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3 years ago

11

Megan is deciding when she needs to leave for school. If it typically takes her 15 minutes to walk to school, which is a reasona

ble estimate of the time she should leave home to get to school by 8:00?

2 answers:

Scrat [10]3 years ago

8 0

Answer:

About 7:45

Step-by-step explanation:

7:40 OR 7:45 because sometimes the traffic light takes a while but 7:45 is best.

Alexeev081 [22]3 years ago

4 0

Answer:

7:45 (a quarter to eight)

Step-by-step explanation:

It takes Megan 15 minutes to walk from home to school, and she get to school by 8:00 am, then she needs to leave home at around 8 hours - 15 minutes = 7 hours + 60 minutes - 15 minutes = 7 hours and 45 minutes.

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Find two positive angles and two negative angles that are coterminal with the given angle and less than 1080 but greater than 72

ivann1987 [24]

Answer:

The correct answer is 450°, 810°, -270° and -630°.

Step-by-step explanation:

According to the given scenario, the calculation of the two positive angles and two negative angles i.e. coterminal is as follows:

The coterminal angles are the angles in which the difference could be 360 degrees or the multiples of 360 degrees

For 90 degrees, the two positive angles are

a. 90° + 360°

= 450°

450° + 360°

= 810°

b. The two negative angles are

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3 0

3 years ago

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Answer:

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Step-by-step explanation:

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7 0

3 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

3 years ago