If an integer is both a square and a cube, it can be of the form:
<span>(<span>a3</span><span>)^2</span></span>
Now,
since a cube can be of the form 7k or 7k+-1(thanks to FoolForMath),
we write
<span><span>a^3</span>=7k</span>
and get the no to be
49k^2
, which is in the form of 7 times something
<span>49<span>k^2</span>=7×(7<span>k^2</span>)</span>
Now put
<span><span>a^3</span>=7k+−1</span>
Square it
and you'll get a number in the form of (7times something +1)
Answer:
The sample size needed if the margin of error of the confidence interval is to be about 0.04 is 18.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
The margin of error is:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
Past studies suggest this proportion will be about 0.15
This means that ![p = 0.15](https://tex.z-dn.net/?f=p%20%3D%200.15)
Find the sample size needed if the margin of error of the confidence interval is to be about 0.04
This is n when M = 0.04. So
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
![0.04 = 1.96\sqrt{\frac{0.15*0.85}{n}}](https://tex.z-dn.net/?f=0.04%20%3D%201.96%5Csqrt%7B%5Cfrac%7B0.15%2A0.85%7D%7Bn%7D%7D)
![0.04\sqrt{n} = 1.96\sqrt{0.15*0.85}](https://tex.z-dn.net/?f=0.04%5Csqrt%7Bn%7D%20%3D%201.96%5Csqrt%7B0.15%2A0.85%7D)
![\sqrt{n} = \frac{1.96\sqrt{0.15*0.85}}{0.04}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%5Csqrt%7B0.15%2A0.85%7D%7D%7B0.04%7D)
![(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.15*0.85}}{0.04})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B1.96%5Csqrt%7B0.15%2A0.85%7D%7D%7B0.04%7D%29%5E%7B2%7D)
![n = 17.5](https://tex.z-dn.net/?f=n%20%3D%2017.5)
Rounding up
The sample size needed if the margin of error of the confidence interval is to be about 0.04 is 18.
Here's how I did it. I got 7.375...
Using the form y1 - y2/x1 - x2, we get:
-2 - (-4)/-4 - 3 => -2 + 4/-7 => 2/-7 => -2/7. The slope is the same, therefore the lines are parallel.
Hope this helps.
For #27 its 3d=144, then you divide 144 by 3 which gives you 48 so its 48 days