weight (pounds) frequency\n13 12\n14 8\n15 8\n16 7\n17 9\n18 10\n19 13\n20 7\nthe frequency - table…

weight (pounds) frequency\n13 12\n14 8\n15 8\n16 7\n17 9\n18 10\n19 13\n20 7\nthe frequency - table summarizes a data set of the weights, rounded to the nearest pound, of 71 tortoises. a weight of 39 pounds is added to the original data set, creating a new data set of the weights, rounded to the nearest pound, of 72 tortoises. which statement best compares the mean and median of the new data set to the mean and median of the original data set?\na) the mean of the new data set is greater than the mean of the original data set, and the median of the new data set is greater than the median of the original data set.\nb) the mean of the new data set is greater than the mean of the original data set, and the medians of the two data sets are equal.\nc) the mean of the new data set is less than the mean of the original data set, and the median of the new data set is less than the median of the original data set.\nd) the mean of the new data set is less than the mean of the original data set, and the medians of the two data sets are equal.\nx - 29=(x - a)(x - 29)\nwhich of the following are solutions to the given equation, where a is a constant and a > 30?\ni. a\nii. a + 1\niii. 29\na) i and ii only\nb) i and iii only\nc) ii and iii only\nd) i, ii, and iii\nin the xy - plane, the graph of the equation y=-x² + 9x - 100 intersects the line y = c at exactly one point. what is the value of c?\na) -\\frac{481}{4}\nb) - 100\nc) -\\frac{319}{4}\nd) -\\frac{9}{2}
Answer
- First problem (comparing mean and median of data - sets):
- Explanation:
- Step1: Understand the effect on the mean
- The original data - set has 71 values. A new value of 39 pounds is added. Since 39 is likely to be larger than the existing values (as no other information about the range of the original data is given, but the values in the frequency table are in the range 13 - 20), the sum of all values in the new data - set will increase. The formula for the mean (\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}). Here, (n) changes from 71 to 72 and (\sum_{i = 1}^{n}x_{i}) increases. So, the mean of the new data - set is greater than the mean of the original data - set.
- Step2: Understand the effect on the median
- For a data - set with (n = 71) (odd number of values), the median is the (\left(\frac{71 + 1}{2}\right)=36^{th}) ordered value. For a data - set with (n=72) (even number of values), the median is the average of the (36^{th}) and (37^{th}) ordered values. Since the new value 39 is large, it will not change the position of the middle - most values in a way that would decrease the median. In fact, if the original data is ordered, the new value will be towards the upper end, and the median of the new data - set is greater than the median of the original data - set.
- Step1: Understand the effect on the mean
- Answer: A. The mean of the new data set is greater than the mean of the original data set, and the median of the new data set is greater than the median of the original data set.
- Explanation:
- Second problem ((x - 29=(x - a)(x - 29))):
- Explanation:
- Step1: Rearrange the equation
- Start with (x - 29=(x - a)(x - 29)). Move all terms to one side: ((x - a)(x - 29)-(x - 29)=0). Factor out ((x - 29)): ((x - 29)[(x - a)-1]=0). So, ((x - 29)(x-(a + 1))=0).
- Step2: Find the solutions
- Using the zero - product property, if (AB = 0), then either (A = 0) or (B = 0). So, (x=29) or (x=a + 1). Since (a>30), (a) is not a solution.
- Step1: Rearrange the equation
- Answer: C. II and III only
- Explanation:
- Third problem ((y=-x^{2}+9x - 100) and (y = c) intersect at one point): [LLM SSE On Failure]