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

Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th
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 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. ∴ Median = 5th observation + 6th observation / 2  = x + 1 + 2x - 13 / 2    = 3x + 12 / 2  But median of data is 24 (given)  ⇒  3x + 12 / 2 = 24 ⇒ 3x - 12 = 48 ⇒ 3x = 60 ⇒ x = 20  ∴ The value of x = 20
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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

Ten observations 6, 14, 15, 17, x+1, 2x -13, -Maths 9th

Description : Ten observations 6, 14, 15, 17, x+1, 2x -13, -Maths 9th

Last Answer : 6, 14, 15, 17, x + 1, 2x -13, 30, 32, 34, 43, Here, n = 10 Since the number of observations is 10 (an even number), therefore, the median = (10/2)th observation + (10/2 + 1)th observation/2 = 5th observation + 6th ... = x + 1 + 2x - 13/2 ⇒ 48 = 3x - 12 ⇒ 3x = 48 + 12 = 60 ⇒ x = 20

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

Insert the missing number: 16/32, 15 /33, 17/31, 14/34, ? (A) 19/35 (B) 19/30 (C) 18/35 (D) 18/30

Description : Insert the missing number: 16/32, 15 /33, 17/31, 14/34, ? (A) 19/35 (B) 19/30 (C) 18/35 (D) 18/30

Last Answer : Answer: D 16/32 then 15/33 means -1/+1, 17/31 means +2/-2, 14/34 means -3/+3 next +4/-4 i.e 18/30

The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Description : The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Last Answer : NEED ANSWER

The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Description : The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Last Answer : (b) Given, mean of 25 observations = 36 ∴ Sum of 25 observations = 36 x 25 = 900 Now, the mean of first 13 observations = 32 ∴ Sum of first 13 observations = 13 x 32 = 416 and the mean of last 13 ... - (Sum of 25 observations) = (520 + 416)-900 = 936 - 900 = 36 Hence, the 13th observation is 36.

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : In conclusion, the range of data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, and 20 is 26.

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : (d) In a given data, maximum value = 32 and minimum value = 6 We know, range of the data = maximum value – minimum value = 32 – 6 = 26 Hence, the range of the given data is 26.

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

The ages of Gaurav and Goutham will be in the ratio 5 : 7 after ten years from now and will be in the ratio 13 : 18 after twelve years from now. Find the ratio of the sum of their ages 10 years hence to the sum of their ages 12 years hence. 1 : 25 : 27 2 : 13 : 15 3 : 14 : 17 4 : 18 : 23 5 : 30 : 31

Description : The ages of Gaurav and Goutham will be in the ratio 5 : 7 after ten years from now and will be in the ratio 13 : 18 after twelve years from now. Find the ratio of the sum of their ages 10 years hence to the sum of their ages 12 ... hence. 1 : 25 : 27 2 : 13 : 15 3 : 14 : 17 4 : 18 : 23 5 : 30 : 31

Last Answer : 5 : 30 : 31

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..

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..

A bag contains 30 tickets numbered from 1 to 30. Five tickets are drawn at random and arranged in ascending order -Maths 9th

Description : A bag contains 30 tickets numbered from 1 to 30. Five tickets are drawn at random and arranged in ascending order -Maths 9th

Last Answer : Total number of ways in which 5 tickets can be drawn = n(S) = 30C5. The tickets are arranged in the form T1, T2, T3 (= 20), T4, T5 Where T1, T2 ∈{1, 2, 3, , 19} and T4, T5 ∈{21, 22, , 30 ... {10 imes9}{2}\) x \(rac{5 imes4 imes3 imes2 imes1}{30 imes29 imes28 imes27 imes26}\) = \(rac{285}{5278}.\)

Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Description : Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Last Answer : Since mean of 20 observations is 17 Sum of the 20 observations = 17 x 20 = 340 New sum of 20 observations = 340 – 40 + 12 = 312 New mean=312 / 20 =15.6

Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Description : Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Last Answer : Since mean of 20 observations is 17 Sum of the 20 observations = 17 x 20 = 340 New sum of 20 observations = 340 – 40 + 12 = 312 New mean=312 / 20 =15.6

If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Description : Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Last Answer : Arranging the data in ascending order, we have 25, 31, 32, 37, 39, 42, 43, 45, 46 Here, number of observations = 9 (odd) Median = value of (9+1 / 2)th Observation = Value of 5th Observation = 39

Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Description : Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Last Answer : Arranging the data in ascending order, we have 25, 31, 32, 37, 39, 42, 43, 45, 46 Here, number of observations = 9 (odd) Median = value of (9+1 / 2)th Observation = Value of 5th Observation = 39

If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3. -Maths 9th

Description : If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3. -Maths 9th

Last Answer : 2x +3y = 13-----(1) xy =6-----(2) 8x³ +27y³ = (2x)³ +(3y)³ = (2x+3y)³ - 3*2x*3y(2x+3y) [using (a+b)³ = a³+b³+3ab(a+b)] = (13)³- 18 * 6 *13 [ using (1) and (2)] = 2197 - 1404 =793

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.

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 : (b) 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.

If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th

Description : If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th

Last Answer : p(x) = (k2 – 14) x2 – 2x – 12 Here a = k2 – 14, b = -2, c = -12 Sum of the zeroes, (α + β) = 1 …[Given] ⇒ − = 1 ⇒ −(−2)2−14 = 1 ⇒ k2 – 14 = 2 ⇒ k2 = 16 ⇒ k = ±4

If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Description : If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Last Answer : x + x + 2 + x + 4 + x + 6 + x + 8 / 5 = 11 5x + 20 = 55 5x = 35 ⇒ x = 7

If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Description : If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Last Answer : x + x + 2 + x + 4 + x + 6 + x + 8 / 5 = 11 5x + 20 = 55 5x = 35 ⇒ x = 7

In a pocket of A, the ratio of Rs.1 coins, 50p coins and 25p coins can be expressed by three consecutive odd prime numbers that are in ascending order. The total value of coins in the bag is Rs 58. If the number of Rs.1, 50p, 25p coins are reversed, find the new total value of coins in the pocket of A? A) Rs 68 B) Rs 43 C) Rs 75 D) Rs 82 E) NONE

Description : In a pocket of A, the ratio of Rs.1 coins, 50p coins and 25p coins can be expressed by three consecutive odd prime numbers that are in ascending order. The total value of coins in the bag is Rs 58. If the number of Rs.1 ... of coins in the pocket of A? A) Rs 68 B) Rs 43 C) Rs 75 D) Rs 82 E) NONE

Last Answer : Answer: D  Since the ratio of the number of Rs. 1, 50p and 25p coins can be represented by 3 consecutive odd numbers that are prime in ascending order, the only possibility for the ratio is 3:5:7. Let the number of ... =100 7k+50 5k+25 3k=1025k (In above we find the value of k). ⇒ 8200p = Rs 82.

24 14 26 ? 28 16 30 a) 13 b) 17 c) 15 d) 11 e) 20

Description : 24 14 26 ? 28 16 30 a) 13 b) 17 c) 15 d) 11 e) 20

Last Answer : The series consists of two series 1 and 2: Series 1 : 24 26 28 30 +2 +2 +2 Series 2 : 14 15 16 +1 +1 Answer: c)

How many hotspots of biodiversity in the world have been identified till date by Norman Myers? (a) 17 (b) 25 (c) 34 (d) 43

Description : How many hotspots of biodiversity in the world have been identified till date by Norman Myers? (a) 17 (b) 25 (c) 34 (d) 43

Last Answer : (c) 34

Compute the value of adding the following two fuzzy integers: A = {(0.3,1), (0.6,2), (1,3), (0.7,4), (0.2,5)} B = {(0.5,11), (1,12), (0.5,13)} Where fuzzy addition is defined as μA+B(z) = maxx+y=z (min(μA(x), μB(x))) Then, f(A+B) is equal to (A) {(0.5,12), (0.6,13), (1,14), (0.7,15), (0.7,16), (1,17), (1,18)} (B) {(0.5,12), (0.6,13), (1,14), (1,15), (1,16), (1,17), (1,18)} (C) {(0.3,12), (0.5,13), (0.5,14), (1,15), (0.7,16), (0.5,17), (0.2,18)} (D) {(0.3,12), (0.5,13), (0.6,14), (1,15), (0.7,16), (0.5,17), (0.2,18)}

Description : Compute the value of adding the following two fuzzy integers: A = {(0.3,1), (0.6,2), (1,3), (0.7,4), (0.2,5)} B = {(0.5,11), (1,12), (0.5,13)} Where fuzzy addition is defined as μA+B(z) = maxx+y=z (min(μA(x), μB( ... ,18)} (D) {(0.3,12), (0.5,13), (0.6,14), (1,15), (0.7,16), (0.5,17), (0.2,18)}

