The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th
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The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Description : The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Last Answer : Here's ur answer..

Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Description : Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Last Answer : Mean of 36 observations = 12 Total of 36 observations = 36 x 12 = 432 Correct sum of 36 observations = 432 – 74 + 47 = 405 Correct mean of 36 observations = 405/ 36 =11.25

Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Description : Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Last Answer : Mean of 36 observations = 12 Total of 36 observations = 36 x 12 = 432 Correct sum of 36 observations = 432 – 74 + 47 = 405 Correct mean of 36 observations = 405/ 36 =11.25

2 male, 6 female, and 8 girls can do a work in 192 hrs, 4 male and 16 girls can do it in 160 hrs, 4 male and 6 female can do it in 240 hrs.10 male and 24 girls can do it in. a) A.87 3/ 11 hrs b) B.78 4 /11 hrs c) C.88 6 /12 hrs d) 80 4 /5 hrs

Description : 2 male, 6 female, and 8 girls can do a work in 192 hrs, 4 male and 16 girls can do it in 160 hrs, 4 male and 6 female can do it in 240 hrs.10 male and 24 girls can do it in. a) A.87 3/ 11 hrs b) B.78 4 /11 hrs c) C.88 6 /12 hrs d) 80 4 /5 hrs

Last Answer : A Let 1 males 1 hr work = x 1 female 1 hour work = y 1 girl 1 hour work = z Then 2x + 6y + 8z = 1/192 ------>1 4x + 16z = 1/160 --------->2 4x + 6y = 1/240 ------------->3 Adding 2 and ... 1/192 + 1/160 =(5+6) / 960 = 11/960 10 male and 24 girls can do work in 960/11 hrs i.e., 87 3 /11 hrs

At normal boiling point, molar entropy of vaporisation is __________ Joule/K°.mole. (A) 72 (B) 92 (C) 142 (D) 192

Description : At normal boiling point, molar entropy of vaporisation is __________ Joule/K°.mole. (A) 72 (B) 92 (C) 142 (D) 192

Last Answer : (B) 92

The area of parallelogram PQRS is 88 cm sq. A perpendicular from S is drawn to intersect PQ at M. If SM = 8 cm, then find the length of PQ. -Maths 9th

Description : The area of parallelogram PQRS is 88 cm sq. A perpendicular from S is drawn to intersect PQ at M. If SM = 8 cm, then find the length of PQ. -Maths 9th

Last Answer : Given area of parallelogram = 88 cm² And SM = 8cm Area of a parallelogram = height × base (Height is the measurement of a perpendicular drawn from one side to other) Here, Area of PQRS = SM × PQ 88cm² = 8cm × PQ 11cm = PQ

Which of the four data sets have more dispersion? (A) 88 91 90 92 89 91 (B) 0 1 1 0 –1 –2 (C) 3 5 2 4 1 5 (D) 0 5 8 10 –2 –8

Description : Which of the four data sets have more dispersion? (A) 88 91 90 92 89 91 (B) 0 1 1 0 –1 –2 (C) 3 5 2 4 1 5 (D) 0 5 8 10 –2 –8

Last Answer : Answer: D

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..

Write each of the following as decimals: (a) 5 / 10 (b) 3 + 7 / 10 (c) 200 + 60 + 5 + 1 / 10 (d) 70 + 8 / 10 (e) 88 / 10 -Maths

Description : Write each of the following as decimals: (a) 5 / 10 (b) 3 + 7 / 10 (c) 200 + 60 + 5 + 1 / 10 (d) 70 + 8 / 10 (e) 88 / 10 -Maths

Last Answer : (a) 5 / 10 = 0.5 (b) 3 + 7 / 10 = 3 + 0.7 = 3.7 (c) 200 + 60 + 5 + 1 / 10 = 265 + 0.1 = 265.1 (d) 70 + 8 / 10 = 70 + 0.8 = 70.8 (e) 88 / 10 = 80 / 10 + 8 / 10 = 8 + 0.8 = 8.8 (f) (g) 3 / 2 = (2 + 1) / ... + 0.5 = 1.5 (h) 2 / 5 = 0.4 (i) 12 / 5 = (10 + 2) / 5 = 10 / 5 + 2 / 5 = 2 + 0.4 = 2.4 (j) (k)

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.

