What is the mean of 12,34,15,3,16? -Maths 9th

What is the mean of 12,34,15,3,16? -Maths 9th
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What is the mean of 12,34,15,3,16? -Maths 9th

Description : What is the mean of 12,34,15,3,16? -Maths 9th

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

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, 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

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.

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

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.

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm

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.

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Last Answer : (d) Given, AD = 34 cm and AB = 30 cm In figure, draw OL ⊥ AB. Since, the perpendicular from the centre of a circle to a chord bisects the chord.

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm

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.

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Last Answer : (d) Given, AD = 34 cm and AB = 30 cm In figure, draw OL ⊥ AB. Since, the perpendicular from the centre of a circle to a chord bisects the chord.

The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Description : The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 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 median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Description : The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Last Answer : According to question find the median of the data

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.

If x2 + 1/x2 = 34, find x3 + 1/x3 - 9. -Maths 9th

Description : If x2 + 1/x2 = 34, find x3 + 1/x3 - 9. -Maths 9th

Last Answer : Solution :-

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

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Description : The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Last Answer : Length of the plank=4m=400cm Breadth=50cm Height=20cm Volume of the plank=L*B*H =400*50*20 =400000cm^3 Length of the pit=16m=1600cm Breadth=12m=1200cm Height=4m=400cm Volume of the pit= L ... *1200*400 =768000000cm^3 Number of planks that can be fitted= 768000000/400000 =1920 planks is the answer.

The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Description : The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Last Answer : Solution of this question

Find the value of 64x3 – 125z3, if 4x – 5z = 16 and xz = 12. -Maths 9th

Description : Find the value of 64x3 – 125z3, if 4x – 5z = 16 and xz = 12. -Maths 9th

Last Answer : answer:

Find the area of a triangle whose sides are 12 cm, 6 cm and 15 cm. -Maths 9th

Description : Find the area of a triangle whose sides are 12 cm, 6 cm and 15 cm. -Maths 9th

Last Answer : Using the formulas A=s(s﹣a)(s﹣b)(s﹣c) s=a+b+c 2Solving forA A=1 4﹣a4+2(ab)2+2(ac)2﹣b4+2(bc)2﹣c4=1 4·﹣124+2·(12·6)2+2·(12·15)2﹣64+2·(6·15)2﹣154≈34.19704cm²

Divide: 12 15 under root 4 by 8 cube root 3. -Maths 9th

Description : Divide: 12 15 under root 4 by 8 cube root 3. -Maths 9th

Last Answer : Solution :-

Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Description : Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Last Answer : We have, Three sides13cm,13cm and 20cm. By using Heron's formula We need to get the semi-perimeter s= 2 a+b+c​ = 2 13+13+20​ = 2 46​ =23 Now, put the heron's formula, s= s(s−a)(s−b)(s−c)​ = 23(23−13)(23−13)(23−20)​ = 23×10×10×3​ =10 23×3​ =83.07cm 2

The lengths of three medians of a triangle are 9 cm, 12 cm and 15 cm. The area (in sq. cm) of this triangle is -Maths 9th

Description : The lengths of three medians of a triangle are 9 cm, 12 cm and 15 cm. The area (in sq. cm) of this triangle is -Maths 9th

Last Answer : (b) 72 cm2Here sm = \(rac{9+12+15}{2}\) = 18 cm, where lengths of medians are m1 = 9 cm, m2 = 12 cm, m3 = 15 cm.∴ Area of triangle = \(rac{4}{3}\sqrt{18(18-9)(18-12)(18-15)}\) cm2= \(rac{4}{3}\sqrt{18 imes9 imes6 imes3}\) cm2 = \(rac{4}{3}\) x 9 x 6 cm2 = 72 cm2.

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

Ratio of earnings of A and B is 3:5. If the earnings of A increase by 25% and those of B decrease by 30%, the new ratio of their earnings become 6:8.What is A’s earnings ? A) 16,000 B) 34,000 C) 15,000 D) 32,000 E) Data inadequate

Description : Ratio of earnings of A and B is 3:5. If the earnings of A increase by 25% and those of B decrease by 30%, the new ratio of their earnings become 6:8.What is A’s earnings ? A) 16,000 B) 34,000 C) 15,000 D) 32,000 E) Data inadequate

Last Answer : Answer: E Data inadequate We cannot derive the earnings from the given details

The population of Balochistan according to 2017 Census is: A. 16 Million B. 12.34 Million C. 18 Million D. None of these

