There are
![\dbinom62](https://tex.z-dn.net/?f=%5Cdbinom62)
ways of selecting two of the six blocks at random. The probability that one of them contains an error is
![\dfrac{\dbinom11\dbinom51}{\dbinom62}=\dfrac5{15}=\dfrac13](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cdbinom11%5Cdbinom51%7D%7B%5Cdbinom62%7D%3D%5Cdfrac5%7B15%7D%3D%5Cdfrac13)
So
![X](https://tex.z-dn.net/?f=X)
has probability mass function
![f_X(x)=\begin{cases}\dfrac13&\text{for }x=1\\\\\dfrac23&\text{for }x=0\end{cases}](https://tex.z-dn.net/?f=f_X%28x%29%3D%5Cbegin%7Bcases%7D%5Cdfrac13%26%5Ctext%7Bfor%20%7Dx%3D1%5C%5C%5C%5C%5Cdfrac23%26%5Ctext%7Bfor%20%7Dx%3D0%5Cend%7Bcases%7D)
These are the only two cases since there is only one error known to exist in the code; any two blocks of code chosen at random must either contain the error or not.
The expected value of finding an error is then
I believe it’s x=2.5 or x=2 1/2 or x=5/2
They are all the same answers in different forms . Hope it’s right
There is actually 2 ways to solve this, I will show you both.
The first is obvious, solve for x in the first one, and plug it into the 2nd one and get the answer
4x + 7 = 12
4x = 5
x = ![\frac{5}{4}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B4%7D%20)
8(
) + 3
2 * 5 + 3
10 + 3
13
The 2nd option is to manipulate the 4x + 7 to be 8x + 3
4x + 7 = 12
start by moving the 7 over
4x = 5
multiply both sides by 2
8x = 10
and add 3 to both sides
8x + 3 = 13
Answer:
on 60!! Hope this help's
Step-by-step explanation:
Answer:
2^{3} +1⋅4−3 = 9
2^{0}+10−4⋅2 = 3
3+5^{2} -6⋅4 = 4
Step-by-step explanation: