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

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

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 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.

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 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 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

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)

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 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.

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 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 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

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 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

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

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)

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.

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.

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:

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 public examination , Raghav scored twice twice Sunitha’s score. If r and s represent the scores of Raghav and Sunitha respectively , Write a linear equation in r and s representing the above situation ? -Maths 9th

Description : In a public examination , Raghav scored twice twice Sunitha’s score. If r and s represent the scores of Raghav and Sunitha respectively , Write a linear equation in r and s representing the above situation ? -Maths 9th

Last Answer : Let raghav and sunita score be r and S respectively. Sunita score=2times raghav score Therefore, s=2r

In a Examination a candidate has to score minimum of 24 marks inorder to clear the exam. The maximum that he can score is 40 marks. Identify the Valid Equivalance values if the student clears the exam. A. 22,23,26 B. 21,39,40 C. 29,30,31 D. 0,15,22

Description : In a Examination a candidate has to score minimum of 24 marks inorder to clear the exam. The maximum that he can score is 40 marks. Identify the Valid Equivalance values if the student clears the exam. A. 22,23,26 B. 21,39,40 C. 29,30,31 D. 0,15,22

Last Answer : C. 29,30,31

Nandita scores 60% marks in five subjects together, viz., Hindi, Science, Mathematics, English and Sanskrit, where in the maximum marks of each subject were 105. How many marks did Nandita score in Science, if she scored 69 marks in Hindi, 62 marks in Sanskrit, 68 marks in Mathematics and 51 marks in English? a) 66 b) 68 c) 55 d) 65 e) None of these

Description : Nandita scores 60% marks in five subjects together, viz., Hindi, Science, Mathematics, English and Sanskrit, where in the maximum marks of each subject were 105. How many marks did Nandita score in Science, if she scored 69 ... and 51 marks in English? a) 66 b) 68 c) 55 d) 65 e) None of these

Last Answer : Answer: D Total of maximum marks of all subjects=105×5=525 75% of 525=525 ×60/100= 315 Obtained marks of foru subjects (Hindi, Sanskrit, mathematics and English) =69+62+68+51=250 So, the obtained marks in Science=315-250=65

A group of students of arithmetic mean of the marks in a test was 63. The brightest 30% of them secured a mean score of 70 and the dullest 15% a mean score of 41. The mean score of remaining 55 % is A) 63.675 B) 61.785 C) 65.181 D) 66.67

Description : A group of students of arithmetic mean of the marks in a test was 63. The brightest 30% of them secured a mean score of 70 and the dullest 15% a mean score of 41. The mean score of remaining 55 % is A) 63.675 B) 61.785 C) 65.181 D) 66.67

Last Answer : C) Let the requird mean score be ‘x’ Then (30*70)+(15*41)+55x = 63*100 2100+615+55x=6300 2715+55x=6300 55x=3585 X=65.181

Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? -Maths 10th

Description : Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? -Maths 10th

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.

If my marks in maths are less than your markes in Maths by `20%` then your marks in Maths are `25%` more than my marks in Maths. (True/False)

Description : If my marks in maths are less than your markes in Maths by `20%` then your marks in Maths are `25%` more than my marks in Maths. (True/False)

Last Answer : If my marks in maths are less than your markes in Maths by `20%` then your marks in Maths are `25%` more than my marks in Maths. (True/False)

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.

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:

What is the inter 1st year ap maths 1a pass marks?

Description : What is the inter 1st year ap maths 1a pass marks?

Last Answer : I am not sure what is being asked here

Give the equations of two lines passing through (4, –2). How many more such lines are there, and why? -Maths 9th

Description : Give the equations of two lines passing through (4, –2). How many more such lines are there, and why? -Maths 9th

Last Answer : Solution :-

The cost of a table is 100 more than half the cost of the chair write the statement linear equation in two varible -Maths 9th

Description : The cost of a table is 100 more than half the cost of the chair write the statement linear equation in two varible -Maths 9th

Last Answer : This answer was deleted by our moderators...

linear equation such that each point on its graph has an ordinate one more than 3×1÷2 times its abscissa -Maths 9th

Description : linear equation such that each point on its graph has an ordinate one more than 3×1÷2 times its abscissa -Maths 9th

Last Answer : This answer was deleted by our moderators...