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

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

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

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

A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Last Answer : NEED ANSWER

A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Last Answer : (b) We arrange the given data into groups like 13-22,23-32 103-112. (since, our data is from 14 to 112). The class width in this case is 9. Now, the given data can be arranged in tabular form as follows. Hence, the number of classes in distribution will be 10.

What is the average of 96 95 92 89 88?

Description : What is the average of 96 95 92 89 88?

Last Answer : 92

In the following partial sequence of mRNA, a mutation of the template DNA results in a change in codon 91 to UAA. The type of mutation is 88 89 90 91 92 93 94 GUC GAC CAG UAG GGC UAA CCG (A) Missene (B) Silent (C) Nonsense (D) Frame shit

Description : In the following partial sequence of mRNA, a mutation of the template DNA results in a change in codon 91 to UAA. The type of mutation is 88 89 90 91 92 93 94 GUC GAC CAG UAG GGC UAA CCG (A) Missene (B) Silent (C) Nonsense (D) Frame shit

Last Answer : Answer : B

Presence of nitrogen in high concentration in contaminated air reduces partial pressure of oxygen in lungs, thereby causing asphyxia (suffocation) leading to death from oxygen deficiency. Concentration of N2 in contaminated air at which it acts as a natural asphyxiate is ≥ __________ percent. (A) 84 (B) 88 (C) 80 (D) 92

Description : Presence of nitrogen in high concentration in contaminated air reduces partial pressure of oxygen in lungs, thereby causing asphyxia (suffocation) leading to death from oxygen deficiency. Concentration of N2 in contaminated air at ... asphyxiate is ≥ __________ percent. (A) 84 (B) 88 (C) 80 (D) 92

Last Answer : (A) 84

In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Description : In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Last Answer : 2 boys, 1 gi(Prl = (15c2×10c1) / 25c3 = 1050/2300 1 boy, 2 girls = (15c1×10c2) / 25c3 = 675/2300 Difference = (1050 - 675)/2300 = 375/2300 = 15/92 Answer: c)

67,74,78,?,92,94 a) 86 b) 85 c) 83 d) 84 e) 88

Description : 67,74,78,?,92,94 a) 86 b) 85 c) 83 d) 84 e) 88

Last Answer : The next term is obtained by adding the unit’s digit to the number. Hence the missing number =78+8=86 Answer: a)

What is the approximate percentage of candidates successful to enrolled from institute A? a) 92 b) 89 c) 86 d) 96 e) 88

Description : What is the approximate percentage of candidates successful to enrolled from institute A? a) 92 b) 89 c) 86 d) 96 e) 88

Last Answer : c) 86

180 sweets are divided among friends A, B, C and D in which B and C are brothers also such that sweets divided between A and B are in the ratio 2 : 3, between B and C in the ratio 2 : 5 and between C and D in ratio 3 : 4. What is the number of sweets received by the brothers together? A) 78 B) 84 C) 92 D) 102 E) 88

Description : 180 sweets are divided among friends A, B, C and D in which B and C are brothers also such that sweets divided between A and B are in the ratio 2 : 3, between B and C in the ratio 2 : 5 and between C ... . What is the number of sweets received by the brothers together? A) 78 B) 84 C) 92 D) 102 E) 88

Last Answer : Answer: B A : B : C : D 2*2*3 : 3*2*3 : 3*5*3 : 3*5*4 4 : 6 : 15 : 20 B and C together = [(6+15)/(4+6+15+20)] * 180 = 84

A number 168 was misread as 24. Find the reading error in per cent. a) 25.7 b) 42.8 c) 42.6 d) 85.7 e) 98.6

Description : A number 168 was misread as 24. Find the reading error in per cent. a) 25.7 b) 42.8 c) 42.6 d) 85.7 e) 98.6

Last Answer : Answer: D Error = 168 – 24 = 144 Therefore, % error = 144/168 [Since, we know decrease% = decrease in value/original value × 100 %]  = 144/168 × 100  = 85.7 %

Calculate the most appropriate unit cost for a distribution division of a multinational company using the following information. Miles travelled 636,500 Tonnes carried 2,479 Number of drivers 20 Hours worked by drivers 35,520 Tonnes miles carried 375,200 Cost incurred 562,800 (a) Rs .88 (b) Rs 1.50 (c) Rs 15.84 (d) Rs28, 140

Description : Calculate the most appropriate unit cost for a distribution division of a multinational company using the following information. Miles travelled 636,500 Tonnes carried 2,479 Number of drivers 20 Hours worked by drivers 35,520 Tonnes miles ... 800 (a) Rs .88 (b) Rs 1.50 (c) Rs 15.84 (d) Rs28, 140

Last Answer : (d) Rs28, 140

When the integers from 1 to 999 are written on a paper, the number of zeros used in writing it is: a) 192 b) 189 c) 200 d) 203 e) 216

Description : When the integers from 1 to 999 are written on a paper, the number of zeros used in writing it is: a) 192 b) 189 c) 200 d) 203 e) 216

Last Answer : From 1 to 10 : 1 zero From 11-100 : 10 zeros (100 has 2 zeros) From 101 – 900 : 8×20 = 160 zeros From 901-999 : we have 18 zeros Total = 1+10+160+18 = 189 zeros Answer: b)

Meeta purchased 35 rings at the rate of Rs. 160 per ring. At what rate per ring should she sell it so that profit earned is 20%? (a) Rs. 180 (b) Rs. 186 (c) Rs. 192 (d) Rs. 194 (e) Rs. 200

Description : Meeta purchased 35 rings at the rate of Rs. 160 per ring. At what rate per ring should she sell it so that profit earned is 20%? (a) Rs. 180 (b) Rs. 186 (c) Rs. 192 (d) Rs. 194 (e) Rs. 200

Last Answer : (c) Rs. 192

Is there a word you always, always misread?

Description : Is there a word you always, always misread?

Last Answer : answer:There’s one I miswrite: ever since I started playing the ACCORDION, I can no longer write ACCORDING. “Accordion to Darwin…”

What funny examples can you recall as a kid when you misread a word?

Description : What funny examples can you recall as a kid when you misread a word?

Last Answer : I used to pronounce “media” as “med-ee-ah” (like medicine) my mother tells me of a childhood friend, who pronounced “grotesque” as “grot-ess-cue”

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

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!!!

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

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!!!

Find current density J when B = 50 x 10-6 units and area dS is 4 units. a) 9.94 b) 8.97 c) 7.92 d) 10.21

Description : Find current density J when B = 50 x 10-6 units and area dS is 4 units. a) 9.94 b) 8.97 c) 7.92 d) 10.21

Last Answer : b) 8.97

Average cost of 10 pens is rupees 200 & average cost of 10 pencils is 100 rupees find the average cost of the all pens and pencils​ -Maths 9th

Description : Average cost of 10 pens is rupees 200 & average cost of 10 pencils is 100 rupees find the average cost of the all pens and pencils​ -Maths 9th

Last Answer : 10 pens of average cost = 200 => 10×200 = 2000 rupees 10 pencils average cost = 100 rupees => 10×100=1000 Pens + Pencils = 20 Total average 2000+1000 = 3000 Rupees 150 is average for average cost of each pen and pencil ✏

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Last Answer : Total number of chances = 23 + 72 + 77 + 28 = 200 Number of chances of coming 2 heads = 72 therefore P( coming 2 heads)= 514 / 642 = 9 / 25

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Last Answer : It is given that coin is tossed 200 times Total number of trials = 200 Number of events for getting less than three tails = 68 + 82 + 30 = 180 Probability of getting less than 3 tails =180 / 200 =9 / 10

