Complete the error interval for the perimeter
WebExample 3: Truncate 4.5891 to one decimal place We delete all the digits after one decimal place: 4.5891 4.5891 truncated to one decimal place is 4.5 WebSep 5, 2024 · For 7.5 cm, LB = 7.25 cm and UB = 7.75 cm. To find the area we multiply. To get the biggest possible answer we need to multiply both upper bounds, and for the smallest possible answer we need to multiply both lower bounds. So max area = 16.5 x 7.75 = 127.875 cm². and min area = 15.5 x 7.25 = 112.375 cm².
Complete the error interval for the perimeter
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WebBack to AQA Maths Higher June 2024 Paper 2 Home Q9: Answers – Paper 2 – June 2024 WebError Intervals Video 377 on www.corbettmaths.com Question 10: A number, x, is 21 when rounded to 2 signiicant igures. Write down the error interval. Question 11: A ...
WebJan 11, 2024 · With this, the interval of the length of each side will be: (5.2cm - 0.05cm) = 5.15cm (as the minimum) (5.2cm + 0.05cm) = 5.25cm (as the maximum) Then the … WebDec 16, 2024 · The Corbettmaths video tutorial on Error Intervals. Videos, worksheets, 5-a-day and much more
WebWhile that is true, it’s an overly simplistic method because it doesn’t indicate how large the difference between two percentages must be to be statistically significant. To evaluate the differences between two percentages, you … WebMeasuring to the nearest meter means the true value could be up to half a meter smaller or larger. The width (w) could be from 5.5m to 6.5m: 5.5 ≤ w < 6.5. The length (l) could be from 7.5m to 8.5m: 7.5 ≤ l < 8.5. The area is width × length: A = w × l. The smallest possible area is: 5.5m × 7.5m = 41.25 m2. The measured area is: 6m × 8m ...
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WebP = perimeter A = area √ = square root Calculator Use Use this square calculator to find the side length, diagonal length, perimeter or area of a geometric square. Given any 1 variable you can calculate the other 3 unknowns. Units: Note that units of length are shown for convenience. They do not affect the calculations. processmaker source codeWebJun 11, 2024 · The perimeter is given as: P = 6.1 cm to 1 decimal place The smallest number that can be approximated to 6.1 is 6.05, while the largest number is less than 6.15 processmaker redditWeb“The total length of the two pipes is 11 metres to the nearest metre.” processmanagehardWebAQA Foundation: November 2024 Paper 1, Q 24 1 Three whole numbers are each rounded to the nearest 10 The sum of the rounded numbers is 50 AQA Foundation: November 2024 Paper 1, Q 24 1 Work out the maximum possible sum for the original three numbers. processmaker transwaterWebThe length of each side of a regular hexagon is 7. 3 cm to 1 decimal place. 1 (a) Complete the error interval for the length of one side. [2 marks] cm length < [2 marks] cm cm length < 1 (b) Complete the error interval for the perimeter. cm 1 (b) Complete the error interval for the perimeter. [1 mark] cm perimeter < [1 mark] cm cm perimeter < … processmaker wiki mailserverWebANSWERS Exercise 1 1. (a) 34.5 (b) 35.5 2. (a) 72.5 (b) 73.5 w 4. (a) m (b) l reha bad herrenalb orthopädieprocessmaker revenue