Five years hence , the age of Ram will be 10 more than the two thirds of Ravi’s age . Assume the present ages of Ram and Ravi as x and y respectively . Express the statement in the form of a linear equation in two variables. -Maths 9th

Description : Five years hence , the age of Ram will be 10 more than the two thirds of Ravi’s age . Assume the present ages of Ram and Ravi as x and y respectively . Express the statement in the form of a linear equation in two variables. -Maths 9th

Last Answer : answer:

Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails. -Maths 9th

Description : Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails. -Maths 9th

Last Answer : (d) \(rac{1}{2}\)Let S be the sample space. Then, S = {HHH, HHT, HTH, HTT, THH, THT,TTH, TTT} ⇒ n(S) = 8 Let A : Event of getting more heads than number of tails. Then, A = {HHH, HHT, HTH, THH} ⇒ n(A) = 4∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{8}\) = \(rac{1}{2}.\)

A man has 6 friends. Number of different ways, he can invite 2 or more for dinner is -Maths 9th

Description : A man has 6 friends. Number of different ways, he can invite 2 or more for dinner is -Maths 9th

Last Answer : He can invite one or more friends by inviting 1 friend, or 2 friends or 3 friends, or all the 6 friends.

Use capital letters, full stops, question marks, commas and inverted commas wherever necessary in the following para-graph. -English 9th

Description : Use capital letters, full stops, question marks, commas and inverted commas wherever necessary in the following para-graph. -English 9th

Last Answer : An arrogant lion was wandering through the jungle one day. He asked the tiger, Who is stronger than you ? You, O! lion, replied the tiger. Who is more fierce than a leopard? asked the lion. You, ... Look , said the lion, there is no need to get mad just because you don't know the answer.

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

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

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Last Answer : Given : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged ... length of coloured tape required = 30 cm (b) The values are : Happiness, beauty, Knowledge.

While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Description : While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Last Answer : (a) Yes , x < y is correct (b) Ð ADB =Ð DBC = y (alternate int. angles) since BC < CD (angle opp. to smaller side is smaller) there for, x < y (c) Truth value

Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. -Maths 9th

Description : Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. -Maths 9th

Last Answer : Edge of cubical tank = 1.5 m ∴ Area of 4 walls = 4 (side)² = 4(1.5)² m² = 4 x 225 = 9 m² Area of floor = (1.5)² = 2.25 m² ∴ Total surface area = 9 + 2.25 = 11.25 m² Edge of square tile = 25 m = 0.25 m² ∴ Area of 1 tile = (0.25)2 = .0625 m²

A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes. -Maths 9th

Description : A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes. -Maths 9th

Last Answer : Side of cube = 4 cm But cutting into 1 cm cubes, we get = 4 x 4 x 4 = 64 Now surface area of one cube = 6 x (1)² = 6 x 1=6 cm² and surface area of 64 cubes = 6 x 64 cm² = 384 cm²

Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Description : Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Last Answer : Let each side of a cube = a cm Then surface area = 6a² cm² and surface area of 3 such cubes = 3 x 6a² = 18a² cm² By placing three cubes side by side we get a cuboid whose ... + 3a²] = 14 a² ∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9

The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m². -Maths 9th

Description : The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m². -Maths 9th

Last Answer : Length of a room (l) = 5m Breadth (b) = 4 m and height (h) = 3 m ∴ Area of 4 walls = 2(l + b) x h = 2(5 + 4) x 3 = 6 x 9 = 54 m² and area of ceiling = l x b = 5 x ... ∴ Total area = 54 + 20 = 74 m2 Rate of white washing = 7.50 per m² ∴ Total cost = ₹74 x 7.50 = ₹555