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zhenek [66]
3 years ago
8

Which classification best represents a triangle with side lengths 6 cm, 10 cm, and 12 cm? 1. acute, because 62 + 102 < 122 2.

acute, because 6 + 10 > 12 3.obtuse, because 62 + 102 < 122 4.obtuse, because 6 + 10 > 12
Mathematics
1 answer:
Yakvenalex [24]3 years ago
7 0

Answer:

The best represents a triangle is obtuse, because 6 + 10 > 12 ⇒ answer 4

Step-by-step explanation:

* Lets talk about some facts in the triangle

- The triangle is formed when the sum of the lengths of the shortest

 two side is greater then the length of the longest side

∵ The lengths of the sides are 6 cm , 10 cm , 12 cm

∵ 6 + 10 > 12

∴ 6 , 10 , 12 are the sides of a triangle

- Two know the type of the triangle use these rules

# The three sides of the triangle have lengths a , b , c where a and b

  are the shortest sides means a , b < c, then

- If a² + b² > c² , the triangle is acute triangle (its 3 angles are acute)

- If a² + b² = c² , the triangle is right triangle (the angle opposite to c is a

 right angle and the other 2 angles are acute angles)

- If a² + b² < c² , the triangle is obtuse triangle (the angle opposite to c is an

 obtuse angle and the other 2 angles are acute angles)

∵ (6)² + (10)² = 36 + 100 = 136

∵ (12)² = 144

∵ 136 < 144

∴ (6)² + (10)² < (12)²

∴ The triangle is obtuse

* The best represents a triangle is obtuse, because 6 + 10 > 12

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Answer:

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Step-by-step explanation:

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3 years ago
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Answer:

Using a formula, the standard error is: 0.052

Using bootstrap, the standard error is: 0.050

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The calculated standard error using the formula is greater than the standard error using bootstrap

Step-by-step explanation:

Given

Sample A                          Sample B

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Solving (a): Standard error using formula

First, calculate the proportion of A

p_A = \frac{x_A}{n_A}

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The standard error is:

SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}

SE_{p_A-p_B} = \sqrt{\frac{0.30 * (1 - 0.30)}{100} + \frac{0.20* (1 - 0.20)}{250}}

SE_{p_A-p_B} = \sqrt{\frac{0.30 * 0.70}{100} + \frac{0.20* 0.80}{250}}

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SE_{p_A-p_B} = \sqrt{0.00274}

SE_{p_A-p_B} = 0.052

Solving (a): Standard error using bootstrapping.

Following the below steps.

  • Open Statkey
  • Under Randomization Hypothesis Tests, select Test for Difference in Proportions
  • Click on Edit data, enter the appropriate data
  • Click on ok to generate samples
  • Click on Generate 1000 samples ---- <em>see attachment for the generated data</em>

From the randomization sample, we have:

Sample A                          Sample B

x_A = 23                              x_B = 57

n_A = 100                             n_B =250

p_A = 0.230                          p_A = 0.228

So, we have:

SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}

SE_{p_A-p_B} = \sqrt{\frac{0.23 * (1 - 0.23)}{100} + \frac{0.228* (1 - 0.228)}{250}}

SE_{p_A-p_B} = \sqrt{\frac{0.1771}{100} + \frac{0.176016}{250}}

SE_{p_A-p_B} = \sqrt{0.001771 + 0.000704064}

SE_{p_A-p_B} = \sqrt{0.002475064}

SE_{p_A-p_B} = 0.050

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