A field is 70 m long and 40 m broad. -Maths 9th

A field is 70 m long and 40 m broad. -Maths 9th
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1 Answer

Answer :

Area of the field on which earth taken out is to be spread = 70 X 40 m2 - 10 X 8 m2  = 2800 m2 - 80 m2 = 2720 m2  Volume of the earth dug out = 10 X 8 X 5 m3 = 400 m3 Rise in level of the field =  Volume of the earth dugout/ Area on which earth taken out is to be spread = 400/ 2720    = 0.147 m = 14.7 cm     
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Related questions

A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Description : A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Last Answer : S(△ABC)=60+80+402=90S(△ABC)=60+80+402=90 ar△ABDar△ABD =90(90−80)(90−60)(90−40)−−−−−−−−−−−−−−−−−−−−−−−√=90(90−80)(90−60)(90−40) =90×10×30×50−−−−−−−−−−−−−−√=90×10×30×50 =30015−−√m2=30015m2 ar□ABCE=2×ar△ABDar◻ABCE=2×ar△ABD =60015−−√m2

A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Description : A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Last Answer : Area of the parallelogram

The sides of a triangular field are 41 m, 40 m and 9 m. -Maths 9th

Description : The sides of a triangular field are 41 m, 40 m and 9 m. -Maths 9th

Last Answer : Let a = 41m, b = 40 m, c = 9 m. s = (a + b + c)/2 = (41 + 40 +9)/2 = 90/2 ⇒ s = 45 m Area of the triangular field = root under( √s(s - a)(s - b)(s -c)) = root under( ... x 5 x 36 ) = 180 m2 = 1800000 cm2 Number of rose beds = Total area / Area needed for one rose bed = 1800000/900 = 2000

Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3 -Maths 9th

Description : Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3 -Maths 9th

Last Answer : The given pit has its length(l) as 8m, width (b)as 6m and depth (h)as 3 m. Volume of cuboidal pit = l×b×h = (8×6×3) = 144 (using formula) Required Volume is 144 m3 Now, Cost of digging per m3 volume = Rs 30 Cost of digging 144 m3 volume = Rs (144×30) = Rs 4320

A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Last Answer : Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 - 3 - 3 = 34 m and breadth of inner-rectangle, FG =15 - 2 - 2 = ... [∴ area of a rectangle = length x breadth] Hence, the largest area where the house can be constructed in 374 m2

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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.

A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Last Answer : Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 - 3 - 3 = 34 m and breadth of inner-rectangle, FG =15 - 2 - 2 = ... [∴ area of a rectangle = length x breadth] Hence, the largest area where the house can be constructed in 374 m2.

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

A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th

Description : A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th

Last Answer : Given: Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m Height of cone, h = 1m Slant height of cone is l, and l2 = (r2+h2) Using given values, l2 = (0.22+12) = (1.04) Or l ... (32.028 12) = Rs.384.336 = Rs.384.34 (approximately) Therefore, the cost of painting all these cones is Rs. 384.34.

In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Description : In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Last Answer : Solution :-

A river 3 m deep and 40 m wide -Maths 9th

Description : A river 3 m deep and 40 m wide -Maths 9th

Last Answer : Length of water canal in one minute = 2 x 1000 m/60 = 100/3 m Volume of water flowing into the sea in one minute = l x b x h = (100 x 40 x 3)/3 = 4000 m 3 = 4000 x 1000 L = 4000000 L ( Since,1 m3 = 1000 L)

The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. -Maths 9th

Description : The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. -Maths 9th

Last Answer : this is the answer

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

A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Description : A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Last Answer : Here, each side of the rhombus = 30 m. Let ABCD be the given rhombus and the diagonal, BD = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m Since, a diagonal divides the rhombus into ... Area of grass for 18 cows to graze = 864 m2 ⇒ Area of grass for 1 cow to graze = 86418 m2 = 48 m2

Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2. -Maths 9th

Description : Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2. -Maths 9th

Last Answer : Sides of the triangle are a=50m,b=65m,c=65m Area of triangle, by Heron's formula =s(s−a)(s−b)(s−c)​where, s=2a+b+c​s=250+65+65​s=90 Area of triangle = 90(40)(25)(25)​Area of triangle = 1500m2 Cost of laying grass = Area ×7 Cost of laying grass =1500×7 Cost of laying grass = Rs 10500

The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

Description : The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

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A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Description : A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Last Answer : NEED ANSWER

Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2. -Maths 9th

Description : Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2. -Maths 9th

Last Answer : Sides of the triangle are a=50m,b=65m,c=65m Area of triangle, by Heron's formula =s(s−a)(s−b)(s−c)​where, s=2a+b+c​s=250+65+65​s=90 Area of triangle = 90(40)(25)(25)​Area of triangle = 1500m2 Cost of laying grass = Area ×7 Cost of laying grass =1500×7 Cost of laying grass = Rs 10500

The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

Description : The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

Last Answer : Let sides of △ are=6x,7x and 8x Perimeter=6x+7x+8x=21x 21x=410 x=20 Sides are 120,140 and 160 m Area =S(S−A)(S−B)(S−C)​ [Heron's Formula] S=2120+140+160​=210 m A=210(210−120)(210−140)(210−160)​=210015​ sq. m

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Description : A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Last Answer : In trapezium ABCD, we draw a perpendicular line CE to the line AB.

