In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only? -Maths 9th

In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only? -Maths 9th
by

1 Answer

Answer :

As each team of 11 players has one goalkeeper and 10 team members, and out of 14 players there are 2 goalkeepers and 12 team members. = 12×112×2 = 132.
by

Related questions

Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers ? -Maths 9th

Description : Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers ? -Maths 9th

Last Answer : The chosen consists of players (6 + 5). ∴ Number of ways of selecting 11 players out of 15 players = n(S) = 15C11 Let A : Event of choosing 6 batsmen of 8 batsmen and 5 bowlers of 7 bowlers Then, n(A) = 8C6 ... 7 imes6}{2}\) x \(rac{4 imes3 imes2 imes1}{15 imes14 imes13 imes12}\) = \(rac{28}{65}.\)

A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be selected so that there are at least -Maths 9th

Description : A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be selected so that there are at least -Maths 9th

Last Answer : The selection of 6 balls, consisting of at least two balls of each color from 5 red and 6 white balls can be made in the following ways: Red balls (5) White balls(6) Number of ways 2 4 5 C 2 ​ × 6 C 4 ​ =150 3 3 5 C 3 ​ × 6 C 3 ​ =200 4 2 5 C 4 ​ × 6 C 2 ​ =75 Total 425

In how many ways can a pack of 52 cards be divided equally among four players in order? -Maths 9th

Description : In how many ways can a pack of 52 cards be divided equally among four players in order? -Maths 9th

Last Answer : Distribution of 52 cards can be equally divided among four players. Hence, number of ways is (13!)4! 52! ​ 4!= (13!) 52! ​

Two players A and B play a game by alternately drawing a card from a well-shuffled pack of playing cards, replacing the card each time after draw. -Maths 9th

Description : Two players A and B play a game by alternately drawing a card from a well-shuffled pack of playing cards, replacing the card each time after draw. -Maths 9th

Last Answer : (a) \(rac{13}{25}\)Let E : Event of drawing a queen in a single draw the pack of 52 cards. As there are 4 queens in a pack of 52 cards,P(E) = \(rac{4}{52}\) = \(rac{1}{13}\)P(\(\bar{E}\)) = P(not ... {25}\). [Sum of a G.P with infinite terms = \(rac{a}{1-r}\) where a = 1st term, r = common ratio.]

In a chess tournament, where participants were to play one game with one another, two players fell ill, having played 6 games each without playing -Maths 9th

Description : In a chess tournament, where participants were to play one game with one another, two players fell ill, having played 6 games each without playing -Maths 9th

Last Answer : answer:

Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

Description : Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

Last Answer : (b) \(rac{2}{5}\)As each horse has equal chance of winning the race, Number of ways in which one of the five horses wins the race = 5C1 ∴ n(S) =5C1 = \(rac{|\underline5}{|\underline4|\underline1}\) 5To find the chance ... n(E) = 2C1 = 2 ∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{2}{5}\).

A committee of 11 members sit at a round table. In how many ways can they be seated if the “President” and the “Secretary” choose to sit together. -Maths 9th

Description : A committee of 11 members sit at a round table. In how many ways can they be seated if the “President” and the “Secretary” choose to sit together. -Maths 9th

Last Answer : ∴ Required number of ways of seating = 9!

From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Description : From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Last Answer : (b) \(rac{7}{10}\)Total number of ways of selecting 2 persons at random out of 5 persons = 5C2 ∴ n(S) = 5C2 = \(rac{|\underline5}{|\underline3|\underline2}\) = \(rac{5 imes4}{2 imes1}\) = 10Let A : Event of selecting ... = 2 3 + 1 = 7 ∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{7}{10}\).

