Answer:
b sides
Step-by-step explanation:
the sides will have markings so you wont need the specific angles
Answer: The test statistic is t= -0.90.
Step-by-step explanation:
Since we have given that
n₁ = 50
n₂ = 25

So, s would be

So, the value of test statistic would be

Hence, the test statistic is t= -0.90.
This is a linear function, of the kind y = mx + b
where x is the number of visit, b is the weight when x = 0, and m is the predicted change of weight for every visit.
m = - 4 ounces / visit, which must be converted to pounds (the negative sign indicates that the change is a decrease)
1 lb = 16 ounces = 4 ounces = 0.25 lb
Then m = - 0.25 lb / visit.
Now, for x = 1, y = 126 => 126 = - 0.25(1) + b => b = 126 + 0.25 = 126.25
Then the function is y = 126.25 - 0.25x
Now round to the nearest tenth:
y = 126.3 - 0.3x
Answer: y = 126.3 - 0.3x
bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.
![\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}](https://tex.z-dn.net/?f=%5Cbf%20cos%5Cleft%5B%20sin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%20%5Cright%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bthen%20we%20can%20say%20that%7D~%5Chfill%20%7D%7Bsin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%5Cimplies%20%5Ctheta%20%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Btherefore%20then%7D~%5Chfill%20%7D%7Bsin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B13%7D%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%7D%7D)
![\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B13%5E2-%28-5%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B144%7D%3Da%5Cimplies%20%5Cpm%2012%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20cos%5Cleft%5B%20sin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20cos%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B%5Cpm%2012%7D%7D%7B13%7D)
le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.