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belka [17]
3 years ago
7

PLS HELP!!!: A meter is about the distance from one hand to the other when your arms are stretched out if 5 friends stood with t

heir arms stretched and fingers touching about how many meters would that be in length??
Mathematics
2 answers:
steposvetlana [31]3 years ago
5 0

Answer:5 meters

Step-by-step explanation:

1 meter=1 person stretched out

5meters=5 people stretched out

Zanzabum3 years ago
4 0

Answer:

5 meters???

Step-by-step explanation:

each friend has arms

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Which function is a horizontal compression of its parent function f(x)=x^7 by a factor of 4?
nasty-shy [4]
Note that what we want is to compress any point (a, b) in the parent function, to (a, b/4).

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                                     (a, a^7).

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                                   \displaystyle{ (a,  \frac{a^7}{4} ).

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Answer: <span>t(x)=1/4x^7
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Simplify the polynomial (4n² + 3n – 5) − (2n² + 3n<br> -<br> + 6)
Veseljchak [2.6K]

Answer:

2n² + 1

Step-by-step explanation:

Given polynomial:

(4n² + 3n – 5) − (2n² + 3n - 6)

Solution:

Removing parentheses,we obtain

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Combine like terms:

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2 years ago
Sleep apnea is a disorder in which there are pauses in breathing during sleep. People with this condition must wake up frequentl
Pachacha [2.7K]

Answer:

a) 0.24356 or 24.36%

b) [102.39 , 105.61]

c) The interpretation of this confidence interval is that in samples of 427 people aged 65 years and older, there is a 99% probability that the number of people that suffers sleep apnea is between 103 and 105.

Step-by-step explanation:

a)

This can be considered a binomial distribution (a person either has sleep apnea or not).

Based on the sample we have the probability of suffering the condition is  

p = 104/427 = 0.24356  

and q (the probability of not suffering the condition) is  

q=1-0.24356=0.75644

So the proportion of people aged 65 years and older who have sleep apnea is 0.24356 or 24.36%

b)

To check if we can <em>approximate this binomial distribution with the Normal distribution</em> we must see that

np ≥ 5 and nq ≥ 5

where n is the sample size. Since

427*0.24356 ≥ 5 and 427*0.75644 ≥ 5  

we can approximate the binomial with a Normal distribution with mean  

np = 427*0.24356 = 104  

and standard deviation  

\large s=\sqrt{npq}=\sqrt{427*0.24356*0.75644}=8.867

The 99% confidence interval (without the continuity correction factor) is given by the interval

\large [\bar x-z^*\frac{s}{\sqrt n}, \bar x+z^*\frac{s}{\sqrt n}]

where

\large \bar x <em>is the sample mean  </em>

<em>s is the sample standard deviation  </em>

<em>n is the sample size </em>

\large z^* <em>is the 0.01 (99%) upper critical value for </em>

<em>the Normal distribution N(0;1). </em>

The value of \large z^* can be found either by using a table or a computer to find it equals to  

\large z^*=2.576

and our 99% confidence interval <em>(without continuity correction) </em>is

\large [104-2.576*\frac{8.867}{\sqrt{427}}, 104+2.576*\frac{8.867}{\sqrt {427}}]=[102.8946,105.1054]

We can now introduce the continuity correction factor. This should be done because <em>we are approximating a discrete distribution (Binomial) with a continuous one (Normal). </em>

This is simply done by widening the interval in 0.5 at each end, so our final 99% confidence interval is

[102.3946 , 105.6054] = [102.39 , 105.61] rounded to 2 decimal places.

c)

The interpretation of this confidence interval is that in samples of 427 people aged 65 years and older, there is a 99% probability that the number of people that suffers sleep apnea is between 103 and 105.

If another study found a 15% of elderly people suffered sleep apnea, that would mean that in a sample of 427 only 64 would have the condition. Since that number is less than 103 by far, that would give us a reason to doubt about the conditions that framed the study (sample size, sampling method, age of people, etc.)

4 0
3 years ago
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