In a medical examination of students of a class, -Maths 9th

1 Answer

Answer :

Total number of students = 10 + 13 + 12 + 5 = 40  P (a student has blood group B) = 12/40 = 3/10

Related questions

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

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

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.

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

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

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.

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

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

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.

Description : The blood groups of 30 students of class VIII are recorded as follows: -Maths 9th

Last Answer : Frequency Distribution Table Blood group O is most common as it has highest frequency, i.e., 12. Blood group AB is rarest as it has lowest frequency, i.e., 3.

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

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:

Description : The number of sincere students (x) in a class -Maths 9th

Last Answer : x = 2 + 2y Sincere students always progress in life as they value time and channelise their talent in productive activities while a careless student always wastes his talent and time.

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.

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

Description : The probability of student A passing examination is 3/7 and of student B passing is 5/7 Assuming the two events “A passes”, -Maths 9th

Last Answer : p1 = P(A) = \(rac{3}{7}\), p2 = P(B) = \(rac{5}{7}\) ∴ q1 = P(\(\bar{A}\)) = 1 - P(A) = 1 - \(rac{3}{7}\) = \(rac{4}{7}\). q2 = P(\(\bar{B}\)) = 1 - P(B) = 1 - \(rac{5}{7}\) = \(rac{2}{7}\) ... passes) = p1 q2 + q1 p2 = \(rac{3}{7}\) x \(rac{2}{7}\) + \(rac{4}{7}\) x \(rac{5}{7}\) = \(rac{26}{49}.\)

Description : In an examination there are 3 multiple choice questions and each question has 4 choices. If a student randomly selects answer for all -Maths 9th

Last Answer : Probability of selecting a correct choice for a question = \(rac{1}{4}\)(∵ Out of 4 choices only one is correct)∴ Probability of answering all the three questions correctly = \(rac{1}{4}\)x \(rac{1}{4}\)x\ ... of not answering all the three questions correctly = 1 - \(rac{1}{64}\) = \(rac{63}{64}\).

Description : A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. -Maths 9th

Last Answer : (c) \(rac{11}{3^5}\)Probability of guessing a correct answer = \(rac{1}{3}\)Probability of guessing an incorrect answer = \(rac{2}{3}\)∴ Probability of guessing 4 or more correct answers = 5C4 \(\big(rac{1}{3}\big)^4\)\(\big(rac{2 ... )^5\) = 5 x \(rac{2}{3^5}\) + \(rac{1}{3^5}\) = \(rac{11}{3^5}\).

Description : In an examination of a certain class, at least `70%` of the students failed in Physics, at least `72%` failed in Chemistry, at least `80%` failed in M

Last Answer : In an examination of a certain class, at least `70%` of the students failed in Physics, at ... 15%` D. Cannot be determined due to insufficient data

Description : Teaching in higher education implies Options: A) presenting the information given in the text book B) asking questions in the class and conducting examinations C) helping students prepare for and pass the examination D) helping students how to learn

Last Answer : D) helping students how to learn 

Description : Students should prefer those teachers who Options: A) dictate notes in the class. B) give important questions before examination. C) can clear their difficulties regarding subject-matter. D) are themselves disciplined. 

Last Answer : C) can clear their difficulties regarding subject-matter. 

Description : The present annual examination system: (A) promotes role learning (B) does not promote good study habits (C) does not encourage students to be regular in class (D) All the above

Last Answer : Answer: D 

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

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!!!

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

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.

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

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

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

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 .

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!!!

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

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.

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

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

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

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 .

Description : A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Volume of milk in 1 glass =πr2h =π×(3.5)2×12=461.58cm2​for 1600 students milk needed is =1600×461.58=738258litre​

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

Description : A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : According to question find the litres of milk is needed to serve 1600 students.

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.

Description : The students of a Vidyalaya were asked to participate in a -Maths 9th

Last Answer : Radius of penholder = r = 3 cm Height of penholder = h = 10.5 cm Total surface area of penholder = πr2 + 2 πrh = πr(r + 2h) = 22/7 x 3(3 + 2 x 10.5) = 66/7 x 24 cm2 Cardboard required for 35 competitors = 35 x 66/7 x 24 = 7920 cm2

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.

Description : To know the opinion of the students about the -Maths 9th

Last Answer : (i) P (a student likes statistics) = Numbers of students who like statistics/Total number of students = 135/200 = 0.675 (ii) P (a students does not like statistics) = Numbers of students who dislikes statistics/Total number of students = 65/200 = 0.325

Description : Teacher asked the students 'Can we write 0.47(recurring) -Maths 9th

Last Answer : Yes, Let x = 0.477777.... ....(i) 10x = 4.77777... .....(ii) Subtracting (i) from (ii), we get 9x = 4.3 or x = 43/90 Scientific temper, knowledge, curosity.

Description : Some students of a school started a campaign against -Maths 9th

Last Answer : Area of rectangle ABCD = 5 x 3 = 15 cm2 Cost incurred on one banner = ₹3 x 15 = ₹45 Cost incurred on 5 banners = 5 x ₹45 = ₹225 Social awareness, caring, cooperative.

Description : 95 students each from 102 schools participated -Maths 9th

Last Answer : 95 x 102 = (100 - 5)(100 + 2) 1002 + (- 5 + 2) x 100 + (- 5 x 2) = 10,000 - 300 – 10 = 9,690 Fraternity, patriotism.

Description : Three students were made to stand on the points P, Q and S -Maths 9th

Last Answer : Coordinates of R are (6, 6). Reasoning, enjoyment, physical fitness.

Description : For her records, a teacher asked the students about their heights. -Maths 9th

Last Answer : Yes, Things which are equal to the same thing are equal to one another. Knowledge, curiosity, truthfulness

Description : A problem in mathematics is given to four students A, B, C and D. Their chances of solving the problem -Maths 9th

Last Answer : Given, P(A) = \(rac{1}{2}\) ⇒ P(\(\bar{A}\)) = P(A not solving the problem) = 1 - \(rac{1}{2}\) = \(rac{1}{2}\)P(B) = \(rac{1}{3}\) ⇒ P(\(\bar{B}\)) = 1 - \(rac{1}{3}\) = \(rac{2}{3}\)P(C) = \ ... 3}{4}\)x \(rac{4}{5}\) = \(rac{1}{5}\)∴ P(problem will be solved) = 1 - \(rac{1}{5}\) = \(rac{4}{5}\).