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Last Answer : Total number of chances = 23 + 72 + 77 + 28 = 200 Number of chances of coming 2 heads = 72 therefore P( coming 2 heads)= 514 / 642 = 9 / 25

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Last Answer : It is given that coin is tossed 200 times Total number of trials = 200 Number of events for getting less than three tails = 68 + 82 + 30 = 180 Probability of getting less than 3 tails =180 / 200 =9 / 10

Over the past 200 working days, -Maths 9th

Description : Over the past 200 working days, -Maths 9th

Last Answer : (i) P (no defective part) = 50/200 = 0.25 (ii) P (at least one defective part) = 1 - P (no defective part) = 1 - 0.25 = 0.75 (iii) P (not more than 5 defective parts) = P (no defective part) + P (1 ... (5 defective parts) = 50/200 + 32/200 + 22/200 + 18/200 + 12/200 + 12/200 = 146/200 = 0.73

The remainder, when x^(200) is divided by x^2 – 3^x + 2 is -Maths 9th

Description : The remainder, when x^(200) is divided by x^2 – 3^x + 2 is -Maths 9th

Last Answer : answer:

Write the following decimals in the place value table. (a) 0.29 (b) 2.08 (c) 19.60 (d) 148.32 (e) 200.812 -Maths 9th

Description : Write the following decimals in the place value table. (a) 0.29 (b) 2.08 (c) 19.60 (d) 148.32 (e) 200.812 -Maths 9th

Last Answer : (a) 0.29 = 0.2 + 0.09 = 2 / 10 + 9 / 100 (b) 2.08 = 2 + 0.08 = 2 + 8 / 100 (c) 19.60 = 19 + 0.60 = 10 + 9 + 6 / 10 (d) 148.32 = 148 + 0.3 + 0. ... / 100 (e) 200.812 = 200 + 0.8 + 0.01 + 0.002 =200 + 8 / 10 + 1 / 100 + 2 / 1000HundredsTensOnesTenthsHundredthsThousandths000290002080019600148320200812

Calculate the energy released by fission of 1 g of `._(92)^(235)U`, assuming that an energy of 200 MeV is released by fission of each atom of `.^(235)

Description : Calculate the energy released by fission of 1 g of `._(92)^(235)U`, assuming that an energy of 200 MeV is released by fission of each atom of `.^(235)

Last Answer : Calculate the energy released by fission of 1 g of `._(92)^(235)U`, assuming that an energy of 200 MeV is ... `= 6.023 xx 10^(26) kg mol^(-1)`)

Number of secondary neutron emitted on fission of an atom of U-235 by slow neutron bombardment is (A) 3 (B) 235 (C) 200 (D) 92

Description : Number of secondary neutron emitted on fission of an atom of U-235 by slow neutron bombardment is (A) 3 (B) 235 (C) 200 (D) 92

Last Answer : (A) 3

Who discovered the nucleus of the cell ? -Maths 9th

Description : Who discovered the nucleus of the cell ? -Maths 9th

Last Answer : answer:

Who discovered Cell ? -Maths 9th

Description : Who discovered Cell ? -Maths 9th

Last Answer : answer:

Write the output of the following SQL command. select round (49.88); (a)49.88 (b)49.8 (c)49.0 (d)50 -Technology

Description : Write the output of the following SQL command. select round (49.88); (a)49.88 (b)49.8 (c)49.0 (d)50 -Technology

Last Answer : The correct answer is : (a) 50

Efficiency of power transmission (η) through a circular pipe is given by (ht - hf )/ht , which has a maximum value of __________ percent. (A) 33:3 (B) 50 (C) 66.6 (D) 88.8

Description : Efficiency of power transmission (η) through a circular pipe is given by (ht - hf )/ht , which has a maximum value of __________ percent. (A) 33:3 (B) 50 (C) 66.6 (D) 88.8

Last Answer : (C) 66.6

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be -Maths 9th

Description : The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be -Maths 9th

Last Answer : NEED ANSWER

There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Last Answer : NEED ANSWER

The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be -Maths 9th

Description : The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be -Maths 9th

Last Answer : Solution of this question