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3 years ago

What segment is the projection of ST on QT? UT QU QT

Mathematics

2 answers:

Likurg_2 [28]3 years ago

3 0

Answer:

Step-by-step explanation:

The projection of ST on QT is found by locating the points on QT that are nearest the endpoints S and T of the original segment. On QT, U is the point closest to S, and T is the point closest to T. Then the projection is the segment between those identified points: UT.

Leto [7]3 years ago

3 0

Answer:

Step-by-step explanation:

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Answer:

$CI = (0.17 - 0.27)\± 1.65\sqrt{\frac{(0.17)*(0.83) + (0.27)*(0.73)}{200}}$

Step-by-step explanation:

Given

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$x_2 = 54$ --- City K

Required

Determine the 90% confidence interval

This is calculated using:

$CI = \bar x \± z\frac{\sigma}{\sqrt n}$

Calculating $\bar x$

$\bar x = \bar x_1 - \bar x_2$

$\bar x = \frac{x_1}{n} - \frac{x_2}{n}$

$\bar x = \frac{34}{200} - \frac{54}{200}$

$\bar x = 0.17 - 0.27$

For a 90% confidence level, the z-score is 1.65. So:

$z = 1.65$

Calculating the standard deviation $\sigma$

$\sigma = \sqrt{(\bar x_1)*(1 - \bar x_1) + (\bar x_2)*(1 - \bar x_2) }$

So:

$\sigma = \sqrt{(0.17)*(1 - 0.17) + (0.27)*(1 - 0.27) }$

$\sigma = \sqrt{(0.17)*(0.83) + (0.27)*(0.73)}$

So:

$CI = (0.17 - 0.27)\± 1.65\frac{\sqrt{(0.17)*(0.83) + (0.27)*(0.73)}}{\sqrt {200}}$

$CI = (0.17 - 0.27)\± 1.65\sqrt{\frac{(0.17)*(0.83) + (0.27)*(0.73)}{200}}$

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