Answer:
Step-by-step explanation:
20% of 260
Or 52 votes
Answer:
The diagonal is 39.54 in
The baton will fit in the box if it is placed in the direction of the diagonal of the cube since:
39.54 inches > 38 inches
Step-by-step explanation:
Answer:
Step-by-step explanation:
The given relation between length and width can be used to write an expression for area. The equation setting that equal to the given area can be solved to find the shed dimensions.
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<h3>Given relation</h3>
Let x represent the width of the shed. Then the length is (2x+3), and the area is ...
A = LW
20 = (2x+3)(x) . . . . . area of the shed
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<h3>Solution</h3>
Completing the square gives ...
2x² +3x +1.125 = 21.125 . . . . . . add 2(9/16) to both sides
2(x +0.75)² = 21.125 . . . . . . . write as a square
x +0.75 = √10.5625 . . . . . divide by 2, take the square root
x = -0.75 +3.25 = 2.50 . . . . . subtract 0.75, keep the positive solution
The width of the shed is 2.5 feet; the length is 2(2.5)+3 = 8 feet.
Answer:
![(D)E[ X ] =np.](https://tex.z-dn.net/?f=%28D%29E%5B%20X%20%5D%20%3Dnp.)
Step-by-step explanation:
Given a binomial experiment with n trials and probability of success p,
![f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq x\leq n](https://tex.z-dn.net/?f=f%28x%29%3D%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7Dn%5C%5Ck%5Cend%7Barray%7D%5Cright%29p%5Ex%281-p%29%5E%7Bn-x%7D%2C%200%5Cleq%20%20x%5Cleq%20n)
![E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}](https://tex.z-dn.net/?f=E%28X%29%3D%5Csum_%7Bx%3D0%7D%5E%7Bn%7Dxf%28x%29%3D%20%5Csum_%7Bx%3D0%7D%5E%7Bn%7Dx%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7Dn%5C%5Ck%5Cend%7Barray%7D%5Cright%29p%5Ex%281-p%29%5E%7Bn-x%7D)
Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:
![E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}](https://tex.z-dn.net/?f=E%28X%29%3D%5Csum_%7Bx%3D1%7D%5E%7Bn%7Dx%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7Dn%5C%5Cx%5Cend%7Barray%7D%5Cright%29p%5Ex%281-p%29%5E%7Bn-x%7D)
Now,
![x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)](https://tex.z-dn.net/?f=x%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7Dn%5C%5Cx%5Cend%7Barray%7D%5Cright%29%3D%20%5Cfrac%7Bxn%21%7D%7Bx%21%28n-x%29%21%7D%3D%5Cfrac%7Bn%21%7D%7B%28x-%29%21%28n-x%29%21%7D%3D%5Cfrac%7Bn%28n-1%29%21%7D%7B%28x-1%29%21%28%28n-1%29-%28x-1%29%29%21%7D%3Dn%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7Dn-1%5C%5Cx-1%5Cend%7Barray%7D%5Cright%29)
Substituting,
![E(X)=\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^x(1-p)^{n-x}](https://tex.z-dn.net/?f=E%28X%29%3D%5Csum_%7Bx%3D1%7D%5E%7Bn%7Dn%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7Dn-1%5C%5Cx-1%5Cend%7Barray%7D%5Cright%29p%5Ex%281-p%29%5E%7Bn-x%7D)
Factoring out the n and one p from the above expression:
![E(X)=np\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^{x-1}(1-p)^{(n-1)-(x-1)}](https://tex.z-dn.net/?f=E%28X%29%3Dnp%5Csum_%7Bx%3D1%7D%5E%7Bn%7Dn%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7Dn-1%5C%5Cx-1%5Cend%7Barray%7D%5Cright%29p%5E%7Bx-1%7D%281-p%29%5E%7B%28n-1%29-%28x-1%29%7D)
Representing k=x-1 in the above gives us:
![E(X)=np\sum_{k=0}^{n}n\left(\begin{array}{c}n-1\\k\end{array}\right)p^{k}(1-p)^{(n-1)-k}](https://tex.z-dn.net/?f=E%28X%29%3Dnp%5Csum_%7Bk%3D0%7D%5E%7Bn%7Dn%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7Dn-1%5C%5Ck%5Cend%7Barray%7D%5Cright%29p%5E%7Bk%7D%281-p%29%5E%7B%28n-1%29-k%7D)
This can then be written by the Binomial Formula as:
![E[ X ] = (np) (p +(1 - p))^{n -1 }= np.](https://tex.z-dn.net/?f=E%5B%20X%20%5D%20%3D%20%28np%29%20%28p%20%2B%281%20-%20p%29%29%5E%7Bn%20-1%20%7D%3D%20np.)