The class mark of the class 90-120 is -Maths 9th

The class mark of the class 90-120 is -Maths 9th
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The class mark of the class 90-120 is -Maths 9th

Description : The class mark of the class 90-120 is -Maths 9th

Last Answer : (b) In a given class 90-120, upper class = 120 and lower class = 90

Find the class mark of the class 100-120. -Maths 9th

Description : Find the class mark of the class 100-120. -Maths 9th

Last Answer : Class mark = (Lower limit + Upper limit)/2 = 100 + 120/2 = 110

The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Description : The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Last Answer : NEED ANSWER

The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Description : The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Last Answer : (b) Since, the difference between mid value is 5. So, the corresponding class to the class mark 20 must have difference 5. Therefore, option (c) and (d) are wrong. Since, the mid value is 20 which can get only, if we take option (b)

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2. -Maths 9th

Description : The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2. -Maths 9th

Last Answer : The roller is in the form of a cylinder of diameter = 84 cm ⇒ Radius of the roller(r) = 842 cm = 42 cm Length of the roller (h) = 120 cm Curved surface area of the ... roller = 31680 cm2 = 3168010000m2 ∴ Area of the playground levelled in 500 revolutions = 500 x 3168010000m2 = 1584m2

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of ₹5000 per m² per year. -Maths 9th

Description : The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of ₹5000 per m² per year. -Maths 9th

Last Answer : Let the sides of the triangular will be a = 122m, b = 12cm, c = 22m Semi-perimeter, s = a+b+c2 (122+120+224)m = 2642 m = 132m The area of the triangular side wall Rent for 1 year (i.e. 12 months) per m2 = ... = Rs. 5000 x 312 = Rent for 3 months for 1320 m2 = Rs. 5000 x 312 x 1320 = Rs. 16,50,000.

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? -Maths 9th

Description : The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? -Maths 9th

Last Answer : A roller is shaped like a cylinder. Let h be the height of the roller and r be the radius. h = Length of roller = 120 cm Radius of the circular end of roller = r = (84/2) cm = 42 cm Now, CSA of roller = 2πrh = ... , we have 2 (22/7) 0.7 h = 4.4 Or h = 1 Therefore, the height of the cylinder is 1 m.

The diameter of a roller is 42 cm and its length is 120 cm. -Maths 9th

Description : The diameter of a roller is 42 cm and its length is 120 cm. -Maths 9th

Last Answer : We have the diameter of a cyclindrial roller = 42 cm ⇒ The radius of cyclindrical roller (r) = 42 / 2 = 21 cm Length of a cyclindrical roller (h) = 120 cm Curved surface of the roller = 2πrh = ... of the playground = Area covered by the roller in 500 complete revolutions = 500 1.584 = 792 m2

The diameter of a roller is 42 cm and its length is 120 cm. -Maths 9th

Description : The diameter of a roller is 42 cm and its length is 120 cm. -Maths 9th

Last Answer : We have the diameter of a cyclindrial roller = 42 cm ⇒ The radius of cyclindrical roller (r) = 42 / 2 = 21 cm Length of a cyclindrical roller (h) = 120 cm Curved surface of the roller = 2πrh = ... of the playground = Area covered by the roller in 500 complete revolutions = 500 1.584 = 792 m2

Two batsman Rahul and Anil while playing a cricket match scored 120 runs. For this, write a linear equation in two variables and draw the graph. -Maths 9th

Description : Two batsman Rahul and Anil while playing a cricket match scored 120 runs. For this, write a linear equation in two variables and draw the graph. -Maths 9th

Last Answer : Solution :-

A tank internally measuring 150 cm × 120 cm × 100 cm has 1281600 cm^3 of water in it. Porous bricks are placed in the water until the tank -Maths 9th

Description : A tank internally measuring 150 cm × 120 cm × 100 cm has 1281600 cm^3 of water in it. Porous bricks are placed in the water until the tank -Maths 9th

Last Answer : Volume of a brick = 20 cm 6 cm 4 cm = 480 cm3. Water absorbed by one brick = (110 480)(110 480) cm3 = 48 cm3. Let x bricks be placed in the water. Then, x bricks absorb 48x cm3 of water. ... tank ⇒ 1281600 + 480xx - 48xx = 150 x 120 x 100 ⇒ 432xx = 1800000 - 1281600 = 518400 ⇒ xx = 1200.