In a rectangular field of dimension 60 m x 50 m, -Maths 9th

Description : In a rectangular field of dimension 60 m x 50 m, -Maths 9th

Last Answer : Area of rectangular field = length x breadth = 60 x 50 = 3,000 m2 Now, a = 50 m, b = 45 m, c = 35 m s = (a + b + c)/2 = (50 + 45 + 35)/2 = 130/2 = 65 m By Heron's formula ... m2 ( approximately ) Hence, the remaining area = Area of rectangle - Area of triangle = 3,000 - 764.85 = 2,235.15 m2

The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, -Maths 9th

Description : The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, -Maths 9th

Last Answer : (c) 67.5 mGiven 2s = a + b + c ⇒ 240 = 78 + 50 + Third side ⇒ Third side = 240 m - 128 m = 112 m.∴ Area of Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\)= \(\sqrt{120(120-78)(120-50)(120-112)}\)= \(\sqrt{120 imes42 imes70 ... (rac{1}{2}\)x b x h ∴ \(rac{1}{2}\) x 50 x h = 1680 ⇒ h = \(rac{1680}{25}\) = 67.5 m.

Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm. -Maths 9th

Description : Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm. -Maths 9th

Last Answer : Length of cuboid (l) = 80 cm Breadth (b) = 40 cm Height (h) = 20 cm (i) ∴ Lateral surface area = 2h(l + b) = 2 x 20(80 + 40) cm² = 40 x 120 = 4800 cm² (ii) Total surface area = 2(lb ... x 40 + 40 x 20 + 20 x 80) cm² = 2(3200 + 800 + 1600) cm² = 5600 x 2 = 11200 cm²

The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Description : The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Last Answer : Inner radius of circular well, r = 3.5/2m = 1.75m Depth of circular well, say h = 10m (i) Inner curved surface area = 2πrh = (2 (22/7 ) 1.75 10) = 110 Therefore, the inner curved surface ... area = Rs (110 40) = Rs.4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Description : PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Last Answer : Here, PQ = SR = 12cm Let PS = x and PS = QR ∴ x + 12 + x +12 = Perimeter 2x + 24 = 40 2x = 16 x = 8 Hence, length of each side of the parallelogram is 12cm , 8cm , 12cm and 8cm.

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

The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Description : The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Last Answer : Let the 8 observations are x1, x2, x3, x4, x5, x6, x7, x8 ∴ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 = 40 × 8 = 320 New mean = 320 + 5 × 8 = 360 / 8 = 45

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

The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

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

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Description : PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Last Answer : Here, PQ = SR = 12cm Let PS = x and PS = QR ∴ x + 12 + x +12 = Perimeter 2x + 24 = 40 2x = 16 x = 8 Hence, length of each side of the parallelogram is 12cm , 8cm , 12cm and 8cm.

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

The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Description : The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Last Answer : Let the 8 observations are x1, x2, x3, x4, x5, x6, x7, x8 ∴ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 = 40 × 8 = 320 New mean = 320 + 5 × 8 = 360 / 8 = 45

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

The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

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

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Last Answer : Cost of painting =

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

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

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Last Answer : Cost of painting =

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.

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.

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

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

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.

In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Description : In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Last Answer : Solution :-

The distance (in km) of 40 engineers -Maths 9th

Description : The distance (in km) of 40 engineers -Maths 9th

Last Answer : Frequency distribution of above data in tabular form is given as : It is observed that 36 engineers out of 40 live at a distance not more than 20 km from their residence.

The length of 40 leaves of a plant are -Maths 9th

Description : The length of 40 leaves of a plant are -Maths 9th

Last Answer : (i) Consider the class 118 - 126 and 127 - 135 The lower limit of 127 - 135 = 127 The upper limit of 118 - 126 = 126 Half of the difference = 127 -126/2 = 0.5 So, the new class interval ... , frequency polygon. (ii) No, this frequency includes all leaves whose length are from 144.5 mm to 153.5 mm.

In a survey of 364 children aged 20-40 months, -Maths 9th

Description : In a survey of 364 children aged 20-40 months, -Maths 9th

Last Answer : Number of children = 364 Number of children not like to eat potato chips = 364 - 91 = 273. The required probability = 273/364 = 0.75

Bulbs are packed in cartons each containing 40 bulbs. -Maths 9th

Description : Bulbs are packed in cartons each containing 40 bulbs. -Maths 9th

Last Answer : (i) P (a carton has no defective bulb) = 400/700 = 4/7 (ii) P (defective bulbs from 2 to 6) = P (2 defective bulbs) + P (3 defective bulbs) + P (4 defective bulbs) + P (5 defective bulbs) + P (6 ... + P (2 defective bulbs) + P (3 defective bulbs) = 400/700 + 180/700 + 48/700 + 41/700 = 669/700

When x^40 + 2 is divided by x^4 + 1, what is the remainder ? -Maths 9th

Description : When x^40 + 2 is divided by x^4 + 1, what is the remainder ? -Maths 9th

Last Answer : answer:

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400