18.95(0.3 + 0.1) is the formula to use.
Let
denote the length of the pond and <span> signify its width. It's recognized that the pond's volume equals the area of its base multiplied by its depth. In this case, the base area can be computed as volume divided by depth, equating to 72000 in³ divided by 24 in, resulting in an area of 3000 in². Given that the area is expressed as x multiplied by y, we come to equation 1, 3000 = x * y. If we have x = 2y, we substitute this into equation 1, leading to 3000 = (2y) * y, simplifying to 2y² = 3000 and consequently y² = 1500, giving y = 38.7 in. Thus, x = 2y yields x = 2 * 38.7 = 77.4 in. The conclusion is that the pond's length is 77.4 in while its width is 38.7 in.
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Set A's standard deviation exceeds that of Set B. To explain, standard deviation reflects variation within data sets. Generally, a dataset with a narrower range will exhibit a smaller standard deviation. For Set A, the range is 25-1 = 24, while for Set B, it's 18-8 = 10. Given that Set A's range is bigger, we would anticipate its standard deviation to also be larger. Standard deviation is calculated as the square root of the average of the squared deviations from the mean. In Set A, the deviations are ±12, ±11, ±10, whereas Set B's deviations are ±5, ±3, ±1. We can reasonably conclude that the value for Set A will be greater without computing the RMS difference. Thus, Set A's standard deviation is larger compared to Set B.
Answer:
D
Step-by-step explanation:
it represents entire F as shown in the diagram
Option C is correct. Step-by-step explanation: We need to identify which options are equivalent to 7/3. By evaluating each option: Option A: 9/49 is incorrect as it cannot be further simplified to yield multiples adequate to match. Option B: 18/42 reduces to 3/7, which does not equal 7/3, hence, incorrect. Option C: 42/18 divides down to 7/3, marking it as the correct choice. Option D: 49/9 also cannot be simplified to produce valid multiples.
