The formula for the general sin function is
![y=Asin(bx+c)](https://tex.z-dn.net/?f=y%3DAsin%28bx%2Bc%29)
where b is the period and c is the phase shift. We are given a period of
![\frac{ \pi }{4}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Cpi%20%7D%7B4%7D%20)
and the formula to solve for b is
![period= \frac{2 \pi }{b}](https://tex.z-dn.net/?f=period%3D%20%5Cfrac%7B2%20%5Cpi%20%7D%7Bb%7D%20)
. We set up to solve for b:
![\frac{ \pi }{4} = \frac{2 \pi }{b}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Cpi%20%7D%7B4%7D%20%3D%20%5Cfrac%7B2%20%5Cpi%20%7D%7Bb%7D%20)
. Cross multiply to get
![b \pi =8 \pi](https://tex.z-dn.net/?f=b%20%5Cpi%20%3D8%20%5Cpi%20)
and b = 8. Now for the phase shift. The formula for that is -c/b, and since we already know b, and we have a phase shift value of pi/2, we solve for c:
![\frac{ \pi }{2}= \frac{-c}{8}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Cpi%20%7D%7B2%7D%3D%20%5Cfrac%7B-c%7D%7B8%7D%20%20)
and
![8 \pi =-2c](https://tex.z-dn.net/?f=8%20%5Cpi%20%3D-2c)
. Therefore,
![c=-4 \pi](https://tex.z-dn.net/?f=c%3D-4%20%5Cpi%20)
. Putting all that together into the equation with the A value of 6, we have
![y=6sin(8x-4\pi)](https://tex.z-dn.net/?f=y%3D6sin%288x-4%5Cpi%29)
. Factoring out the 8 then, the final equation is
![y=6sin(8(x- \frac{\pi}{2} ))](https://tex.z-dn.net/?f=y%3D6sin%288%28x-%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%29%29)
, third choice down.
Answer:(x + 6) + (y + 2)^2 = 13
Step-by-step explanation:
The centre of the circle is halfway along the diameter = (-6, -2)
r^2 = (-4 - -6)^2 + ( 1 - -2)^2 = 2^2 + 3^2 = 13
So circle equation is (x - -6)^2 + (y - -2)^2 = 13
Answer:
polygon y is circumscribed about polygon x. i think
Step-by-step explanation:
Answer:
![\large \boxed{\mathrm{34.2}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Cmathrm%7B34.2%7D%7D)
Step-by-step explanation:
![\sf B= arcsin (\frac{b \times sin(A)}{a} )](https://tex.z-dn.net/?f=%5Csf%20B%3D%20arcsin%20%28%5Cfrac%7Bb%20%5Ctimes%20sin%28A%29%7D%7Ba%7D%20%29)
![\sf B= arcsin (\frac{7 \times sin(40\°)}{8} )](https://tex.z-dn.net/?f=%5Csf%20B%3D%20arcsin%20%28%5Cfrac%7B7%20%5Ctimes%20sin%2840%5C%C2%B0%29%7D%7B8%7D%20%29)
![\sf B = 0.59733 \ rad = 34.225\°](https://tex.z-dn.net/?f=%5Csf%20B%20%3D%200.59733%20%5C%20rad%20%3D%2034.225%5C%C2%B0)
Answer:
Std deviation would change but IQR would not change.
Step-by-step explanation:
Given that after entering the test scores from her Statistics class of 25 students, the instructor calculated some statistics of the scores. Upon checking, she discovered that she had entered the top score as 46, but it should have been 56.
When arranged in ascending order the last entry would be 56 instead of 46.
Hence this would not affect the middle entry or quartiles
i.e. in other words there would not be any change in the Interquartile range and Quartile 1 and Quartile 3 are not changed
But standard error would change. this is because total of all entries now increase by 10, so average would increase by 10/25 = 0.40
This in turn would change the mean deviation x-average for each entry and hence square of deviations. Thus variance would change and also standard deviation.