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irina [24]
3 years ago
7

If two out of every 10 high school students are licensed drivers, about how many students would be expected to have driver's lic

enses in a high school with 615 students?
Mathematics
2 answers:
g100num [7]3 years ago
6 0

Answer:62 students

Step-by-step explanation:

I am Lyosha [343]3 years ago
6 0

Answer:

idk

Step-by-step explanation:

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Dayquan decided to run some laps around the track. If the diameter of the track is 75 feet, how far around the track did he run?
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Answer: 235.5 feet

Step-by-step explanation:

In this case, we have to calculate the circumference of the track. The circumference will be calculated as:

= 2πr

Diameter = 75 feet.

Radius = 75/2 = 37.5 feet

Circumference = 2πr = 2 × 3.14 × 37.5 = 235.5feet

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A manufacturer samples 100 wires for quality testing. 4 wires were found defective. if 750 were produced in 1 hour how many shou
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30 wires would be defective

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4 years ago
Evaluate this exponential expression. (0.125)2/3
Artemon [7]
The answer should be 0.25


5 0
3 years ago
Read 2 more answers
Solve the literal equation ab+c=d in terms of a
Tanzania [10]
ab+c=d 


ab=d-c
 
  |Subtract \ c \ from \ both \ sides| 

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3 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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