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)

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

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)

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

80 bulbs are selected at random from a lot -Maths 9th

Description : 80 bulbs are selected at random from a lot -Maths 9th

Last Answer : Number of bulbs having life less than 900 hours = 10 + 12 + 23 = 45 P (a bulb has life less than 900 hours) = 45/80 = 9/16

1500 families with 2 children were selected -Maths 9th

Description : 1500 families with 2 children were selected -Maths 9th

Last Answer : (i) P (a family having 2 girls) = Number of families having 2 girls/Total number of families = 475/1500 = 19/60 (ii) P (a family having 1 girl) = Number of families having 1 girl/Total number of ... families = 211/1500 Sum of probabilities = 475/1500 + 814/1500 + 211/1500 = 1500/1500 = 1

An Insurance company selected 2000 drivers -Maths 9th

Description : An Insurance company selected 2000 drivers -Maths 9th

Last Answer : Total number of drivers = 2000 (i) Number of drivers who are 18-29 years old and have exactly 3 accidents in one year is 61 So, P (driver is 18-29 years old with exactly 3 accidents) = 61/2000 = 0.0305 ~ 0. ... = 440 + 505 + 360 = 1305 So, P (drivers with no accident) = 1305/2000 = 0.6525 = 0.653

In a group there are 3 women and 3 men. 4 people are selected at random from this group -Maths 9th

Description : In a group there are 3 women and 3 men. 4 people are selected at random from this group -Maths 9th

Last Answer : A : Selected 3 women and 1 man B : Selected 1 women and 3 men S : Selected 4 people from 6 people (3 + 3) Then n(A) = 3C3 3C1, n(B) = 3C1 3C3, n(S) = 6C4∴ Required probability = P(A) + P(B) = \(rac{ ... 3C_1}{^6C_4}\) + \(rac{^3C_1 imes^3C_3}{^6C_4}\)= \(rac{2 imes1 imes3}{15}\) = \(rac{2}{5}.\)

What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Description : What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Last Answer : (a) \(rac{1}{25}\) Let us assume S as the sample space in all questions. S means the set denoting the total number of outcomes possible. Let S = {1, 2, 3, , 100} be the sample space. Then, n(S) = 100 Let A : ... ∴Required probability P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{100}\) = \(rac{1}{25}\)

If three natural numbersfrom 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is -Maths 9th

Description : If three natural numbersfrom 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is -Maths 9th

Last Answer : (c) \(rac{4}{1155}\)Let n(S) = Number of ways of selecting 3 numbers from 100 numbers = 100C3 Let E : Event of selecting three numbers divisible by both 2 and 3 from numbers 1 to 100 = Event of selecting three ... C_3}{^{100}C_3}\) = \(rac{16 imes15 imes14}{100 imes99 imes98}\) = \(rac{4}{1155}\).

A management institute has six senior professors and four junior professors. Three professors are selected at random -Maths 9th

Description : A management institute has six senior professors and four junior professors. Three professors are selected at random -Maths 9th

Last Answer : (a) \(rac{5}{6}\)P(At least one junior professor is selected) = P(Selecting 1 Junior) P(Selecting 2 Seniors) + P(Selecting 2 Junior) P(Selecting 1 Senior) + P(Selecting all 3 Juniors)∴ Required probability = \(rac{^4C_1 imes^ ... }{30}\) = \(rac{15+9+1}{30}\) = \(rac{25}{30}\) = \(rac{5}{6}\).

A box contains 100 balls numbers from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability -Maths 9th

Description : A box contains 100 balls numbers from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability -Maths 9th

Last Answer : (d) \(rac{1}{4}\)The box contains 100 balls numbered from 1 to 100. Therefore, there are 50 even and 50 odd numbered balls. The sum of the three numbers drawn will be odd, if all three are odd or one is even and 2 are odd. ∴ Required probability = P(odd) × P(odd) × P(odd) + P(even) × P(odd) × P(odd)

A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Description : A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Last Answer : answer:

Three letters are randomly selected from the 26 capital letters of the English alphabet. -Maths 9th

Description : Three letters are randomly selected from the 26 capital letters of the English alphabet. -Maths 9th

Last Answer : (d) \(rac{36}{1001}\)Required probability = \(rac{P( ext{2 blue balls}) imes{P}( ext{2 red balls})}{P( ext{4 balls out of 14 balls})}\) + \(rac{P( ext{2 green balls}) imes{P}( ext{2 black balls})}{P( ext{4 out of 14 balls})}\)

Why are there no goalkeepers in american rugby?

Description : Why are there no goalkeepers in american rugby?

Last Answer : There isn’t a goalkeeper in Rugby Union, why should there be one in American Rugby? After all, it’s not Gaelic football.

Why do Goalkeepers need to bounce the ball?

Description : Why do Goalkeepers need to bounce the ball?

