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ycow [4]
3 years ago
12

Suppose 4-year-olds in a certain country average 3 hours a day unsupervised and that most of the unsupervised children live in r

ural areas, considered safe. Suppose that the standard deviation is 1.8 hours and the amount of time spent alone is normally distributed. We randomly survey one 4-year-old living in a rural area. We are interested in the amount of time the child spends alone per day. Part (a) In words, define the random variable X.
A) the number of 4-year-old children that live in rural areas

B) the time (in hours) a child spends unsupervised per day

C)the number of people that live in rural areas

D)the time (in hours) a 4-year-old spends unsupervised per day

E)the time (in hours) a 4-year-old spends unsupervised per week

Part I: Give the distribution of X.

X ~ ____ (____,_____)

Part II:

Find the probability that the child spends less than 1 hour per day unsupervised.

Write the probability statement.

P_________

What is the probability? (Round your answer to four decimal places.) _____
Mathematics
1 answer:
natta225 [31]3 years ago
4 0

Answer:

D)the time (in hours) a 4-year-old spends unsupervised per day

Step-by-step explanation:

Given that 4-year-olds in a certain country average 3 hours a day unsupervised and that most of the unsupervised children live in rural areas, considered safe

X - the time (in hours) a 4-year-old spends unsupervised per day

X is N(3,1.8)

P(X

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Find the exact value of sin(cos^-1(4/5))
boyakko [2]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2762144

_______________


Let  \mathsf{\theta=cos^{-1}\!\left(\dfrac{4}{5}\right).}


\mathsf{0\le \theta\le\pi,}  because that is the range of the inverse cosine funcition.


Also,

\mathsf{cos\,\theta=cos\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{5}}\\\\\\ \mathsf{5\,cos\,\theta=4}


Square both sides and apply the fundamental trigonometric identity:

\mathsf{(5\,cos\,\theta)^2=4^2}\\\\
\mathsf{5^2\,cos^2\,\theta=4^2}\\\\
\mathsf{25\,cos^2\,\theta=16\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{25\cdot (1-sin^2\,\theta)=16}

\mathsf{25-25\,sin^2\,\theta=16}\\\\
\mathsf{25-16=25\,sin^2\,\theta}\\\\
\mathsf{9=25\,sin^2\,\theta}\\\\
\mathsf{sin^2\,\theta=\dfrac{9}{25}}


\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{9}{25}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{3^2}{5^2}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{3}{5}}


But \mathsf{0\le \theta\le\pi,} which means \theta lies either in the 1st or the 2nd quadrant. So \mathsf{sin\,\theta} is a positive number:

\mathsf{sin\,\theta=\dfrac{3}{5}}\\\\\\
\therefore~~\mathsf{sin\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]=\dfrac{3}{5}\qquad\quad\checkmark}


I hope this helps. =)


Tags:  <em>inverse trigonometric function cosine sine cos sin trig trigonometry</em>

3 0
3 years ago
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Plz help i dont have much time
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Answer:

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1. 5

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5. 0

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Step-by-step explanation:

hope this is what u need

if it's right pls mark brainleist

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60$ for leroy and $150 for mom.

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