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faltersainse [42]

3 years ago

6

you are flying a kite at an angle of 70 degrees. you have let out a total of 400ft of string and are holding the reel steady 4 f

eet above the ground. a friend watching the kite estimates that the angle of elevation to the kite is 85 degrees. how far from your friend are you standing?

2 answers:

sp2606 [1]3 years ago

5 0

Answer:

170.466 feet

Step-by-step explanation:

Given that the kite had an angle of elevation of 70 degrees and 4 ft above the ground.

Using sine values, we find that the height of the kite above the reel would be

$h = 400 sin 70=379.88$

Total height from the earth = 379.88+4 = 383.88 ft.

The friend estimates the angle of elevations as 85 degrees

i.e. if d is the distance from the kite position, then we have

d=383.88 cot 85

=33.586 ft

Distance of you from the kite $= 400 cos 70 = 136.8 ft$

Distnce between friend and you = 136.88+33.586

= 170.466 feet

natta225 [31]3 years ago

3 0

Answer: 170 ft away

First, find the height of the kite from the ground:

h + 4 = 400 sin 70
h + 4 = 379.88 ft (from the ground)

Then, my distance to the directly downwards of where the kite is flying is:

d = 400 cos 70
= 136.8 ft

Then the distance of my friend from me:

D = 136.8 ft + (379.87) tan 85
= 170 ft

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Zigmanuir [339]

Answer:

The probability that all 4 selected workers will be from the day shift is, = 0.0198

The probability that all 4 selected workers will be from the same shift is = 0.0278

The probability that at least two different shifts will be represented among the selected workers is = 0.9722

The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5256

Step-by-step explanation:

Given that:

A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A quality control consultant is to select 4 of these workers for in-depth interviews:

The number of selections result in all 4 workers coming from the day shift is :

$(^n _r) = (^{10} _4)$

$=\dfrac{(10!)}{4!(10-4)!}$

= 210

The probability that all 5 selected workers will be from the day shift is,

$\begin{array}{c}\\P\left( {{\rm{all \ 4 \ selected \ workers\ will \ be \ from \ the \ day \ shift}}} \right) = \frac{{\left( \begin{array}{l}\\10\\\\4\\\end{array} \right)}}{{\left( \begin{array}{l}\\24\\\\4\\\end{array} \right)}}\\\\ = \frac{{210}}{{10626}}\\\\ = 0.0198\\\end{array}$

(b) The probability that all 4 selected workers will be from the same shift is calculated as follows:

P( all 4 selected workers will be) $= \dfrac{ (^{10}_4) }{(^{24}_4)}+\dfrac{ (^{8}_4) }{(^{24}_4)} + \dfrac{ (^{6}_4) }{(^{24}_4)}$

where;

$(^{8}_4) } = \dfrac{8!}{4!(8-4)!} = 70$

$(^{6}_4) } = \dfrac{6!}{4!(6-4)!} = 15$

∴ P( all 4 selected workers is ) $=\dfrac{210+70+15}{10626}$

The probability that all 4 selected workers will be from the same shift is = 0.0278

(c) What is the probability that at least two different shifts will be represented among the selected workers?

P ( at least two different shifts will be represented among the selected workers) $= 1-\dfrac{ (^{10}_4) }{(^{24}_4)}+\dfrac{ (^{8}_4) }{(^{24}_4)} + \dfrac{ (^{6}_4) }{(^{24}_4)}$

$=1 - \dfrac{210+70+15}{10626}$

= 1 - 0.0278

The probability that at least two different shifts will be represented among the selected workers is = 0.9722

(d)What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

The probability that at least one of the shifts will be unrepresented in the sample of workers is:

$P(AUBUC) = \dfrac{(^{6+8}_4)}{(^{24}_4)}+ \dfrac{(^{10+6}_4)}{(^{24}_4)}+ \dfrac{(^{10+8}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0$

$P(AUBUC) = \dfrac{(^{14}_4)}{(^{24}_4)}+ \dfrac{(^{16}_4)}{(^{24}_4)}+ \dfrac{(^{18}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0$

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The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5256

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Answer:

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Step-by-step explanation:

We have the following info given:

$\bar X = 35.6$ represent the sampel mean for the age of customers

$\sigma = 5$ represent the population standard deviation

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$p_v = P(z>2.007) = 0.0224$

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