Amy must form a three-letter arrangement using only letters from the word dice. She cannot use a letter more than once in the arrangement. (Her arrangement doesn't need to be a valid word.) The probability that Amy forms an arrangement that begins and ends with vowels is . The probability that Amy forms an arrangement whose letters are in alphabetical order (ascending or descending) is .

Answer #1:

Answer #2:

Step-by-step explanation:

__ __ __

There are 4 × 3 × 2 possibilities. For the first question, "The probability that Amy forms an arrangement that begins and ends with vowels is _____." To have a her word end in a vowel, then her word will have to end in either 'e', or 'i'. For this to be possible there are two possibilities for the last letter. Then the middle letter will have three possibilities because one letter will have been used for the last one. Finally, there will be two more possibilities for the letter at the end. That's 2 × 3 × 2 possibilities.

To get a word whose arrangement is in alphabetical order either backwards or forwards I just listed possibilities:

CDE

CDI

CEI

IDC

IEC

IED

There are six in all which means the answer to problem two is six out of twenty four.

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.

what is the approximate angle of acb if abc is 25cm and the length of segment bc is 15

step-by-step explanation: