Mean = Sum of all observation/Total number of observations = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60/15 = 822/10 = 54.8 Arranging the data in ascending order: 39, 40, 40, 41, 42, 46, 48, 52, 52, 54, 60, 62, 96, 98 Number of observations = 15 15 is an odd number. TherRead more
Mean = Sum of all observation/Total number of observations
= 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60/15
= 822/10 = 54.8
Arranging the data in ascending order:
39, 40, 40, 41, 42, 46, 48, 52, 52, 54, 60, 62, 96, 98
Number of observations = 15
15 is an odd number. Therefore, the mode = 8th[(n+1/2)th] term of the observation.
Hence,
Median = 8th term = 52
Mode is the observation having maximum frequency. Here, the number 52 occurs maximum (3) times.
Hence, the mode of the observations is 52.
Number of observations = 10 10 is an even number. Therefore, the modian = Mean of 5th[(n/2)th] term and 6th[(n/2 + 1)th] term. Hence, Median = 1/2(5th observation + 6th observation) ⇒ 63 = 1/2 [(x) + (x + 2)] ⇒ 63 = 1/2 [2x + 2] ⇒ 63 = x + 1 ⇒ x = 62 Hence, the valueof x is 62.
Number of observations = 10
10 is an even number. Therefore, the modian = Mean of 5th[(n/2)th] term and 6th[(n/2 + 1)th] term.
Hence,
Median = 1/2(5th observation + 6th observation)
⇒ 63 = 1/2 [(x) + (x + 2)]
⇒ 63 = 1/2 [2x + 2]
⇒ 63 = x + 1
⇒ x = 62
Hence, the valueof x is 62.
Arranging the data in ascending order: 14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28 Mode is the observation having maximum frequency. Here, the number 14 occurs maximum (4) times. Hence, the mode of the observations is 14.
Arranging the data in ascending order:
14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Mode is the observation having maximum frequency. Here, the number 14 occurs maximum (4) times.
Hence, the mode of the observations is 14.
(i) Lenth of plastic box l = 1.5 m, breadth b = 1.25 m height h = 65 cm = 0.65 m Area of sheet required for making a plastic box = total surface area of box - Area of top of box = 2(lb + bh + hl) - lb = 2(1.5 × 1.25 + 1.25 × 0.65 + 0.65 × 1.5) - 1.5 × 1.25m² = 2(1.875 + 0.8125 + 0.975) - 1.875 m² =Read more
(i) Lenth of plastic box l = 1.5 m, breadth b = 1.25 m height h = 65 cm = 0.65 m
Area of sheet required for making a plastic box = total surface area of box – Area of top of box
= 2(lb + bh + hl) – lb = 2(1.5 × 1.25 + 1.25 × 0.65 + 0.65 × 1.5) – 1.5 × 1.25m²
= 2(1.875 + 0.8125 + 0.975) – 1.875 m²
= 2(3.6625) – 1.875 m²
= 7.325 – 1.875 = 5.45 m²
Hence, the area of the sheet required for making the box is 5.45 m².
(ii) Total cost of cheet = Rs 20 × 5.45 = Rs 109.
Lenght of room l = 5 m, breadth b = 4 m and height h = 3 m Area of four walls and ceilling = Total surface area of room - area of floor = 2(lb + bh + hl) - lb = 2(5 × 4 + 4 × 3 + 3 × 5) - 5 × 4 m² = 2(20 + 12 + 15) - 20 m² = 2(47) - 20 m² = 94 - 20 = 74 m² Therefore, the area of four walls and ceillRead more
Lenght of room l = 5 m, breadth b = 4 m and height h = 3 m
Area of four walls and ceilling = Total surface area of room – area of floor
= 2(lb + bh + hl) – lb
= 2(5 × 4 + 4 × 3 + 3 × 5) – 5 × 4 m²
= 2(20 + 12 + 15) – 20 m²
= 2(47) – 20 m²
= 94 – 20 = 74 m²
Therefore, the area of four walls and ceilling = 74 m²
Hence, the cost of white washing the walls of walls of the room and the ceilling
= Rs 7.50 × 74 = Rs. 555.00
In a mathematics test given to 15 students, the following marks (out of 100) are recorded: 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60 Find the mean, median and mode of this data.