The minimum compressive strength of 1st class bricks should be (A) 75 kg/cm2 (B) 90 kg/cm2 (C) 100 kg/cm2 (D) 120 kg/cm2

Description : The minimum compressive strength of 1st class bricks should be (A) 75 kg/cm2 (B) 90 kg/cm2 (C) 100 kg/cm2 (D) 120 kg/cm2

Last Answer : Answer: Option C

The minimum compressive strength of 2nd class bricks should be (A) 75 kg/cm2 (B) 90 kg/cm2 (C) 100 kg/cm2 (D) 120 kg/cm

Description : The minimum compressive strength of 2nd class bricks should be (A) 75 kg/cm2 (B) 90 kg/cm2 (C) 100 kg/cm2 (D) 120 kg/cm

Last Answer : Answer: Option A

Ashwini has a certain number of Pencils she wishes to distribute in the class. when she tries to distribute 7 pencils to each student, 3 pencils are left out. if the class had 3 less students, each student would have got 2 pencils more, and no pencils would have been left out. How many pencils does Ashwini have? a) 90 b) 120 c) 140 d) 160 e) None of these

Description : Ashwini has a certain number of Pencils she wishes to distribute in the class. when she tries to distribute 7 pencils to each student, 3 pencils are left out. if the class had 3 less students, each student would have ... How many pencils does Ashwini have? a) 90 b) 120 c) 140 d) 160 e) None of these

Last Answer : Let number of students be X. so, total Pencils = 7X + 3 7X + 3 = (7+2) (X-3) or., 7X + 3 = 9X – 27 or X = 15 total Pencils = 7X + 3 = 108 Answer: e)

he minimum compressive strength of first class brick should be a) 75 kg per centimeter square b) 90 kg per centimeter square c) 100 kg per centimeter square* d) 120 kg per centimeter square

Description : he minimum compressive strength of first class brick should be a) 75 kg per centimeter square b) 90 kg per centimeter square c) 100 kg per centimeter square* d) 120 kg per centimeter square

Last Answer : c) 100 kg per centimeter square*

The average score in a bank examination of 13 students of a class is 50. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark was 35 and the maximum mark was 100. It is also known that none of the students failed. If each of the top five scorers had distinct integral scores, the maximum possible score of the topper is. A) 85 B) 90 C) 95 D) 100

Description : The average score in a bank examination of 13 students of a class is 50. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark ... integral scores, the maximum possible score of the topper is. A) 85 B) 90 C) 95 D) 100

Last Answer : D Average = Sum of observations/Number of observations Given, average score in a bank examination of 13 students of a class is 50. Sum of total scores = 13 50= 650 Given, if the scores of the top five students are not ... b + c + d + e = 290 ⇒46+ 47 + 48+ 49 +e = 290 ⇒e = 100

The average marks of a class of 90 students is 126. Out Of them, 4 scores zero, first 60 students scored an average of 116, next 24 scored an average of 118. What is the mark obtained by the remaining student in the class? A) 750 B) 862 C) 774 D) 875

Description : The average marks of a class of 90 students is 126. Out Of them, 4 scores zero, first 60 students scored an average of 116, next 24 scored an average of 118. What is the mark obtained by the remaining student in the class? A) 750 B) 862 C) 774 D) 875

Last Answer : C) According to the question, 90*126=(4*0)+(60*116)+(24*118)+2y 11340=6960+2832+2y 2y=1548 y=774

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 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 area of the parallelogram ABCD is 90 cm2. -Maths 9th

