Answer:
B:942$
Step-by-step explanation:
(pi*20^2)/2 is approximately 628.32
628.32*1.5=942.48
Answer:
The answer to your question is AC = 9.9
Step-by-step explanation:
To solve this problem use trigonometric functions. The trigonometric function that relates the hypotenuse and the adjacent side is cosine.
cos α = 
Solve for hypotenuse
Substitution
hypotenuse = 
Simplification and result
hypotenuse = 9.85
When you have 3 choices for each of 6 spins, the number of possible "words" is
3^6 = 729
The number of permutations of 6 things that are 3 groups of 2 is
6!/(2!×2!×2!) = 720/8 = 90
A) The probability of a word containing two of each of the letters is 90/729 = 10/81
The number of permutations of 6 things from two groups of different sizes is
(2 and 4) : 6!/(2!×4!) = 15
(3 and 3) : 6!/(3!×3!) = 20
(4 and 2) : 15
(5 and 1) : 6
(6 and 0) : 1
B) The number of ways there can be at least 2 "a"s and no "b"s is
15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.
_____
These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
The probability of one head and one tail is 2/3.
<u>Step-by-step explanation</u>:
- The possibilities for flipping two fair coins are {T,T}, {H,H}, {H,T}, {T,H}
- Given the case that at least one coin lands on a head, So the total possibilities are {H,H}, {H,T}, {T,H} = 3 possibilities
- Required event is 1 head and 1 tail= {H,T}, {T,H} = 2 possibilities
To calculate the probability of one head and one tail,
Probability = required events / Total events
Probability = 2/3
