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

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

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

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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

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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.

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

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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

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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

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IfP (event E) = 0.47, then find P(not E). -Maths 9th

Description : IfP (event E) = 0.47, then find P(not E). -Maths 9th

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Last Answer : 22

Assume postal rates for 'light letters' are: $0.25 up to 10 grams; $0.35 up to 50 grams; $0.45 up to 75 grams; $0.55 up to 100 grams. Which test inputs (in grams) would be selected using boundary value analysis? A. 0, 9,19, 49, 50, 74, 75, 99,100 B. 10, 50, 75,100, 250, 1000 C. 0, 1,10,11, 50, 51, 75, 76,100,101 D. 25, 26, 35, 36, 45, 46, 55, 56

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Description : The average age of 160 boys in a class is 58 yrs. The average group of 30 boys in the class is 42 yrs and the average of another group of 50 boys in the class is 36 years. What is the average age of the remaining boys? A) 72.58 B) 74.25 C) 77.75 D) 75.68 

Last Answer : C) Total age of 160 boys = 160* 58= 9280 total age of 30 boys = 30 * 42= 1260 total age of next 50boys = 50 * 36= 1800 average of the remaining boys = [(9280-{1260+1800})/[160 - (30 + 50)] =>9280-3060/80 =>6220/80 =77.75yrs

A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th

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Description : In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. -Maths 9th

Last Answer : NEED ANSWER

(36)^7/2 - (36)^5/2 ÷ (36)^3/2 -Maths 9th

Description : (36)^7/2 - (36)^5/2 ÷ (36)^3/2 -Maths 9th

Last Answer : NEED ANSWER

In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. -Maths 9th

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Description : (36)^7/2 - (36)^5/2 ÷ (36)^3/2 -Maths 9th

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In a survey of 364 children aged 19-36 months, -Maths 9th

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