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Leni [432]
3 years ago
12

WILL GIVE BRAINLIEST ANSWER!!

Mathematics
2 answers:
Natasha2012 [34]3 years ago
4 0
Check the picture below.

now, bear in mind that, there are 60minutes in 1 degree, so 23' is just 23/60 degrees or about 0.38 degrees.

and 29' is just 29/60 degrees or about 0.48 degrees.

so 16°23' is about 16.38°, and 49°29' is about 49.48°

\bf tan(16.38^o)=\cfrac{200}{a+b}\implies a+b=\cfrac{200}{tan(16.38^o)}
\\\\\\
\boxed{b=\cfrac{200}{tan(16.38^o)}-a}
\\\\\\
tan(49.48^o)=\cfrac{200}{b}\implies \boxed{b=\cfrac{200}{tan(49.48^o)}}\\\\
-------------------------------\\\\
\cfrac{200}{tan(49.48^o)}=\cfrac{200}{tan(16.38^o)}-a\\\\\\a=\cfrac{200}{tan(16.38^o)}-\cfrac{200}{tan(49.48^o)}

make sure your calculator is in Degree mode.

Verizon [17]3 years ago
4 0

The distance covered by the boat is \boxed{489.67{\text{ feet}}}.

Further explanation:

The Pythagorean formula can be expressed as,

\boxed{{H^2} = {P^2} + {B^2}}.

Here, H represents the hypotenuse, P represents the perpendicular and B represents the base.

The formula for tan of angle a can be expressed as

\boxed{\tan a = \frac{P}{B}}

Explanation:

The perpendicular AB. The length of AB is 200{\text{ feet}}.

The angle of depression is \angle ACB = {14^ \circ }52'.

One degree has 60 minutes.

{1^ \circ } = 60'

\begin{aligned}\angle ACB &= 49 + \frac{{29}}{{60}}\\&= 49 + 0.48\\&= 49.48\\\end{aligned}

The angle ADB is \angle ADB = {16^ \circ }23'.

\begin{aligned}\angle ADB&= {16^ \circ }23' \\&= 16 + \frac{{23}}{{60}}\\&= {16.38^ \circ }\\\end{aligned}

In triangle ABC.

\begin{aligned}\tan\left( {{{49.48}^\circ }} \right)&=\frac{{200}}{{BC}}\\1.13&= \frac{{200}}{{BC}}\\BC &= \frac{{200}}{{1.13}}\\BC &= 177{\text{ feet}}\\\end{aligned}

In triangle ABD.

\begin{aligned}\tan \left( {{{16.38}^ \circ }} \right)&= \frac{{200}}{{BD}}\\0.30&= \frac{{200}}{{BD}}\\BD&= \frac{{200}}{{0.30}}\\BD &= 666.67{\text{ feet}}\\\end{aligned}

The distance boat can travel can be obtained as follows,

\begin{aligned}DC& = BD - BC\\&= 666.67 - 177\\&= 489.67{\text{ feet}}\\\end{aligned}

The distance covered by the boat is \boxed{489.67{\text{ feet}}}.

Kindly refer to the image attached.

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Trigonometry

Keywords: perpendicular, person watching boat, top, lighthouse, angle of depression, angle of elevation, 200 feet tall, travel, sides, right angle triangle, triangle, altitudes, hypotenuse, on the triangle, hypotenuse, trigonometric functions.

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(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

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And the confidence interval for the difference of means would be given by:

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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We have the following data given:

\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by \alpha=1-0.93=0.07 and \alpha/2 =0.035. And the critical value would be given by:  

z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811  

The confidence interval is given by:

(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

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