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2 years ago

Lines AB and CD are straight lines. Find x and y. Give reasons to justify your solutions. I need both answers.

Mathematics

1 answer:

Kazeer [188]2 years ago

4 0

Answer:

1) x = 14°, y = 5°

2) x = 18.5°, y = 37°

Step-by-step explanation:

1) ∠AOC and ∠BOD are vertical angles, then ∠BOD = 25°

∠MOD = ∠MOB + ∠BOD = 90°

3x + 23° + 25° = 90°

3x = 90° - 23° - 25°

x = 42°/3

x = 14°

∠LOB = ∠LOM + ∠MOB = 90°

5y + 3x + 23° = 90°

5y = 90° - 23° - 3(14°)

y = 25°/5

y = 5°

2) ∠AOC and ∠BOD are vertical angles, then ∠BOD = 16°

∠EOB = ∠EOD + ∠DOB = 90°

2y + 16° = 90°

y = (90° - 16°)/2

y = 37°

∠DOF = ∠BOF + ∠DOB = 90°

4x + 16° = 90°

x = (90° - 16°)/4

x = 18.5°

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$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

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$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

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