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serg [7]
3 years ago
15

What is 7 as a percent

Mathematics
2 answers:
cluponka [151]3 years ago
8 0
7 as a percent is 70%. 7/10 = 70%
Lelechka [254]3 years ago
3 0
The answer would be .07

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AC = 2AB

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State the degree: 11m^3n^2p
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<h3>Answer:  6</h3>

Explanation:

Using the rule that x = x^1, we can rewrite the p as p^1

So 11m^3n^2p is the same as 11m^3n^2p^1

The exponents are: 3, 2, 1

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Which properties justify the steps taken to solve the system?
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X=-1

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6x-12y

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6 0
2 years ago
As indicated below, write the equations of the line passing through the point (2, 2) and perpendicular to the line whose equatio
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Step-by-step explanation:

y = x

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7 0
3 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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