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Tutor profile: Qichen H.

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Qichen H.
Graduate Student at UChicago
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Questions

Subject: Geometry

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Question:

The lengths of two sides of a triangle are 3cm and 4cm. All angles in the triangle are less than 90 degrees. What is the range of possible lengths for the third side?

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Qichen H.
Answer:

Answer: $$\sqrt{7}$$ cm < the length of the third side < 5 cm Explanation: The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side. If the 2 sides (3cm and 4cm) are the 2 shorter sides of the triangle, then the length of the third side should be less than the sum of these 2 sides (the length of the third side < 3 cm+4 cm = 7 cm). If the 4cm side is the longest side of the triangle, then the length of the third side should be greater than the difference between the 2 other sides (the length of the third side > 4 cm - 3 cm = 1 cm). Thus, 1 cm < the length of the third side < 7 cm should be the general case. However, the question indicates that all angles in the triangle are less than 90 degrees. According to the Pythagorean Theorem, the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other 2 sides: $$a^{2}+b^{2}+c^{2}$$ a = side of the right triangle b = side of the right triangle c = hypotenuse Thus, If the 2 sides (3cm and 4cm) are the 2 shorter sides of the triangle, the third side of the triangle should be less than $$\sqrt{3^{2}+4^{2}} = 5 cm$$. If the 4cm side is the longest side of the triangle, then the length of the third side should be greater than $$\sqrt{4^{2}-3^{2}} = \sqrt{7} cm$$. Thus, the range of possible lengths for the third side is from $$\sqrt{7}$$ cm to 5 cm (exclusive).

Subject: Statistics

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Question:

If n<1, which of the following sets has a greater standard deviation? n, 1, 2 or 1, 2, 3

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Qichen H.
Answer:

Answer: The relationship cannot be determined from the information given Explanation: First, we need to know the definition and calculation of the standard deviation. The standard deviation measures the dispersion of a dataset relative to its mean. It is calculated as the square root of the variance. The formula for the standard deviation ($$ \sigma$$) is: $$ \sigma= \sqrt{\frac{\sum_{i=1}^N (x_i - \overline{x})^2}{N}}$$ Where: $$x_i$$ = value of the $$i^{\text{th}}$$ point in the data set. $$\overline{x}$$ = the mean value of the data set N = the number of data points in the data set For this question, we need to consider 3 cases: n=0, n<0, and 0<n<1 Method 1: We can solve this question by calculation: set n=0, n=-1, and n=0.5 as 3 examples for the 3 cases. Plug them into the standatd deviation formula: $$ \sigma= \sqrt{\frac{\sum_{i=1}^N (x_i - \overline{x})^2}{N}}$$ and then compare the results to the standard deviation of the set 1, 2, 3 (the standard deviation of the set 1, 2, 3 = $$\sqrt{\frac{(1-2)^{2}+(2-2)^{2}+(3-2)^{2}}{3}}$$ = $$\sqrt\frac{2}{3}$$) However, it is easier to solve this question by the definition of the standard deviation. Method 2: By definition, the standard deviation measures the dispersion of a dataset relative to its mean. A lower standard deviation indicates that the values tend to be closer to their mean, while a higher standard deviation indicates that the values tend to be farther away from their mean (more spread out). When n = 0, the two data set 0, 1, 2 and 1, 2, 3 have exactly the same standard deviation because 0 and 2 are both 1 point away from their mean “1”, whereas 1 and 3 are both 1 point away from their mean “2”. When n < 0, the data set n, 1, 2 is more spread out compared to the data set 1, 2, 3. Thus, the data set n, 1, 2 has a greater standard deviation when n < 0. When 0<n<1, the data set n, 1, 2 is less spread out compared to the data set 1, 2, 3. Thus, the data set 1, 2, 3 has a greater standard deviation when 0<n<1. Therefore, without knowing whether n=0, n<0, or 0<n<1, we cannot determine which sets have a greater standard deviation Tips: We can also estimate the standard deviation with the range rule. The range rule indicates that the standard deviation of a sample is approximately equal to $$\frac{1}{4}$$ of the range of the data. Thus, the larger the range of the data, the greater the standard deviation of the data.

Subject: Algebra

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Question:

In a classroom, the number of chairs is equal to 6 times the number of tables. The weight of each table is 9 times the weight of each chair. The total weight of the tables is 12,000 pounds. What is the total weight of all the tables and chairs in the classroom?

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Qichen H.
Answer:

Answer: 20,000 pounds Method 1: The ratio of the number of chairs and tables is 6:1 The ratio of the weight of chairs and tables is 1:9 Thus, the ratio of the total weight of chairs and tables is 6:9 = 2:3 (because total weight = number $$\times$$ weight) In other words, the total weight of the chairs is $$\frac{2}{3}$$ times the total weight of the tables The total weight of the chairs = 12,000 $$\times \frac{2}{3}$$ = 8,000 Finally, the total weight of all the tables and chairs = 12,000 + 8,000 = 20,000 Method 2 Set the number of chairs as X, then the number of tables is $$\frac{1}{6}$$ X. Set the weight of a chair as Y, then the weight of a table is 9Y. Thus, the total weight of the chairs is XY and the total weight of the tables is $$\frac{1}{6} \times 9Y$$ = 1.5 XY The total weight of the tables is 1.5XY = 12,000, then the total weight of the chairs is XY = $$\frac{12,000}{1.5}$$ = 8,000 Finally, the total weight of all the tables and chairs = 12,000 + 8,000 = 20,000

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