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

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

Here the marks are out of 50 , so we find its percentage (i.e. out of 100)
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The table shows the marks obtained by a student in unit tests out of 50 : -Maths 9th

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)

The percentage of marks obtained by a student in monthly unit tests are given below. -Maths 9th

Description : The percentage of marks obtained by a student in monthly unit tests are given below. -Maths 9th

Last Answer : (i) Number of tests in which the student scored more than 70% marks = 3 ∴ P(more than 70% marks) = 3/6 = 1/2 (ii) Number of tests in which the student scored less than 70% marks = 3 ∴ P(less ... ) Number of tests in which the student scored at least 60% marks = 5 ∴ P(at least 60% marks) = 5/6

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 .

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 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 .

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

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!!!

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!!!

The average marks obtained by a student in Tamil, english, maths, science, and social together is 65% above the average mark obtained in maths and science together. How many more marks exceeded by the average of maths and science together than the sum of the tamil and english together? A) 50 B) 65 C) Data insufficient D) None of these

Description : The average marks obtained by a student in Tamil, english, maths, science, and social together is 65% above the average mark obtained in maths and science together. How many more marks exceeded by the average of ... of the tamil and english together? A) 50 B) 65 C) Data insufficient D) None of these

Last Answer : C) According to the question  There is no other information about the marks in any one of the subject. The data gives is insufficient to answer the question.

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:

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 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.

What are the characteristics of Continuous and Comprehensive Evaluation ? (a) It increases the workload on students by taking multiple tests. (b) It replaces marks with grades. (c) It evaluates every aspect of the student. (d) It helps in reducing examination phobia. Select the correct answer from the codes given below: (A) (a), (b), (c) and (d) (B) (b) and (d) (C) (a), (b) and (c) (D) (b), (c) and (d)

Description : What are the characteristics of Continuous and Comprehensive Evaluation ? (a) It increases the workload on students by taking multiple tests. (b) It replaces marks with grades. (c) It evaluates every aspect of the student. (d) It helps in ... ) and (d) (C) (a), (b) and (c) (D) (b), (c) and (d)

Last Answer : Answer: D

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

Consider the following database Student:Roll No. First Name Last Name Class Marks Obtained(%) Scholarship Awarded -Technology

Description : Consider the following database Student:Roll No. First Name Last Name Class Marks Obtained(%) Scholarship Awarded -Technology

Last Answer : 1. 95.42. Mohan Garg and scholarship awarded is 50000.3. 54. 6

The percentage of marks obtained by a student in a monthly test is as follows: What is the probability that the student gets more than 75% marks in a

Description : The percentage of marks obtained by a student in a monthly test is as follows: What is the probability that the student gets more than 75% marks in a

Last Answer : The percentage of marks obtained by a student in a monthly test is as follows: What is the ... student gets more than 75% marks in a test?

In a post graduate examination the marks obtained by a student is 75 per paper. If he had obtained 33 marks more in Evs paper & 27 more marks in science paper, then his average per paper is increased by 3 marks. Then how many papers were there in exam? A) 10 B) 12 C) 14 D) 20

Description : In a post graduate examination the marks obtained by a student is 75 per paper. If he had obtained 33 marks more in Evs paper & 27 more marks in science paper, then his average per paper is increased by 3 marks. Then how many papers were there in exam? A) 10 B) 12 C) 14 D) 20

Last Answer : Answer: D)  Let the number of paper be A. Then total marks earned him = 75A from questions, 75A + 33+ 27= 78A 3A = 60=> A=20 = number of subjects

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

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

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Last Answer : NEED ANSWER

There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Last Answer : Find the mean of the given number is

In the table “Student”, Priya wanted to increase the Marks (Column Name: Marks) of those students -Technology

Description : In the table “Student”, Priya wanted to increase the Marks (Column Name: Marks) of those students -Technology

Last Answer : Correct SQL statement :UPDATE Student SET Marks = Marks +5 WHERE Marks

In Marks column of ‘Student’ table, for Rollnumber 2, the Class Teacher entered the marks as 45. -Technology

Description : In Marks column of ‘Student’ table, for Rollnumber 2, the Class Teacher entered the marks as 45. -Technology

Last Answer : UPDATE Student SET Marks = 50 WHERE Roll number = 2;

Consider the table STUDENT given below: RollNo Name Class DOB Gender City Marks 1 Anand XI 6/6/97 M Agra 430 -Technology

Description : Consider the table STUDENT given below: RollNo Name Class DOB Gender City Marks 1 Anand XI 6/6/97 M Agra 430 -Technology

Last Answer : (1)b. Both (iii) and (iv)(2) b.RollnoNameClassDOBGenderCityMarks6734MaakhiyNehaGeetPreetiXIXXIXII12/12/948/12/956/5/978/8/95FFFFDubaiMoscowAgraMumbai256324470492