Description : The population of Balochistan according to 2017 Census is: A. 16 Million B. 12.34 Million C. 18 Million D. None of these

Last Answer : ANSWER: B

The population of Balochistan according to 2017 Census is: A. 16 Million B. 12.34 Million C. 18 Million D. None of these

Description : The population of Balochistan according to 2017 Census is: A. 16 Million B. 12.34 Million C. 18 Million D. None of these

Last Answer : ANSWER: B

The population of Balochistan according to 2017 Census is: A. 16 Million B. 12.34 Million C. 18 Million D. None of these

Description : The population of Balochistan according to 2017 Census is: A. 16 Million B. 12.34 Million C. 18 Million D. None of these

Last Answer : ANSWER: B

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

Description : A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

Last Answer : NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula Ex 12.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths ... = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m S

A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. -Maths 9th

Description : A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. -Maths 9th

Last Answer : Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm Formula for Surface area of hemispherical bowl = 2πr2 = 2 (22/7) (5.25)2 = 173.25 Surface area of hemispherical ... of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.

For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Description : For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Description : For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th

Description : Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th

Last Answer : Volume of rectangular box=lbh=16(64)=1024cm3 Volume of sphere=34​πr3=33.5238cm3 16 sphere=16(33.5238)=536.3808 Volume of liquid=1024−536.3808=488cm3

Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th

Description : Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th

Last Answer : According to question find the volume of this liquid.

Find the value of (16) to the power 3/4 /(16) to the power 1/4. -Maths 9th

Description : Find the value of (16) to the power 3/4 /(16) to the power 1/4. -Maths 9th

Last Answer : Solution :-

If 4 to the power 2x-1 - 16 to the power x-1 = 348,Find the value of x. -Maths 9th

Description : If 4 to the power 2x-1 - 16 to the power x-1 = 348,Find the value of x. -Maths 9th

Last Answer : Solution :-

In Fig. 10.33, if OA = 10cm, AB = 16 cm and OD perpendicular to AB. Find the value of CD. -Maths 9th

Description : In Fig. 10.33, if OA = 10cm, AB = 16 cm and OD perpendicular to AB. Find the value of CD. -Maths 9th

Last Answer : Solution :- As OD is perpendicular to AB ⇒ AC = AB (Perpendicular from the centre to the chord bisects the chord) ∴ AC = AB/2 = 8cm In right △OCA, OA2 = AC2 + OC2 (102) = 82 + OC2 OC2 = 100 - 64 OC2 = 36 ... = 6cm CD = OD - OC = 10 - 6 = 4cm [∴ OA = OD = 10cm (radii)]

If the side of a rhombus is 10 cm and the diagonal is 16 cm,...... -Maths 9th

Description : If the side of a rhombus is 10 cm and the diagonal is 16 cm,...... -Maths 9th

Last Answer : True. AC = 16 cm BD = ? and AB = 10 cm As the diagonals of a rhombus bisect each other at 90° ∴ OA = 1/2AC = 1/2 x 16 = 8cm OB = 1/2 BD ∴ OA2 + OB2 = AB2 82 + OB2 = 102 ⇒ OB2 = 100 - 64 OB2 = 36 ... ∴ BD = 2 x OB = 2 x 6 = 12 cm Area of rhombus = 1/2 AC x BD = 1/2 x 16 x 12 = 96cm 2

A floral design on a floor is made up of 16 tiles which are triangular, ... -Maths 9th

Description : A floral design on a floor is made up of 16 tiles which are triangular, ... -Maths 9th

Last Answer : Measures of the sides of the triangular tile are 28 cm, 9 cm and 35 cm. Let a = 28 cm, b = 9 cm, c = 35 cm Semi-perimeter, s = (a + b + c)/2 = (28 + 9 + 35)/2 = 36 cm ∴ Area of one ... 1411.2 cm2 Hence, cost of polishing the tiles at the rate of ₹ 1/2 per cm2 = ₹ 1/2 x 1411.2 = ₹ 705.60

ABCD is a parallelogram AE pependicular to DC CF perpendixular to AD AB =16 m ,AE =8m ,CF =10m ,fimd AD -Maths 9th

Description : ABCD is a parallelogram AE pependicular to DC CF perpendixular to AD AB =16 m ,AE =8m ,CF =10m ,fimd AD -Maths 9th

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

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!

If loge (x^2 – 16) < loge (4x – 11), then -Maths 9th

Description : If loge (x^2 – 16) < loge (4x – 11), then -Maths 9th

Last Answer : answer: