Step-by-step explanation:
There are 12 games in the population. You need to use a random number generator to choose 2 of these games.
RandomSample[{1,2,3,4,5,6,7,8,9,10,11,12},2]
Let's say the first sample you get is {1,5}. That corresponds to game times of 8 minutes and 7 minutes. The mean game time for that sample is 7.5 minutes. So the first row in your table would be:
![\left[\begin{array}{ccc}Sample&List\ of\ Game\ Times&Mean\ Game\ Time\\1&8,7&7.5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DSample%26List%5C%20of%5C%20Game%5C%20Times%26Mean%5C%20Game%5C%20Time%5C%5C1%268%2C7%267.5%5Cend%7Barray%7D%5Cright%5D)
Answer:
P(A∣D) = 0.667
Step-by-step explanation:
We are given;
P(A) = 3P(B)
P(D|A) = 0.03
P(D|B) = 0.045
Now, we want to find P(A∣D) which is the posterior probability that a computer comes from factory A when given that it is defective.
Using Bayes' Rule and Law of Total Probability, we will get;
P(A∣D) = [P(A) * P(D|A)]/[(P(A) * P(D|A)) + (P(B) * P(D|B))]
Plugging in the relevant values, we have;
P(A∣D) = [3P(B) * 0.03]/[(3P(B) * 0.03) + (P(B) * 0.045)]
P(A∣D) = [P(B)/P(B)] [0.09]/[0.09 + 0.045]
P(B) will cancel out to give;
P(A∣D) = 0.09/0.135
P(A∣D) = 0.667
Answer:
y+x is greater than or equal to -1
Step-by-step explanation:
Take 1960 as 0 then find out next year's
- 1970=10
- 1980=20
- 1990=30
- 2000=40
- 2005=45
- 2010=50
Also
Divide each population by one thousand i.e turn to decimal .
This we done to create the plot on graph easily
The graph has been attached
for the below points
- (0,9.706)
- (10,10.652)
- (20,10.798)
- (30,10.847)
- (40,11.353)
- (45,11.478)
- (50,11.576)
First, turn 
into an improper fraction with a denominator of 4, which matches 
.
Now, just add the numerators.
