now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.
let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.
![\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%3D3-i%5Cimplies%20%26x-3%2Bi%3D0%5C%5C%20x%3D3%2Bi%5Cimplies%20%26x-3-i%3D0%5C%5C%20x%3D4i%5Cimplies%20%26x-4i%3D0%5C%5C%20x%3D-4i%5Cimplies%20%26x%2B4i%3D0%20%5Cend%7Bcases%7D%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28x-3%2Bi%29%28x-3-i%29%28x-4i%29%28x%2B4i%29%3D%5Cstackrel%7By%7D%7B0%7D%20%5C%5C%5B2em%5D%20%5Cunderset%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%5B%28x-3%29%2Bi%5D%5B%28x-3%29-i%5D%7D%5Cunderset%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%5Bx-4i%5D%5Bx%2B4i%5D%7D%3D0)
![\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5B%28x-3%29%5E2-i%5E2%5D%5Bx%5E2-%284i%29%5E2%5D%3Dy%5Cimplies%20%5B%28x-3%29%5E2-%28-1%29%5D%5Bx%5E2-%284%5E2i%5E2%29%5D%3D0%20%5C%5C%5B2em%5D%20%5B%28x-3%29%5E2-%28-1%29%5D%5Bx%5E2-%2816%28-1%29%29%5D%3D0%5Cimplies%20%5B%28x-3%29%5E2%2B1%5D%5Bx%5E2%2B16%5D%3D0%20%5C%5C%5B2em%5D%20%5B%28x%5E2-6x%2B9%29%2B1%5D%5Bx%5E2%2B16%5D%3Dy%5Cimplies%20%28x%5E2-6x%2B10%29%28x%5E2%2B16%29%3D0%20%5C%5C%5C%5C%5C%5C%20x%5E4-6x%5E3%2B10x%5E2%2B16x%5E2-96x%2B160%3D0%20%5C%5C%5C%5C%5C%5C%20x%5E4-6x%5E3%2B26x%5E2-96x%2B160%3D0%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%204%7D%7D%7B4%28x%5E4-6x%5E3%2B26x%5E2-96x%2B160%29%3D4%280%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%204x%5E4-24x%5E3%2B104x%5E2-384x%2B640%3Dy~%5Chfill)
Answer:
B. The mean amount of 
emitted by the new fuel is actually lower than 89 kg but they fall to conclude it is lower than 89 kg
Step-by-step explanation:
A Type II error is the failure to reject a false null hypothesis.
Given the null and alternate hypothesis of a fuel company which wants to test a new type of gasoline designed to have lower emissions.:
where 
is the mean amount of emitted by burning one gallon of this new gasoline.
If the null hypothesis is false, then:
A rejection of the alternate hypothesis above will be a Type II error.
Therefore:
The Type II error is: (B) The mean amount of emitted by the new fuel is actually lower than 89 kg but they fall to conclude it is lower than 89 kg.
Answer:
ok ill try
Step-by-step explanation:
Answer:
Step-by-step explanation:
Just remember area equals length times width it’s like a window to a door etc!!
Answer:
R is not collinear.
S is collinear.
Step-by-step explanation:
The equation of the graph is given by y =x +2 ........ (1).
Now, the points P and Q having coordinates (5,7) and (-1,1) respectively lie on the graph of the equation (1) because those points satisfy the equation.
Now, if we consider another point R having coordinates (1,-1), then it does not lie on the graph of equation (1) as it does not satisfy the equation.{If you put x=1, then y will be 3 but not -1}
Therefore, P, Q, R are not collinear. (Answer)
Again if we consider the point S(-3,-1), it satisfies equation (1).{If you put x=-3 in the equation (1), y will be -1}.
Therefore, P, Q, and S lie on the same straight line. and they are collinear. (Answer)
