Answer: The middle number in a data set when the data are put in order.
Hello there
When we multiply a function by -1, the graphs flips over the x-axis.
Answer:
The probability of winning a jackpot is
The probability of winning the pick 5 game is ![P_a = 0.00001](https://tex.z-dn.net/?f=P_a%20%3D%20%200.00001)
The earning of the lottery organisation if the game were to be runed for no profit is
$10 000
Step-by-step explanation:
From the question
The sample size is n= 37
The number of selection is ![r = 5](https://tex.z-dn.net/?f=r%20%3D%20%205)
Now the number of way by which these five selection can be made is mathematically represented as
![\left n} \atop {}} \right.C_r = \frac{n!}{(n-r)!r! }](https://tex.z-dn.net/?f=%5Cleft%20n%7D%20%5Catop%20%7B%7D%7D%20%5Cright.C_r%20%20%20%3D%20%5Cfrac%7Bn%21%7D%7B%28n-r%29%21r%21%20%7D)
Now substituting values
![\left n} \atop {}} \right.C_r = \frac{37!}{(37-5)!5! }](https://tex.z-dn.net/?f=%5Cleft%20n%7D%20%5Catop%20%7B%7D%7D%20%5Cright.C_r%20%20%20%3D%20%5Cfrac%7B37%21%7D%7B%2837-5%29%215%21%20%7D)
![\left n} \atop {}} \right.C_r = 333333.3](https://tex.z-dn.net/?f=%5Cleft%20n%7D%20%5Catop%20%7B%7D%7D%20%5Cright.C_r%20%20%20%3D%20333333.3)
Now the probability of winning a jackpot from any of the way of selecting 5 whole number from 37 is mathematically evaluated as
![P = \frac{1}{333333.3}](https://tex.z-dn.net/?f=P%20%20%3D%20%5Cfrac%7B1%7D%7B333333.3%7D)
Now the number of ways of selecting 5 whole number from 0 to 9 with repetition is mathematically evaluated as
![k = 10^5](https://tex.z-dn.net/?f=k%20%20%3D%20%2010%5E5)
Now the probability of winning the game is
![P_a = \frac{1}{10^5}](https://tex.z-dn.net/?f=P_a%20%3D%20%20%5Cfrac%7B1%7D%7B10%5E5%7D)
![P_a = 0.00001](https://tex.z-dn.net/?f=P_a%20%3D%20%200.00001)
We are told that for a $1 ticket that the pick 5 game returns $50 , 000
Generally the expected value is mathematically represented as
![E(X) = x * P(X =x )](https://tex.z-dn.net/?f=E%28X%29%20%3D%20%20x%20%20%2A%20P%28X%20%3Dx%20%29)
In this question the expected value is $1
So
![1 = x * 0.00001](https://tex.z-dn.net/?f=1%20%3D%20%20x%20%2A%200.00001)
So ![x = \frac{1}{0.00001}](https://tex.z-dn.net/?f=x%20%3D%20%20%20%5Cfrac%7B1%7D%7B0.00001%7D)
$10 000
Answer:
Scale factor of 1/2
Step-by-step explanation:
Answer:
The probability that the animal chosen is brown-haired is 0.6333.
Step-by-step explanation:
Denote the events as follows:
<em>A</em> : a brown-haired rodent
<em>B</em> : Litter 1
The information provided is:
![P (A|B) =\frac{2}{3}\\\\P(A|B^{c})=\frac{3}{5}](https://tex.z-dn.net/?f=P%20%28A%7CB%29%20%3D%5Cfrac%7B2%7D%7B3%7D%5C%5C%5C%5CP%28A%7CB%5E%7Bc%7D%29%3D%5Cfrac%7B3%7D%7B5%7D)
The probability of selecting any of the two litters is equal, i.e.
![P(B)=P(B^{c})=\frac{1}{2}](https://tex.z-dn.net/?f=P%28B%29%3DP%28B%5E%7Bc%7D%29%3D%5Cfrac%7B1%7D%7B2%7D)
According to the law of total probability:
![P(X)=P(X|Y_{1})P(Y_{1})+P(X|Y_{2})P(Y_{2})+...+P(X|Y_{n})P(Y_{n})](https://tex.z-dn.net/?f=P%28X%29%3DP%28X%7CY_%7B1%7D%29P%28Y_%7B1%7D%29%2BP%28X%7CY_%7B2%7D%29P%28Y_%7B2%7D%29%2B...%2BP%28X%7CY_%7Bn%7D%29P%28Y_%7Bn%7D%29)
Compute the total probability of event <em>A</em> as follows:
![P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})](https://tex.z-dn.net/?f=P%28A%29%3DP%28A%7CB%29P%28B%29%2BP%28A%7CB%5E%7Bc%7D%29P%28B%5E%7Bc%7D%29)
![=[\frac{2}{3}\times\frac{1}{2}]+[\frac{3}{5}\times\frac{1}{2}]\\\\=\frac{1}{3}+\frac{3}{10}\\\\=\frac{10+9}{30}\\\\=\frac{19}{30}\\\\=0.6333](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B2%7D%7B3%7D%5Ctimes%5Cfrac%7B1%7D%7B2%7D%5D%2B%5B%5Cfrac%7B3%7D%7B5%7D%5Ctimes%5Cfrac%7B1%7D%7B2%7D%5D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B3%7D%2B%5Cfrac%7B3%7D%7B10%7D%5C%5C%5C%5C%3D%5Cfrac%7B10%2B9%7D%7B30%7D%5C%5C%5C%5C%3D%5Cfrac%7B19%7D%7B30%7D%5C%5C%5C%5C%3D0.6333)
Thus, the probability that the animal chosen is brown-haired is 0.6333.