Answer:
Step-by-step explanation:
Given
Represent Boys with B and Girls with G
Required
Find the probability or having 1 boy 2 girls
Since the order is not important, the probability is calculated as follows;
Substitute 
for P(B) and P(G)
<em>Hence, the fractional probability is </em>
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Your answer is the second option, she should choose the rectangular tiles because the total cost will be $8 less.
To find this answer we need to first find the total cost for using square tiles, and the cost for using rectangular tiles, and compare them. We can do this by finding the area of each tile individually, calculating how many tiles we would need, and multiplying this by the cost for one tile:
Square tiles:
The area of one square tile is 1/2 × 1/2 = 1/4 ft. Therefore we need 40 ÷ 1/4 = 160 tiles. If each tile costs $0.45, this means the total cost will be $0.45 × 160 = $72
Rectangular tiles:
The area of one rectangular tile is 2 × 1/4 = 2/4 = 1/2 ft. Thus we need 40 ÷ 1/2 = 80 tiles. Each tile costs $0.80, so the total cost will be 80 × $0.80 = $64.
This shows us that the rectangular tiles will be cheaper by $8.
I hope this helps! Let me know if you have any questions :)
<span>25: 2 × 2 × 13 65: 5 × 13</span>
First, let me show you some notation.
To show a matrix is an inverse of another matrix, we write
-1 is not an exponent. It just shows that a matrix is an inverse of another matrix.
For a 2x2 matrix, we can get the inverse by first making b and c negatives and swap the positions of a and d.
Then multiply each entry in the matrix by 1 divided by the determinant.
![\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]^{-1} = \frac{1}{ad - bc}\left[\begin{array}{ccc}d&{-b}\\{-c}&a\end{array}\right] = \\ \\ \\ \left[\begin{array}{ccc}d(\frac{1}{ad-bc})&{-b}(\frac{1}{ad-bc}) \\ {-c}(\frac{1}{ad-bc}) &a(\frac{1}{ad-bc}) \end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5E%7B-1%7D%20%3D%20%0A%20%20%5Cfrac%7B1%7D%7Bad%20-%20bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dd%26%7B-b%7D%5C%5C%7B-c%7D%26a%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5C%5C%20%20%5C%5C%20%5C%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dd%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%26%7B-b%7D%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%20%5C%5C%20%7B-c%7D%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%20%26a%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%20%5Cend%7Barray%7D%5Cright%5D)
I hope this helped!
