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3 years ago

14

You can draw a quadrilateral with two sets of parallel lines and no right angles. true false

2 answers:

kolezko [41]3 years ago

7 0

Answer:

it is true i just took the test and guess what it was aced!

Step-by-step explanation:

yuradex [85]3 years ago

5 0

False. When a quadrilateral has two sets of parallel sides ,like I drew below, all four corners form right angles.

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What is the area of this polygon?

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2 years ago

Standard Error from a Formula and a Bootstrap Distribution Sample A has a count of 30 successes with and Sample B has a count of

tia_tia [17]

Answer:

Using a formula, the standard error is: 0.052

Using bootstrap, the standard error is: 0.050

Comparison:

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

Step-by-step explanation:

Given

Sample A Sample B

$x_A = 30$ $x_B = 50$

$n_A = 100$ $n_B =250$

Solving (a): Standard error using formula

First, calculate the proportion of A

$p_A = \frac{x_A}{n_A}$

$p_A = \frac{30}{100}$

$p_A = 0.30$

The proportion of B

$p_B = \frac{x_B}{n_B}$

$p_B = \frac{50}{250}$

$p_B = 0.20$

The standard error is:

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

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

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

$SE_{p_A-p_B} = \sqrt{\frac{0.21}{100} + \frac{0.16}{250}}$

$SE_{p_A-p_B} = \sqrt{0.0021+ 0.00064}$

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

$SE_{p_A-p_B} = 0.052$

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:

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

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

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

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

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

$SE_{p_A-p_B} = 0.050$

5 0

2 years ago