Graph 1 which is shown in the image below
I’m pretty sure it’s 1640 or A.
Answer: 1,000,000,000*10
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Answer:
c. x ≈ 1.71, y ≈ 6.29, z ≈ 60.71
Step-by-step explanation:
You are at a bit of a disadvantage with this problem, because both the hint and the answer choices are incorrect.
The matrix equation is ...
![AX=B\\\\\left[\begin{array}{ccc}-1&1&0\\4&3&0\\1&-7&1\end{array}\right]\left[\begin{array}{c}x&y&z\end{array}\right]=\left[\begin{array}{c}8&12&15\end{array}\right]](https://tex.z-dn.net/?f=AX%3DB%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%261%260%5C%5C4%263%260%5C%5C1%26-7%261%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%26z%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%2612%2615%5Cend%7Barray%7D%5Cright%5D)
Contrary to what the hint is telling you, the multiplication by the inverse must be on the left:

Using appropriate tools to compute the inverse, this becomes ...
![\left[\begin{array}{c}x&y&z\end{array}\right]=\dfrac{1}{7}\left[\begin{array}{ccc}-3&1&0\\4&1&0\\31&6&7\end{array}\right]\left[\begin{array}{c}8&12&15\end{array}\right]\\\\\left[\begin{array}{c}x&y&z\end{array}\right]=\dfrac{1}{7}\left[\begin{array}{c}-12&44&425\end{array}\right]\approx\left[\begin{array}{c}-1.714&6.286&60.714\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%26z%5Cend%7Barray%7D%5Cright%5D%3D%5Cdfrac%7B1%7D%7B7%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%261%260%5C%5C4%261%260%5C%5C31%266%267%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%2612%2615%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%26z%5Cend%7Barray%7D%5Cright%5D%3D%5Cdfrac%7B1%7D%7B7%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-12%2644%26425%5Cend%7Barray%7D%5Cright%5D%5Capprox%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1.714%266.286%2660.714%5Cend%7Barray%7D%5Cright%5D)
The closest match is choice C.
<span>Interval between 2.6 minutes late and 8.2 minutes late is within 2
standard deviations of the mean (k =8.2â’5.4/1.4 =5.4â’2.6/1.4 = 2). By Chebyshev’s Theorem, at least 1 â’1/2^2 =3/4 of all flights, i.e. at least 75% of all flights arrive
anywhere between 2.6 minutes late and 8.2 minutes late.</span>