This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Complete the calculation of class boundaries. The class boundaries are calculated by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class. • For 0-9: Lower boundary = 0 - 0.5 = -0.5, Upper boundary = 9 + 0.5 = 9.5. • For 10-19: Lower boundary = 10 - 0.5 = 9.5, Upper boundary = 19 + 0.5 = 19.5. • For 20-29: Lower boundary = 20 - 0.5 = 19.5, Upper boundary = 29 + 0.5 = 29.5. • For 30-39: Lower boundary = 30 - 0.5 = 29.5, Upper boundary = 39 + 0.5 = 39.5. • For 40-49: Lower boundary = 40 - 0.5 = 39.5, Upper boundary = 49 + 0.5 = 49.5. Step 2: Determine the scale for the horizontal axis. The horizontal axis of a histogram represents the continuous data, marked by the class boundaries. The scale should cover the range from the lowest lower boundary to the highest upper boundary. The range of the class boundaries is from -0.5 to 49.5. The horizontal scale should be marked with these class boundaries. The interval between each major mark is 10 units (the class width). The horizontal scale for the histogram should be marked with the following values: -0.5, 9.5, 19.5, 29.5, 39.5, 49.5 Each interval on the horizontal axis represents a class width of 10 units. For example, if using graph paper, you might choose 2 cm to represent 10 units. Step 3: Clarify the role of frequency. In a histogram, the frequency determines the height of each bar (or the frequency density if class widths are unequal), which is represented on the vertical axis. The horizontal axis represents the data values (class boundaries) and is not directly "multiplied by frequency." The scale for the horizontal axis is based on the class boundaries: from -0.5 to 49.5 with intervals of 10 units. Frequency determines the height of the bars on the vertical axis.