Description : The area of the parallelogram ABCD is 90 cm2. -Maths 9th

Last Answer : Given, area of parallelogram, ABCD = 90 cm2 1.We know that, parallelograms on the same base and between the same parallel are equal in areas. Here, parallelograms ABCD and ABEF are on same base AB and between the same parallels AB ... (ABEF) = 1/2 x 90 = 45 cm2 [∴ ar (ABEF) = 90 cm2, from part (i)]

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 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 area of the parallelogram ABCD is 90 cm2. -Maths 9th

Description : The area of the parallelogram ABCD is 90 cm2. -Maths 9th

Last Answer : Given, area of parallelogram, ABCD = 90 cm2 1.We know that, parallelograms on the same base and between the same parallel are equal in areas. Here, parallelograms ABCD and ABEF are on same base AB and between the same parallels AB ... (ABEF) = 1/2 x 90 = 45 cm2 [∴ ar (ABEF) = 90 cm2, from part (i)]

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Description : A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Last Answer : NEED ANSWER

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Description : A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Last Answer : In trapezium ABCD, we draw a perpendicular line CE to the line AB.

PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of △ASR = 90 cm2. Find this statement is true or false. -Maths 9th

Description : PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of △ASR = 90 cm2. Find this statement is true or false. -Maths 9th

Last Answer : Solution :- As diagonal of the parallelogram divides it into two triangles of equal area. Since, area (△SRQ ) = 1/2 area(PQRS) area (△SRQ ) = 1/2 x 180 ... = 90 cm2 (Given) This is not possible unless area (△SRQ ) = area (△ASR ) So, the given statement is false.

case study questions class 9 maths surface area and volume -Maths 9th

Description : case study questions class 9 maths surface area and volume -Maths 9th

Last Answer : Q. Read the source or text given below and answer the following questions: A conical circus tent has to be made with a cloth that is 5m wide, whose height is 24m, and the radius of the base is 7m. ... 4. Find the Curved Surface Area Answers: 1. Curved Surface Area 2. 550 m 3. Rs.3850 4.110m²

All the theorems of circles class 9 -Maths 9th

Description : All the theorems of circles class 9 -Maths 9th

Last Answer : The Most important theorems (10) in Circle are Below ... Note: The derivations are also given below. Theorem: Equal chords of a circle subtend equal angles at the centre. Theorem: This is the ... If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.

How should i study maths main chapters like Lines and Angle,Triangles class 9 ?? -Maths 9th

Description : How should i study maths main chapters like Lines and Angle,Triangles class 9 ?? -Maths 9th

Last Answer : You should study them thoroughly so that you won't find them difficult when you are preparing for competitive exams like Olympiad s,etc. Study other reference books along with ncert textbook. Hope my and will help you.

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.

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

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

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

In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Description : In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 Number of students having blood group ‘B’ = 12 Required probability =12 / 40 = 3 / 10

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)

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

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

How should i study maths main chapters like Lines and Angle,Triangles class 9 ?? -Maths 9th

Description : How should i study maths main chapters like Lines and Angle,Triangles class 9 ?? -Maths 9th

Last Answer : You should study them thoroughly so that you won't find them difficult when you are preparing for competitive exams like Olympiad s,etc. Study other reference books along with ncert textbook. Hope my and will help you.

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.

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

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

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

In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Description : In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 Number of students having blood group ‘B’ = 12 Required probability =12 / 40 = 3 / 10

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)

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

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

In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is -Maths 9th

Description : In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is -Maths 9th

Last Answer : NEED ANSWER

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

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 : NEED ANSWER

If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Description : If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Last Answer : NEED ANSWER

In the class intervals 10-20, 20-30, the number 20 is included in -Maths 9th

Description : In the class intervals 10-20, 20-30, the number 20 is included in -Maths 9th

Last Answer : NEED ANSWER

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

Is it correct to say that in a histogram, the area of each rectangle is proportional to the class size of the corresponding class interval? If not, correct the statement. -Maths 9th

Description : Is it correct to say that in a histogram, the area of each rectangle is proportional to the class size of the corresponding class interval? If not, correct the statement. -Maths 9th

Last Answer : It is not correct. Because in a histogram, the area of each rectangle is proportional to the corresponding frequency of its class.

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