Mean = Sum of all observation/Total number of observations = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60/15 = 822/10 = 54.8 Arranging the data in ascending order: 39, 40, 40, 41, 42, 46, 48, 52, 52, 54, 60, 62, 96, 98 Number of observations = 15 15 is an odd number. TherRead more
Mean = Sum of all observation/Total number of observations
See less= 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60/15
= 822/10 = 54.8
Arranging the data in ascending order:
39, 40, 40, 41, 42, 46, 48, 52, 52, 54, 60, 62, 96, 98
Number of observations = 15
15 is an odd number. Therefore, the mode = 8th[(n+1/2)th] term of the observation.
Hence,
Median = 8th term = 52
Mode is the observation having maximum frequency. Here, the number 52 occurs maximum (3) times.
Hence, the mode of the observations is 52.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x. 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95.
Number of observations = 10 10 is an even number. Therefore, the modian = Mean of 5th[(n/2)th] term and 6th[(n/2 + 1)th] term. Hence, Median = 1/2(5th observation + 6th observation) ⇒ 63 = 1/2 [(x) + (x + 2)] ⇒ 63 = 1/2 [2x + 2] ⇒ 63 = x + 1 ⇒ x = 62 Hence, the valueof x is 62.
Number of observations = 10
See less10 is an even number. Therefore, the modian = Mean of 5th[(n/2)th] term and 6th[(n/2 + 1)th] term.
Hence,
Median = 1/2(5th observation + 6th observation)
⇒ 63 = 1/2 [(x) + (x + 2)]
⇒ 63 = 1/2 [2x + 2]
⇒ 63 = x + 1
⇒ x = 62
Hence, the valueof x is 62.
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Arranging the data in ascending order: 14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28 Mode is the observation having maximum frequency. Here, the number 14 occurs maximum (4) times. Hence, the mode of the observations is 14.
Arranging the data in ascending order:
See less14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Mode is the observation having maximum frequency. Here, the number 14 occurs maximum (4) times.
Hence, the mode of the observations is 14.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:
(i) Lenth of plastic box l = 1.5 m, breadth b = 1.25 m height h = 65 cm = 0.65 m Area of sheet required for making a plastic box = total surface area of box - Area of top of box = 2(lb + bh + hl) - lb = 2(1.5 × 1.25 + 1.25 × 0.65 + 0.65 × 1.5) - 1.5 × 1.25m² = 2(1.875 + 0.8125 + 0.975) - 1.875 m² =Read more
(i) Lenth of plastic box l = 1.5 m, breadth b = 1.25 m height h = 65 cm = 0.65 m
See lessArea of sheet required for making a plastic box = total surface area of box – Area of top of box
= 2(lb + bh + hl) – lb = 2(1.5 × 1.25 + 1.25 × 0.65 + 0.65 × 1.5) – 1.5 × 1.25m²
= 2(1.875 + 0.8125 + 0.975) – 1.875 m²
= 2(3.6625) – 1.875 m²
= 7.325 – 1.875 = 5.45 m²
Hence, the area of the sheet required for making the box is 5.45 m².
(ii) Total cost of cheet = Rs 20 × 5.45 = Rs 109.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m².
Lenght of room l = 5 m, breadth b = 4 m and height h = 3 m Area of four walls and ceilling = Total surface area of room - area of floor = 2(lb + bh + hl) - lb = 2(5 × 4 + 4 × 3 + 3 × 5) - 5 × 4 m² = 2(20 + 12 + 15) - 20 m² = 2(47) - 20 m² = 94 - 20 = 74 m² Therefore, the area of four walls and ceillRead more
Lenght of room l = 5 m, breadth b = 4 m and height h = 3 m
See lessArea of four walls and ceilling = Total surface area of room – area of floor
= 2(lb + bh + hl) – lb
= 2(5 × 4 + 4 × 3 + 3 × 5) – 5 × 4 m²
= 2(20 + 12 + 15) – 20 m²
= 2(47) – 20 m²
= 94 – 20 = 74 m²
Therefore, the area of four walls and ceilling = 74 m²
Hence, the cost of white washing the walls of walls of the room and the ceilling
= Rs 7.50 × 74 = Rs. 555.00