Consider the table Student(stuid, name, course, marks). Which one of the following two queries is correct to find the highest marks student in course 5? Q.1. Select S.stuid From student S Where not exists (select * from student e where e course='5' and e marks ≥ s marks) Q.2. Select S.stuid From student S Where s.marks > any (select distinct marks from student S where s.couse = 5) (A) Q.1 (B) Q.2 (C) Both Q.1 and Q.2 (D) Neither Q.1 nor Q.2

Description : Consider the table Student(stuid, name, course, marks). Which one of the following two queries is correct to find the highest marks student in course 5? Q.1. Select S.stuid From student S Where not exists (select * from student ... ) Q.1 (B) Q.2 (C) Both Q.1 and Q.2 (D) Neither Q.1 nor Q.2

Last Answer : (B) Q.2 Explanation: First query gives stuid of students whose marks are greater than all students taking course 5. Second query gives stuid of students whose marks are greater than any student taking ... comparison is between maximum of marks by any student in course 5. So the answer is option D.

In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

Description : In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

Last Answer : For Mean : As we know that ⇒ x̅ = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60 / 15 x̅ = 822 / 15 = 54.8 ⇒ x̅ = 54.8 For Median : First of ... 52 For Mode : Make a frequency table for given data : Here, the marks 52 has the maximum frequency '3'. ∴ Mode = 52

In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

Description : In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

Last Answer : For Mean : As we know that ⇒ x̅ = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60 / 15 x̅ = 822 / 15 = 54.8 ⇒ x̅ = 54.8 For Median : First of ... 52 For Mode : Make a frequency table for given data : Here, the marks 52 has the maximum frequency '3'. ∴ Mode = 52

In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : NEED ANSWER

In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : Median will be a good representative of the data, because 1.each value occurs once. 2.the data is influenced by extreme values.

how i can score more marks in sa1 -Maths 9th

Description : how i can score more marks in sa1 -Maths 9th

Last Answer : u need to focus on ncert book than iinterfaces onlin if ur weak in thes u should do these math example science pratical social explanation surf internet on ur topic hindi open up with others inspeeaking english same

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.

how i can score more marks in sa1 -Maths 9th

Description : how i can score more marks in sa1 -Maths 9th

Last Answer : u need to focus on ncert book than iinterfaces onlin if ur weak in thes u should do these math example science pratical social explanation surf internet on ur topic hindi open up with others inspeeaking english same

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.

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 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 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 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.

The class marks of a continuous distribution are: -Maths 9th

Description : The class marks of a continuous distribution are: -Maths 9th

Last Answer : It is not correct because the difference between two consecutive marks should be equal to the class size.

If the class marks in frequency distribution are -Maths 9th

Description : If the class marks in frequency distribution are -Maths 9th

Last Answer : The class size of the distribution is = 40.5 - 33.5 = 7 The required class of the class mark 33.5 is [33.5 - 7/2] - [33.5 + 7/2], i.e., 30 - 37.

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

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.

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

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.

The number of students in four classes A, B, C, D and their respective mean marks obtained by each of the class are given below : Image The combined mean of the marks of four classes together will be : (A) 32 (B) 50 (C) 20 (D) 15

Description : The number of students in four classes A, B, C, D and their respective mean marks obtained by each of the class are given below :   The combined mean of the marks of four classes together will be : (A) 32 (B) 50 (C) 20 (D) 15 

Last Answer : Answer: A

The average of marks obtained by 60 students in a computer examination is 18. If the average marks of passed students is 20and that of the failed students is 8, what is the number of students who passed the examination? A) 100 B) 75 C) 50 D) 85

Description : The average of marks obtained by 60 students in a computer examination is 18. If the average marks of passed students is 20and that of the failed students is 8, what is the number of students who passed the examination? A) 100 B) 75 C) 50 D) 85 

Last Answer : C Let the number of passed students be x. Then total marks = 60 × 18= 20x + (60– x) × 8 1080= 20x + 480– 8x 12x = 600 ∴ x = 50 ∴ number of passed students = 50

Rupesh secures 495 marks out of 750 in his annual examination. Find the percentage of marks obtained by him. -Maths

Description : Rupesh secures 495 marks out of 750 in his annual examination. Find the percentage of marks obtained by him. -Maths

Last Answer : answer:

In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. -Maths 10th

Description : In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. -Maths 10th

Last Answer : Let P has obtained x in mathematics and y in science. Then by the problem, x+y=28.......(1). If P would have got 3 marks more in mathematics, then P would have got (x+3) in mathematics ... P obtained 12 marks in mathematics and 16 in science. or, P obtained 9 marks in mathematics and 19 in science.

A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Description : The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Last Answer : x̅ = 1 / n (Σxi) ⇒ 55 = x1 + x2 + .......+ ⇒ x7 / 7 ⇒ x1 + x2 + ...... + x7 = 55 × 7 = 385 x1 + x2 + ...... + x6 = 52 + 54 + 55 + 53 + 56 + 54 = 324 ∴ x7 = 385 - 324 = 61 kg ∴ weight of the seventh student is 61 kg.

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)