Note that fiona cannot sell decimal number of cupcakes or muffins. That's why the options A and D are false.
Consider all remaining options:
B. 40 cupcakes and 80 muffins will cost $(40·3+80·2)=$(120+160)=$280, but Fiona Follies sold at least $300 worth of cupcakes and muffins. Thus, this option is false.
C. 60 cupcakes and 70 muffins will cost $(60·3+70·2)=$(180+140)=$320. Now consider expenses. $(60·0.75+70·0.5)=$(45+35)=$80. This option is true.
E. 80 cupcakes and 80 muffins will cost $(80·3+80·2)=$(240+160)=$400. Now consider expenses. $(80·0.75+80·0.5)=$(60+40)=$100 (exactly $100). This option is true.
Answer: correct options are C and E.
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.
Answer:
4 + 8x = 6
Step-by-step explanation:
4 more (add)
quotient (multiply) of an unknown number (x) and 8
