Answer:
f(x) = 4x^2 + 2x - 4.
Step-by-step explanation:
Let the quadratic function be y = f(x) = ax^2 + bx + c.
For the point (-2, 8) ( x = -2 when y = 8) we have:
a(-2)^2 + (-2)b + c = 8
4a - 2b + c = 8 For (0, -4) we have:
0 + 0 + c = -4 so c = -4. For (4, 68) we have:
16a + 4b + c = 68
So we have 2 systems of equations in a and b ( plugging in c = -4):
4a - 2b - 4 = 8
16a + 4b - 4 = 68
4a - 2b = 12
16a + 4b = 72 Multiplying 4a - 2b = 12 by 2 we get:
8a - 4b = 24
Adding the last 2 equations:
24a = 96
a = 4
Now plugging a = 4 and c = -4 in the first equation:
4(4) - 2b - 4 = 8
-2b = 8 - 16 + 4 = -4
b = 2.
Answer:
0.01 is alpha level
Step-by-step explanation:
Given that a survey of a magazine of 1500 adults revealed that 12% chose chocolate pie.
i.e. p = proportion of people who like chocolate pie = 0.12
Margin of error = ±0.03
The value of q =1-\hat p , =1-0.12 = 0.88
The sample size n is given as 1500. Therefore, n=1500
The confidence level is given as 99%
Hence alpha = 1- confidence level
=1-0.99=0.01
Hence answer is 0.01
The answer is,
![\frac{10935}{59048}](https://tex.z-dn.net/?f=%20%5Cfrac%7B10935%7D%7B59048%7D%20)
Also equivalent to about, 0.185188.
Answer:
C
Step-by-step explanation:
The general rule for the quadratic function is
![y=ax^2+bx+c](https://tex.z-dn.net/?f=y%3Dax%5E2%2Bbx%2Bc)
Use the data from the table:
![y(50)=130\Rightarrow 130=a\cdot 50^2+b\cdot 50+c\\ \\y(70)=130\Rightarrow 130=a\cdot 70^2+b\cdot 70+c\\ \\y(90)=200\Rightarrow 200=a\cdot 90^2+b\cdot 90+c](https://tex.z-dn.net/?f=y%2850%29%3D130%5CRightarrow%20130%3Da%5Ccdot%2050%5E2%2Bb%5Ccdot%2050%2Bc%5C%5C%20%5C%5Cy%2870%29%3D130%5CRightarrow%20130%3Da%5Ccdot%2070%5E2%2Bb%5Ccdot%2070%2Bc%5C%5C%20%5C%5Cy%2890%29%3D200%5CRightarrow%20200%3Da%5Ccdot%2090%5E2%2Bb%5Ccdot%2090%2Bc)
We get the system of three equations:
![\left\{\begin{array}{l}2500a+50b+c=130\\ \\4900a+70b+c=130\\ \\8100a+90b+c=200\end{array}\right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7D2500a%2B50b%2Bc%3D130%5C%5C%20%5C%5C4900a%2B70b%2Bc%3D130%5C%5C%20%5C%5C8100a%2B90b%2Bc%3D200%5Cend%7Barray%7D%5Cright.)
Subtract these equations:
![\left\{\begin{array}{l}4900a+70b+c-2500a-50b-c=130-130\\ \\8100a+90b+c-2500a-50b-c=200-130\end{array}\right.\Rightarrow \left\{\begin{array}{l}2400a+20b=0\\ \\5600a+40b=70\end{array}\right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7D4900a%2B70b%2Bc-2500a-50b-c%3D130-130%5C%5C%20%5C%5C8100a%2B90b%2Bc-2500a-50b-c%3D200-130%5Cend%7Barray%7D%5Cright.%5CRightarrow%20%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7D2400a%2B20b%3D0%5C%5C%20%5C%5C5600a%2B40b%3D70%5Cend%7Barray%7D%5Cright.)
From the first equation
![b=-120a](https://tex.z-dn.net/?f=b%3D-120a)
Substitute it into the second equation:
![5600a+40\cdot (-120a)=70\Rightarrow 800a=70,\\ \\ a=\dfrac{7}{80},\\ \\ b=-120\cdot \dfrac{7}{80}=-\dfrac{21}{2}=-10.5](https://tex.z-dn.net/?f=5600a%2B40%5Ccdot%20%28-120a%29%3D70%5CRightarrow%20800a%3D70%2C%5C%5C%20%5C%5C%20a%3D%5Cdfrac%7B7%7D%7B80%7D%2C%5C%5C%20%5C%5C%20b%3D-120%5Ccdot%20%5Cdfrac%7B7%7D%7B80%7D%3D-%5Cdfrac%7B21%7D%7B2%7D%3D-10.5)
So,
![2500\cdot \dfrac{7}{80}+50\cdot (-10.5)+c=130\Rightarrow 218.75-525+c=130\\ \\c=130-218.75+525=436.25](https://tex.z-dn.net/?f=2500%5Ccdot%20%5Cdfrac%7B7%7D%7B80%7D%2B50%5Ccdot%20%28-10.5%29%2Bc%3D130%5CRightarrow%20218.75-525%2Bc%3D130%5C%5C%20%5C%5Cc%3D130-218.75%2B525%3D436.25)
The quadratic function is
![y=\dfrac{7}{80}x^2-10.5x+436.25\\ \\y=0.0875x^2-10.5x+436.25](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B7%7D%7B80%7Dx%5E2-10.5x%2B436.25%5C%5C%20%5C%5Cy%3D0.0875x%5E2-10.5x%2B436.25)
Answer:
![\theta = 82.2^\circ](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2082.2%5E%5Ccirc)
Step-by-step explanation:
![Let \ \angle BAC = \theta](https://tex.z-dn.net/?f=Let%20%5C%20%20%5Cangle%20BAC%20%3D%20%5Ctheta)
Opposite = BC ,
Adjacent = AB = x = 3 ,
Hypotenuse = AC = y = 22
<em><u>Using trigonometric ratios.</u></em>
![sin \theta = \frac{opposite}{hypotenuse }\\\\cos \theta = \frac{adjacent}{hypotenuse}\\\\tan \theta = \frac{opposite}{adjacent}](https://tex.z-dn.net/?f=sin%20%5Ctheta%20%3D%20%5Cfrac%7Bopposite%7D%7Bhypotenuse%20%7D%5C%5C%5C%5Ccos%20%5Ctheta%20%3D%20%5Cfrac%7Badjacent%7D%7Bhypotenuse%7D%5C%5C%5C%5Ctan%20%5Ctheta%20%3D%20%5Cfrac%7Bopposite%7D%7Badjacent%7D)
Since we have adjacent and hypotenuse we use cosine's ratio
to find the angle.
![cos \theta = \frac{x}{y} = \frac{3}{22} \\\\\theta = cos^{-1} \frac{3}{22} = 82.16 = 82.2^\circ](https://tex.z-dn.net/?f=cos%20%5Ctheta%20%3D%20%5Cfrac%7Bx%7D%7By%7D%20%3D%20%5Cfrac%7B3%7D%7B22%7D%20%5C%5C%5C%5C%5Ctheta%20%3D%20cos%5E%7B-1%7D%20%5Cfrac%7B3%7D%7B22%7D%20%3D%2082.16%20%3D%2082.2%5E%5Ccirc)