40%=0.4
0.4+1=1.4
4.5(1.4)=6.3
The answer is $6.30
Answer:
Jayce should not use either option. Option 1 is likely to be based so that paperbacks are overrepresented, while Option 2 is likely to be biased so that e-books are overrepresented.
Step-by-step explanation:
<em>Given:</em>
<em>Jayce volunteers for the local library. The librarian wants to find out whether the patrons prefer paperback books or e-books. Jayce cannot decide which method to use for polling the patrons.</em>
<em>Option 1: Poll every fifth patron who enters the library on Paperback Lovers Day.</em>
<em>Option 2: Poll every third patron who enters the library on Technology Appreciation Day.</em>
<em />
<em>Since, Option 1: Poll every fifth patron who enters the library on Paperback Lovers Day. It biased because since it paperback lovers day thus, it would be likely that most people like the paperback which make it biased.</em>
<em>Since, Option 2: Poll every third patron who enters the library on Technology Appreciation Day. It biased because since it technology appreciation day thus, it would be likely that most people don't care about these stuff since people coming in are for technology appreciation.</em>
<em />
<em />
<em>Therefore, the answer is:</em>
Jayce should not use either option. Option 1 is likely to be based so that paperbacks are overrepresented, while Option 2 is likely to be biased so that e-books are overrepresented.
<u><em>Kavinsky</em></u>
Complete Question
A survey is planned to estimate the proportion of voters who support a proposed gun control law. The estimate should be within a margin of error of ±5% with 90% confidence, and we do not have any prior knowledge about the proportion who might support the law. How many people need to be included in the sample?
Round your answer up to the nearest integer.
sample size = _____
Answer:
The sample size is 
Step-by-step explanation:
From the question we are told that
The margin of error is E = 5% = 0.05
Generally given that there was no prior knowledge about the proportion of who might support the law, we will assume the sample proportion to be
From the question we are told the confidence level is 90% , hence the level of significance is
=> 
Generally from the normal distribution table the critical value of 
is
Generally the sample size is mathematically represented as
![n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )](https://tex.z-dn.net/?f=n%20%3D%20%5B%5Cfrac%7BZ_%7B%5Cfrac%7B%5Calpha%20%7D%7B2%7D%20%7D%7D%7BE%7D%20%5D%5E2%20%2A%20%5C%5E%20p%20%281%20-%20%5C%5E%20p%20%29%20)
=> ![n = [\frac{1.645 }{0.05} ]^2 * 0.5 (1 - 0.5 )](https://tex.z-dn.net/?f=n%20%3D%20%5B%5Cfrac%7B1.645%20%7D%7B0.05%7D%20%5D%5E2%20%2A%200.5%20%281%20-%200.5%20%29%20)
=>
Answer:
half
Step-by-step explanation:
Answer:2 2/9
Step-by-step explanation:
3 1/3= 10/3, 10/3 x 2/3= 20/9= 2 2/9
