1. LI = 12
2. TV = 8.75, X = 9.75
3. No, it would be if AD = 15, DB = 12 and AE = 10, EC = 8.
4. No, CE needs to = 1/2 of AE, not 1/3.
5. Yes, DB and EC are each 1/3 of AB and AC.
6. X = 15
7. X = 9
8. X = 8
Answer:
56 1/4
Step-by-step explanation:
hope this helps, good luck!
Answer:
236 cubic units (to 3 significant figures)
Step-by-step explanation:
volume of a cylinder = r²h (where r = radius and h = height)
Therefore, volume = x 5² x 3
= x 25 x 3
= 75
= 236 cubic units (to 3 significant figures)
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.
C = 36 pi units = 2 pi r. Thus, r = 36 pi / (2 pi) = 18 units.
Thus, A = pi*r^2 = pi*(18 units)^2