Using the principle of binomial probability, the probability of having exactly two black cards is 0.3125
- <em>Probability</em><em> </em><em>of</em><em> </em><em>having</em><em> </em><em>a</em><em> </em><em>black</em><em> </em><em>card</em><em> </em><em>=</em><em> </em><em>26</em><em>/</em><em>52</em><em> </em><em>=</em><em> </em><em>1</em><em>/</em><em>2</em><em> </em>
<u>Using the binomial probability relation</u> :
- P(x = x) = nCx * p^x * q^(n-x)
- <em>p</em><em> </em><em>=</em><em> </em><em>probability</em><em> </em><em>of</em><em> </em><em>success</em><em> </em><em>=</em><em> </em><em>1</em><em>/</em><em>2</em><em> </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em>
- <em>Number of</em><em> </em><em>picks</em><em>,</em><em> </em><em>n</em><em> </em><em>=</em><em> </em><em>5</em>
- <em>q</em><em> </em><em>=</em><em> </em><em>1</em><em> </em><em>-</em><em> </em><em>p</em><em> </em><em>=</em><em> </em><em>0.5</em><em> </em>
- <em>x</em><em> </em><em>=</em><em> </em><em>2</em><em> </em>
Hence,
P(x = 2) = 5C2 × 0.5² × 0.5³
P(x = 2) = 10 × 0.25 × 0.125
P(x = 2) = 0.3125
Hence. Probability of having exactly 2 black cards is 0.3125
Learn more : brainly.com/question/12474772
Answer:
45.75
Step-by-step explanation:
you divided 274.50 by 6 and get the answer
Answer:
![m \times H=\left[\begin{array}{c c c}\boxed{-9} & \boxed{36} & \boxed{-\dfrac{9}{2}}\end{array}\right]](https://tex.z-dn.net/?f=m%20%5Ctimes%20H%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%20c%7D%5Cboxed%7B-9%7D%20%26%20%5Cboxed%7B36%7D%20%26%20%5Cboxed%7B-%5Cdfrac%7B9%7D%7B2%7D%7D%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
<u>Calculate the value of m</u>
Given:
![3\left[\begin{array}{c c}-1 & 2 \\4 & 8\end{array}\right]=\dfrac{2}{3}m \times \left[\begin{array}{c c}-1 & 2 \\4 & 8\end{array}\right]](https://tex.z-dn.net/?f=3%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%7D-1%20%26%202%20%5C%5C4%20%26%208%5Cend%7Barray%7D%5Cright%5D%3D%5Cdfrac%7B2%7D%7B3%7Dm%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%7D-1%20%26%202%20%5C%5C4%20%26%208%5Cend%7Barray%7D%5Cright%5D)
Therefore:



<u>Calculate the value of H</u>
Given:
![\left(H+ \left[\begin{array}{c c c}1 & 4 & -2\end{array}\right]\right)+\left[\begin{array}{c c c}3 & 2 & -6\end{array}\right]=\left[\begin{array}{c c c}-2 & 8 & -1\end{array}\right]+\left(\left[\begin{array}{c c c}1 & 4 & -2\end{array}\right]+\left[\begin{array}{c c c}3 & 2 & -6\end{array}\right]\right)](https://tex.z-dn.net/?f=%5Cleft%28H%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%20c%7D1%20%26%204%20%26%20-2%5Cend%7Barray%7D%5Cright%5D%5Cright%29%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%20c%7D3%20%26%202%20%26%20-6%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%20c%7D-2%20%26%208%20%26%20-1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%20c%7D1%20%26%204%20%26%20-2%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%20c%7D3%20%26%202%20%26%20-6%5Cend%7Barray%7D%5Cright%5D%5Cright%29)
Therefore:
![\implies H= \left[\begin{array}{c c c}-2 & 8 & -1\end{array}\right]](https://tex.z-dn.net/?f=%5Cimplies%20H%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%20c%7D-2%20%26%208%20%26%20-1%5Cend%7Barray%7D%5Cright%5D)
<u />
<u>Calculating m × H</u>
<u />
<u />![\implies m \times H=\dfrac{9}{2} \times \left[\begin{array}{c c c}-2 & 8 & -1\end{array}\right]](https://tex.z-dn.net/?f=%5Cimplies%20m%20%5Ctimes%20H%3D%5Cdfrac%7B9%7D%7B2%7D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%20c%7D-2%20%26%208%20%26%20-1%5Cend%7Barray%7D%5Cright%5D)
<u />![\implies m \times H=\left[\begin{array}{c c c}\dfrac{9}{2}(-2) & \dfrac{9}{2}(8) & \dfrac{9}{2}(-1)\end{array}\right]](https://tex.z-dn.net/?f=%5Cimplies%20m%20%5Ctimes%20H%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%20c%7D%5Cdfrac%7B9%7D%7B2%7D%28-2%29%20%26%20%5Cdfrac%7B9%7D%7B2%7D%288%29%20%26%20%5Cdfrac%7B9%7D%7B2%7D%28-1%29%5Cend%7Barray%7D%5Cright%5D)
<u />
Answer:
(7,19) is a solution for only the first equation (2x-y= -5), so it is not a solution to the systemmmm
Step-by-step explanation:
have a nice dayyy