Last Answer : Not necessarily....they are supposed to catch the ball, bouncing, grabbing, kicking the ball...it really does not matter, as long as the ball steers clear from the goal. When they are passing the ball ... from his team, there is no big rule on that either. Hope my answer was helpful and insightful.

In how many ways can a mixed doubles game be arranged from amongst 8 married couples if no husband and wife play in the same game? -Maths 9th

Description : In how many ways can a mixed doubles game be arranged from amongst 8 married couples if no husband and wife play in the same game? -Maths 9th

Last Answer : For mixed doubles, we have 2 men and 2 women. 2 men out of 9 can be selected in 9C8 ways Out of 9 women, 2 will be there wife 30 only 7 are remaining for selecting 2 Now out of them 2 possible combinations can be made. So, ... =3024 Hence the answer is 3024.

A bag contains a white and b black balls. Two players A and B alternately draw a ball from the bag -Maths 9th

Description : A bag contains a white and b black balls. Two players A and B alternately draw a ball from the bag -Maths 9th

Last Answer : (c) 2 : 1Let W denote the event of drawing a white ball at any draw and B that of drawing a black ball. Then, P (W) = \(rac{a}{a+b},\) P(B) = \(rac{b}{a+b}\)∴ P (A wins the game) = P (W or BBW or BBBBW or ... ... the given condition,\(rac{a+b}{a+2b}\) = 3. \(rac{b}{a+2b}\) ⇒ a = 2b ⇒ a : b = 2 : 1.

How many players play in each team of Baseball

Description : How many players play in each team of Baseball

Last Answer : 9

7 persons enter an elevator on the ground floor of a 11 storey hotel. Any one of them can leave the elevator at any of the 10 floors. -Maths 9th

Description : 7 persons enter an elevator on the ground floor of a 11 storey hotel. Any one of them can leave the elevator at any of the 10 floors. -Maths 9th

Last Answer : answer:

Four friends have 7 shirts, 6 pants and 8 ties. In how many ways can they wear them? -Maths 9th

Description : Four friends have 7 shirts, 6 pants and 8 ties. In how many ways can they wear them? -Maths 9th

Last Answer : 7 shirts can be worn by 4 friends in 7P4 ways. Similarly, 6 pants and 8 ties can be worn by 4 friends in 6P4 and 8P4 ways respectively. ∴ Total number of ways in which 7 shirts, 6 pants and 8 ties can be worn by 4 friends = 7P4 × 6P4 × 8P4 = 7! 3!

7 relatives of a man comprise 4 ladies and 3 gentlemen; his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways -Maths 9th

Description : 7 relatives of a man comprise 4 ladies and 3 gentlemen; his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways -Maths 9th

Last Answer : The possible cases are: Case I : A man invites (3 ladies) and woman invites (3 gentlemen) 4 C 3 4 C 3 =16 Case II : A man invites (2 ladies,1 gentleman) and woman invites (2 gentleman, 1 lady) ( 4 ... and woman invites (3 ladies) 3 C 3 3 C 3 =1 Total number of ways =16+324+144+1=485

In how many ways can a pack of 52 cards be divided into 4 sets, three of them having 16 cards each and the fourth just 4 cards? -Maths 9th

Description : In how many ways can a pack of 52 cards be divided into 4 sets, three of them having 16 cards each and the fourth just 4 cards? -Maths 9th

Last Answer : First we divide 52 cards into two sets which contains 1 and 51 cards respectively is 1! 51! 52! Now 51 cards can be divided equally in three sets each contains 17 cards (Here order of sets is not important) in 3!(17!) ... ways Hence, the required number of ways = 1! 51! 52! 3! (17!) 3 51!

The points scored by a basketball team in a series of matches are as follows: -Maths 9th

Description : The points scored by a basketball team in a series of matches are as follows: -Maths 9th

Last Answer : Arranging the data in ascending order, we get 2, 5, 7, 7, 8, 10, 10, 10, 14, 17, 18, 24, 25, 27, 28, 48 Since number of observations (n) = 16, which is even. Therefore, median is the mean of (n/2) ... ∴ Median = 10 + 14/2 = 12 As 10 occurs most frequently, i.e., three times. So, the mode is 10.