Last Answer : (D) {(0.3,12), (0.5,13), (0.6,14), (1,15), (0.7,16), (0.5,17), (0.2,18)} 

If the polynomial x^19 + x^17 + x^13 + x^11 + x^7 + x^5 + x^3 is divided by (x^2 + 1), then the remainder is : -Maths 9th

Description : If the polynomial x^19 + x^17 + x^13 + x^11 + x^7 + x^5 + x^3 is divided by (x^2 + 1), then the remainder is : -Maths 9th

Last Answer : answer:

Which of the following couldnot be an Ethernet unicast destination? A) 43:7B:6C:DE:10:00 B) 44:AA:C1:23:45:32 C) 46:56:21:1A:DE:F4 D) 48:32:21:21:4D:34

Description : Which of the following couldnot be an Ethernet unicast destination? A) 43:7B:6C:DE:10:00 B) 44:AA:C1:23:45:32 C) 46:56:21:1A:DE:F4 D) 48:32:21:21:4D:34

Last Answer : 43:7B:6C:DE:10:0

When Samantha was 10 years old, Luis was 15 years old. When Samantha was 13 years old, Luis was 18 years old. When Samantha was 17 years old, Luis was 22 years old. How old is Luis when Samantha is 34 years old?

Description : When Samantha was 10 years old, Luis was 15 years old. When Samantha was 13 years old, Luis was 18 years old. When Samantha was 17 years old, Luis was 22 years old. How old is Luis when Samantha is 34 years old?

Last Answer : Based upon the other ages, Luis is 5 years older than Samantha, therefore, when Samantha is 34, Luis would be 39. 34+5=39

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 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.

A boat covers 14 kms in upstream and 20 kms downstream in 7 hours. Also it covers 22 kms upstream and 34 kms downstream in 10 hours. Find the speed of the boat in still water and of that the stream. -Maths 9th

Description : A boat covers 14 kms in upstream and 20 kms downstream in 7 hours. Also it covers 22 kms upstream and 34 kms downstream in 10 hours. Find the speed of the boat in still water and of that the stream. -Maths 9th

Last Answer : Given, The boat covers 14 km upstream and 20km downstream . at time 7 hours also cover 22km ups. and 34km dwn in10 hours total speed = total distance/total time :.total distance = 14+20+22+34=90km and total time=7+10= ... =90/17 => 5.294km/h => 5.294km/h the speed of boat in still water is 5.29 km/h

The line L is given by x/5 + y/b = 1 passes through the point (13, 32). The line K is parallel to L and has the equation -Maths 9th

Description : The line L is given by x/5 + y/b = 1 passes through the point (13, 32). The line K is parallel to L and has the equation -Maths 9th

Last Answer : (a) 45º The equations of the given lines are: A\(x\) + By = A + B ⇒ By = -A\(x\) + (A + B) ⇒ y = \(-rac{A}{B}x\) + \(rac{(A+B)}{B}\) ....(i)and A(\(x\) - y) + B(\(x\) ... (ii) = m2 = \(rac{(A+B)}{B-A}\)Let θ be the angle between both the lines, then∴ tan θ = 1 ⇒ θ = tan-1 (1) = 45°.

lf the series 4,5,8,13,14,17,22,........ is continued in the same pattern, which one of the following is not a term of this series? (A) 31 (B) 32 (C) 33 (D) 35

Description : lf the series 4,5,8,13,14,17,22,........ is continued in the same pattern, which one of the following is not a term of this series? (A) 31 (B) 32 (C) 33 (D) 35

Last Answer : Answer: C

There are three events A, B and C , one of which must and only can happen. If the odds are 8:3 against A, 5:2 against B, then the odds against C must be: e) 13:7 f) 3:2 g) 43:77 h) 43:34

Description : There are three events A, B and C , one of which must and only can happen. If the odds are 8:3 against A, 5:2 against B, then the odds against C must be: e) 13:7 f) 3:2 g) 43:77 h) 43:34

Last Answer : Answer : d

Five years ago, the age of Arun was 4 times the age of Sarmi after 10 years, Arun will be twice as old as Sarmi. Find the Present ages of Arun and sarmi? a) 30 years, 10.5 years b) 32 years, 11.5 years c) 34 years, 14.5 years d) 35 years, 12.5 years e) None of these

Description : Five years ago, the age of Arun was 4 times the age of Sarmi after 10 years, Arun will be twice as old as Sarmi. Find the Present ages of Arun and sarmi? a) 30 years, 10.5 years b) 32 years, 11.5 years c) 34 years, 14.5 years d) 35 years, 12.5 years e) None of these

