two -Maths 9th

Last Answer : (a) We know that, a median of a triangle is a line segment joining a vertex to the mid-point of the opposite side. Thus, a median of a triangle divides it into two triangles of equal area.

The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Description : The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Last Answer : According to question the area of ΔGBC = area of the quadrilateral AFGE.

Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Description : Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Last Answer : Here's ur answer Hope it has helped u..

The median of a triangle divides it into two -Maths 9th

Description : The median of a triangle divides it into two -Maths 9th

Last Answer : (a) We know that, a median of a triangle is a line segment joining a vertex to the mid-point of the opposite side. Thus, a median of a triangle divides it into two triangles of equal area.

The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Description : The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Last Answer : According to question the area of ΔGBC = area of the quadrilateral AFGE.

A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Description : A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Last Answer : NEED ANSWER

A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Description : A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Last Answer : The child does not understand, that data has to be arranged in ascending or descending order before finding the median.

Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side -Maths 9th

Description : Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side -Maths 9th

Last Answer : Solution :-

Prove that median of a triangle divides it into two triangles of equal area. -Maths 9th

Description : Prove that median of a triangle divides it into two triangles of equal area. -Maths 9th

Last Answer : Solution :-

A child says that the median of 3, 14, 18, 20, 5 is 18. -Maths 9th

Description : A child says that the median of 3, 14, 18, 20, 5 is 18. -Maths 9th

Last Answer : In order to find the median, data has to be arranged in ascending or descending order before finding the median.

Find the median of the numbers: 4, 4, 5, 7, 6, 7, 7, 12, 3. -Maths 9th

Description : Find the median of the numbers: 4, 4, 5, 7, 6, 7, 7, 12, 3. -Maths 9th

Last Answer : Arranging the data in ascending order, we get 3, 4, 4, 5, 6, 7, 7, 7, 12 Here, n = 9 Median = (9 +1)th/2 observation = 5th observation = 6.

If AD is the median of DABC, then -Maths 9th

Description : If AD is the median of DABC, then -Maths 9th

Last Answer : answer:

The median of a trapezoid cuts the trapezoid into two regions whose areas are in the ratio 1 : 2. Compute the ratio of the smaller base of the -Maths 9th

Description : The median of a trapezoid cuts the trapezoid into two regions whose areas are in the ratio 1 : 2. Compute the ratio of the smaller base of the -Maths 9th

Last Answer : answer:

Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). -Maths 9th

Description : Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). -Maths 9th

Last Answer : (b) a = √2b Let D be the mid-point of BC. Then D ≡ \(\bigg(rac{a+0}{2},rac{0}{2}\bigg)\)i.e. \(\bigg(rac{a}{2},0\bigg)\)Let E be the mid-point of AC, thenE = \(\bigg(rac{a+0}{2},rac{0+b}{2}\bigg)\) = \(\bigg ... \(rac{b}{a}.\)∴ From (i), \(rac{-2b}{a}\) x \(rac{b}{a}\) = -1⇒ 2b2 = a2 ⇒ a = √2 .

If A (-2, 4), B (0, 0) and C (4, 2) are the vertices of triangle ABC, then find the length of the median through the vertex A. -Maths 9th

Description : If A (-2, 4), B (0, 0) and C (4, 2) are the vertices of triangle ABC, then find the length of the median through the vertex A. -Maths 9th

Last Answer : D=slid ht of BC D≅(20+4​,20+2​) =(2,1) ∴ Length of median = Light of AD =root(−2−2)2+(4−1)2​=root42+32​=5 hope it helps thank u

What is the mode of the following data values 54 78 26 36 26?

Description : What is the mode of the following data values 54 78 26 36 26?

Last Answer : Feel Free to Answer

What is the median of 84 86 87 84 86 89 90 87 87 84 86.?

Description : What is the median of 84 86 87 84 86 89 90 87 87 84 86.?

Last Answer : Need answer

The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, -Maths 9th

Description : The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, -Maths 9th

Last Answer : (c) 67.5 mGiven 2s = a + b + c ⇒ 240 = 78 + 50 + Third side ⇒ Third side = 240 m - 128 m = 112 m.∴ Area of Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\)= \(\sqrt{120(120-78)(120-50)(120-112)}\)= \(\sqrt{120 imes42 imes70 ... (rac{1}{2}\)x b x h ∴ \(rac{1}{2}\) x 50 x h = 1680 ⇒ h = \(rac{1680}{25}\) = 67.5 m.

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

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