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB=6cm, BC=8cm, CD=12cm and DA=14 cm. Now. Join AC. We have, ABC is a right angled triangle at B. Now, AC2=AB2+BC2 [by Pythagoras theorem]Now, AC2=AB2+BC2 ... =24(1+6-√)cm2=24+246=24(1+6)cm2 Hence, the area of quadrilateral is 241+6-√−−−−−−√cm2241+6cm2 .

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB = 6 cm, BC = 8 cm, CD = 12 cm and DA = 14 cm. Now, join AC.

A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Description : A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Last Answer : (c)\(\bigg(rac{147 imes\sqrt3}{4}\bigg)\) cm2Take the trapezoid ABCD in the semi-circle with centre O such that AD = DC = CB. Now, complete the circle and draw an identical trapezoid in the other semicircle also. Then, ADCBEF ... {1}{2}\)x \(rac{3\sqrt3}{2}\) x 49 = \(rac{147 imes\sqrt3}{4}\) cm2.

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Description : A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Last Answer : Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m Non-parallel sides are AD = 13 m and BC = 14 m. We draw BE || AD, such that BE = 13 m. ... = 112 m2 So, area of the field = area of ∆BCE + area of parallelogram ABED = 84 m2 + 112 m2 = 196 m2

The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th

Description : The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th

Last Answer : Slant height of conical tomb, l = 25m Base radius, r = diameter/2 = 14/2 m = 7m CSA of conical tomb = πrl = (22/7)×7×25 = 550 CSA of conical tomb= 550m2 Cost of white-washing 550 m2 area, which is Rs (210×550)/100 = Rs. 1155 Therefore, cost will be Rs. 1155 while white-washing tomb.

Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

Description : Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

Last Answer : Slant height of cone, l = 14 cm Let the radius of the cone be r. (i) We know, CSA of cone = πrl Given: Curved surface area of a cone is 308 cm2 (308 ) = (22/7) r 14 308 = 44 r r = 308 ... Total surface area of cone = 308+(22/7) 72 = 308+154 Therefore, the total surface area of the cone is 462 cm2.

The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Description : The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Last Answer : Height of cylinder, h = 14cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that, formula to find Curved surface area of cylinder is 2πrh. So 2πrh =88 cm2 (r is the ... 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm.

If X^3-1/x^3=14 then find x-1/x=? -Maths 9th

Description : If X^3-1/x^3=14 then find x-1/x=? -Maths 9th

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

Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Description : Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Last Answer : Here, radius of hemispherical dome (r) = 14 m Surface area of dome = 2πr2 = 2 × 22 / 7 × 14 × 14 = 1232 m2 Hence , total surface area to be whitewashed from outside is 1232 m2

Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Description : Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Last Answer : 14 repeat maximum number of times (4 times) in the given data. Mode = 14

Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Description : Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Last Answer : Here, the arranged data is 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 Total number of observations = 10 Here, 10 is an even number , therefore median will be the mean of (10 / 2)th and (10 / 2 + 1)th observation. ... ⇒ 3x + 12 / 2 = 24 ⇒ 3x - 12 = 48 ⇒ 3x = 60 ⇒ x = 20 ∴ The value of x = 20

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

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.

If X^3-1/x^3=14 then find x-1/x=? -Maths 9th

Description : If X^3-1/x^3=14 then find x-1/x=? -Maths 9th

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

Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Description : Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Last Answer : Here, radius of hemispherical dome (r) = 14 m Surface area of dome = 2πr2 = 2 × 22 / 7 × 14 × 14 = 1232 m2 Hence , total surface area to be whitewashed from outside is 1232 m2

Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Description : Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Last Answer : 14 repeat maximum number of times (4 times) in the given data. Mode = 14

Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Description : Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Last Answer : Here, the arranged data is 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 Total number of observations = 10 Here, 10 is an even number , therefore median will be the mean of (10 / 2)th and (10 / 2 + 1)th observation. ... ⇒ 3x + 12 / 2 = 24 ⇒ 3x - 12 = 48 ⇒ 3x = 60 ⇒ x = 20 ∴ The value of x = 20

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

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.

The mode of given data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15,17 and 15 is -Maths 9th

Description : The mode of given data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15,17 and 15 is -Maths 9th

Last Answer : We first arrange the given data in ascending order as follows 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20 From above, we see that 15 occurs most frequently i.e., 5 times. Hence, the mode of the given data is 15.