Answer:
![A^{-1} = \frac{1}{66} \left[\begin{array}{cc}-2&-5\\6&-18\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7B66%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%26-5%5C%5C6%26-18%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Given
![A = \left[\begin{array}{cc}-18&5\\-6&-2\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-18%265%5C%5C-6%26-2%5Cend%7Barray%7D%5Cright%5D)
Required
Determine the inverse
A matric is of the form:
![A = \left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
First, we need to calculate the determinant (D)

By comparison, we have:




The inverse is then represented as:
![A^{-1} = \frac{1}{D} \left[\begin{array}{cc}d&-b\\-c&a\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7BD%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dd%26-b%5C%5C-c%26a%5Cend%7Barray%7D%5Cright%5D)
This gives:
![A^{-1} = \frac{1}{66} \left[\begin{array}{cc}-2&-5\\6&-18\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7B66%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%26-5%5C%5C6%26-18%5Cend%7Barray%7D%5Cright%5D)
Sample proportion: 190/425 = 0.45
ME = 1.96*sqrt[0.45*0.55/425] = 0.047
-----
95% CI: 0.45-0.047 < p < 0.45+0.047
Answer:
#3
Step-by-step explanation:
The y-intercept (the point where the line intersects with the y-axis) is -2, and the slope is 2/3 (two units up every three units to the right).
According to the image:
the right hexagonal pyramid surface area is given by the following formula:
S = 3√3 / 2) a² + 3a √h²+3a²/4
a = the base edge
h = the height of the pyramid
looking at the image a =6m, h =21m - 7m= 14m
so the surface area is S = 3√3 / 2) 6² + 3*6√14²+3*6²/4 =93.53+ 14.93=108.46 m²
for the another one
the figure is a prism hexagonal
its surface area is
SA= 6ah +3√3 a²
a= base edge, h =height
in our case a =6m and h = 7m
so SA= 6*6*7 + 3√3 * 6²= 252 + 187 = 439 m²
finally, the surface area of the complete image is
ST= S + SA = 108.46 m² + 439 m² = 547.52 m²
it is so approximate
to <span>B. 615.53 m2
look at the image</span>
The answer to your problem is 10m minus 20n or 10m-20n for short.