Last Answer : Five years ago, x – 5/y – 5 = 4/1 x – 4y = – 15 —-(i) After 10 years, x + 10/y + 10 = 2/1 x – 2y = 10 —–(ii) From eqn (i) and (ii) => x = 35, y = 12.5 Answer: d)

f the mass is of 10 Kg, find the natural frequency in Hz of the free longitudinal vibrations. The displacement is 0.01mm. a) 44.14 b) 49.85 c) 43.43 d) 46.34

Description : f the mass is of 10 Kg, find the natural frequency in Hz of the free longitudinal vibrations. The displacement is 0.01mm. a) 44.14 b) 49.85 c) 43.43 d) 46.34

Last Answer : b) 49.85

Find the natural frequency in Hz of the free longitudinal vibrations if the displacement is 2mm. a) 11.14 b) 12.38 c) 11.43 d) 11.34

Description : Find the natural frequency in Hz of the free longitudinal vibrations if the displacement is 2mm. a) 11.14 b) 12.38 c) 11.43 d) 11.34

Last Answer : a) 11.14

How many biosphere reserves are present in India ? A. 41 B. 34 C. 14 D. 43

Description : How many biosphere reserves are present in India ? A. 41 B. 34 C. 14 D. 43

Last Answer : C. 14

If the mean of the observations x, x + 3, x + 5, x + 7and x + 10 is 9, then mean of the last three observations is -Maths 9th

Description : If the mean of the observations x, x + 3, x + 5, x + 7and x + 10 is 9, then mean of the last three observations is -Maths 9th

Last Answer : NEED ANSWER

If the mean of the observations x, x + 3, x + 5, x + 7and x + 10 is 9, then mean of the last three observations is -Maths 9th

Description : If the mean of the observations x, x + 3, x + 5, x + 7and x + 10 is 9, then mean of the last three observations is -Maths 9th

Last Answer : According to question find the mean of the last three observations

Three pipes P, Q and R can fill a tank from empty to full in 40 minutes, 15 minutes, and 30 minutes respectively. When the tank is empty, all the three pipes are opened. P, Q and R discharge chemical solutions X, Y and Z respectively. What is the proportion of the solution Y in the liquid in the tank after 5 minutes? a) 8 / 15 b) 7 / 15 c) 8 / 17 d) 6 / 13 e) 8 / 13

Description : Three pipes P, Q and R can fill a tank from empty to full in 40 minutes, 15 minutes, and 30 minutes respectively. When the tank is empty, all the three pipes are opened. P, Q and R discharge chemical solutions X, Y and Z respectively. ... 5 minutes? a) 8 / 15 b) 7 / 15 c) 8 / 17 d) 6 / 13 e) 8 / 13

Last Answer : a Part of the tank filled by pipe P in 1 minute = 1 / 40 Part of the tank filled by pipe Q in 1 minute = 1 / 15 Part of the tank filled by pipe R in 1 minute = 1/ 30 Here we have to find the proportion of ... together in 5 minute = 5 1/8 = 5/ 8 Required proportion = (1/3) / ( 5/8) = 8 / 15

Let A and B be two fuzzy integers defined as: A={(1,0.3), (2,0.6), (3,1), (4,0.7), (5,0.2)} B={(10,0.5), (11,1), (12,0.5)} Using fuzzy arithmetic operation given by Image (A) {(11,0.8), (13,1), (15,1)} (B) {(11,0.3), (12,0.5), (13,1), (14,1), (15,1), (16,0.5), (17,0.2)} (C) {(11,0.3), (12,0.5), (13,0.6), (14,1), (15,1), (16,0.5), (17,0.2)} (D) {(11,0.3), (12,0.5), (13,0.6), (14,1), (15,0.7), (16,0.5), (17,0.2)}

Description : Let A and B be two fuzzy integers defined as: A={(1,0.3), (2,0.6), (3,1), (4,0.7), (5,0.2)} B={(10,0.5), (11,1), (12,0.5)} Using fuzzy arithmetic operation given by (A) {(11,0.8), (13,1), (15,1)} ( ... ,0.2)} (D) {(11,0.3), (12,0.5), (13,0.6), (14,1), (15,0.7), (16,0.5), (17,0.2)}

Last Answer : (D) {(11,0.3), (12,0.5), (13,0.6), (14,1), (15,0.7), (16,0.5), (17,0.2)}