24???????????????????????????????????????????????????????????????????????
Answer:
E. P(W | H)
Step-by-step explanation:
What each of these probabilities mean:
P(H): Probability of the game being at home
P(W): Probability of the game being a win.
P(H and W): Probability of the game being at home and being a win.
P(H|W): Probability of a win being at home.
P(W|H): Probability of winning a home game.
Which of the following probabilities do you need to find in order to determine the proportion of at-home games that were wins?
This is the probability of winning a home game. So the answer is:
E. P(W | H)
Loge 70.81 = a
Log to the base e is also written as ln
ln 70.81 = a
Answer:
The mode and Median would be lower than the mean.
Step-by-step explanation:
The distribution data is positively skewed which means that most of the data would be towards the lower side. In this situation the mean is higher than the mode and median. Dr. Hammer has distribution data which is positively skewed so the mode and median will be lower than the mean.
To add monomials, you have to look at the variables that are accompanied by their coefficients. In the given problem, (–4c2 + 7cd + 8d) + (–3d + 8c2 + 4cd), you can combine both cd ut nt cd and c² and cd and d and d and c² because they have different variables.
<span>(–4c2 + 7cd + 8d) + (–3d + 8c2 + 4cd)
(-4c</span>² + 8c²) + (7cd + 4cd) + (8d - 3d)
4c² + 11cd + 5d
