16x^2 + 25y^2 + 160x - 200y + 400 = 0 Rearrange and regroup.
(16x^2 + 160x) + (25y^2 - 200y ) = 0-400. Group the xs together and the ys together.
16(X^2 + 10x) + 25(y^2-8y) = -400. Factorising.
We are going to use completing the square method.
Coefficient of x in the first expression = 10.
Half of it = 1/2 * 10 = 5. (Note this value)
Square it = 5^2 = 25. (Note this value)
Coefficient of y in the second expression = -8.
Half of it = 1/2 * -8 = -4. (Note this value)
Square it = (-4)^2 = 16. (Note this value)
We are going to carry out a manipulation of completing the square with the values
25 and 16. By adding and substracting it.
16(X^2 + 10x) + 25(y^2-8y) = -400
16(X^2 + 10x + 25 -25) + 25(y^2-8y + 16 -16) = -400
Note that +25 - 25 = 0. +16 -16 = 0. So the equation is not altered.
16(X^2 + 10x + 25) -16(25) + 25(y^2-8y + 16) -25(16) = -400
16(X^2 + 10x + 25) + 25(y^2-8y + 16) = -400 +16(25) + 25(16) Transferring the terms -16(25) and -25(16)
to other side of equation. And 16*25 = 400
16(X^2 + 10x + 25) + 25(y^2-8y + 16) = 25(16)
16(X^2 + 10x + 25) + 25(y^2-8y + 16) = 400
We now complete the square by using the value when coefficient was halved.
16(x-5)^2 + 25(y-4)^2 = 400
Divide both sides of the equation by 400
(16(x-5)^2)/400 + (25(y-4)^2)/400 = 400/400 Note also that, 16*25 = 400.
((x-5)^2)/25 + ((y-4)^2)/16 = 1
((x-5)^2)/(5^2) + ((y-4)^2)/(4^2) = 1
Comparing to the general format of an ellipse.
((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) = 1
Coordinates of the center = (h,k).
Comparing with above (x-5) = (x - h) , h = 5.
Comparing with above (y-k) = (y - k) , k = 4.
Therefore center = (h,k) = (5,4).
Sorry the answer came a little late. Cheers.
When the x-intercept solution of a system of two-variable equations is plugged in, the equations yield the same value. Thus, for the points to be solutions, F(x) = G(x)
The only option that shows this is C, where
F(8) = G(8) and F(24) = G(24)
A repeating decimal is when the number repeat or go on forever and do not change
a terminating decimal is one that has an end to it's numbers
example of repeating: .303030303030303.....
example of terminating: .2427
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.